when the crate falls off in what direction is it moving relative to the ground

6 Applications of Newton'southward Laws

half dozen.ii Friction

Learning Objectives

Past the end of the section, you will be able to:

Describe the full general characteristics of friction
List the various types of friction
Calculate the magnitude of static and kinetic friction, and utilise these in problems involving Newton's laws of movement

When a body is in motion, information technology has resistance because the torso interacts with its surround. This resistance is a force of friction. Friction opposes relative move betwixt systems in contact merely also allows usa to move, a concept that becomes obvious if you try to walk on ice. Friction is a common even so complex force, and its beliefs withal not completely understood. Still, it is possible to understand the circumstances in which it behaves.

Static and Kinetic Friction

The basic definition of friction is relatively simple to state.

Friction

Friction is a force that opposes relative motility between systems in contact.

There are several forms of friction. One of the simpler characteristics of sliding friction is that it is parallel to the contact surfaces between systems and is always in a management that opposes motion or attempted movement of the systems relative to each other. If ii systems are in contact and moving relative to one another, then the friction between them is called kinetic friction. For case, friction slows a hockey puck sliding on ice. When objects are stationary, static friction can deed between them; the static friction is usually greater than the kinetic friction between two objects.

Static and Kinetic Friction

If two systems are in contact and stationary relative to one some other, then the friction betwixt them is called static friction. If ii systems are in contact and moving relative to one another, and then the friction betwixt them is chosen kinetic friction.

Imagine, for example, trying to slide a heavy crate across a physical floor—you might push very difficult on the crate and not move it at all. This means that the static friction responds to what you lot do—information technology increases to exist equal to and in the reverse management of your push. If y'all finally push hard enough, the crate seems to slip suddenly and starts to move. Now static friction gives way to kinetic friction. One time in motility, it is easier to keep it in motion than it was to become it started, indicating that the kinetic frictional force is less than the static frictional force. If yous add together mass to the crate, say by placing a box on elevation of it, you demand to push fifty-fifty harder to get it started and also to continue it moving. Furthermore, if you oiled the concrete you lot would find it easier to get the crate started and proceed it going (equally you might expect).

(Effigy) is a crude pictorial representation of how friction occurs at the interface between two objects. Close-up inspection of these surfaces shows them to be crude. Thus, when you push to get an object moving (in this case, a crate), you lot must raise the object until it can skip along with only the tips of the surface hit, breaking off the points, or both. A considerable force can be resisted by friction with no apparent movement. The harder the surfaces are pushed together (such as if another box is placed on the crate), the more force is needed to movement them. Role of the friction is due to adhesive forces between the surface molecules of the two objects, which explains the dependence of friction on the nature of the substances. For example, safe-soled shoes sideslip less than those with leather soles. Adhesion varies with substances in contact and is a complicated attribute of surface physics. In one case an object is moving, there are fewer points of contact (fewer molecules adhering), and so less force is required to keep the object moving. At small but nonzero speeds, friction is nearly independent of speed.

The figure shows a crate on a flat surface. A black arrow points toward the right, away from the crate, and is labeled as the direction of motion or attempted motion. A red arrow pointing toward the left is located near the bottom left corner of the crate, at the interface between that corner and the supporting surface and is labeled as f. A magnified view of a bottom corner of the crate and the supporting surface shows that the roughness in the two surfaces leads to small gaps between them. There is direct contact only at a few points. — **Figure half-dozen.10** Frictional forces, such as
$\overset{\to }{f},$

always oppose move or attempted movement between objects in contact. Friction arises in part considering of the roughness of the surfaces in contact, as seen in the expanded view. For the object to move, it must rising to where the peaks of the superlative surface tin skip along the bottom surface. Thus, a force is required just to set the object in movement. Some of the peaks volition be broken off, also requiring a force to maintain motion. Much of the friction is really due to bonny forces between molecules making up the two objects, so that even perfectly smooth surfaces are not friction-costless. (In fact, perfectly smooth, clean surfaces of similar materials would attach, forming a bond chosen a "common cold weld.")

The magnitude of the frictional forcefulness has two forms: i for static situations (static friction), the other for situations involving motion (kinetic friction). What follows is an approximate empirical (experimentally determined) model only. These equations for static and kinetic friction are not vector equations.

Magnitude of Static Friction

The magnitude of static friction

${f}_{\text{s}}$

${f}_{\text{s}}\le {\mu }_{\text{s}}N,$

where

${\mu }_{\text{s}}$

is the coefficient of static friction and North is the magnitude of the normal forcefulness.

The symbol

$\le$

means less than or equal to, implying that static friction can accept a maximum value of

${\mu }_{\text{s}}N.$

Static friction is a responsive forcefulness that increases to be equal and opposite to whatever force is exerted, upwardly to its maximum limit. Once the practical force exceeds

${f}_{\text{s}}\text{(max),}$

the object moves. Thus,

${f}_{\text{s}}(\text{max})={\mu }_{\text{s}}N.$

Magnitude of Kinetic Friction

The magnitude of kinetic friction

${f}_{\text{k}}$

is given by

${f}_{\text{k}}={\mu }_{\text{k}}N,$

where

${\mu }_{\text{k}}$

is the coefficient of kinetic friction.

A system in which

${f}_{\text{k}}={\mu }_{\text{k}}N$

is described as a organisation in which friction behaves simply. The transition from static friction to kinetic friction is illustrated in (Effigy).

(a) The figure shows a block on a horizontal surface. The situation is that of impending motion. The following forces are shown: N vertically up, w vertically down, F to the right, f sub s to the left. Vectors N and w are the same size. Vectors F and f sub s are the same size. (b) The figure shows a block on a horizontal surface. The motion is to the right. The situation is that of friction behaving simply. The following forces are shown: N vertically up, w vertically down, F to the right, f sub k to the left. Vectors N and w are the same size. Vectors F is larger than f sub s. (c) A graph of the magnitude of the friction force f as a function of the applied force F is shown. In the interval from 0 to when the magnitude of f equals f sub s max, the graph is a straight line described by f sub s equals F. This is the static region, and f sub s max equals mu sub s times N. For values of F larger than this maximum value of f, the graph drops a bit then flattens out to a somewhat noisy but constant on average value. This is the kinetic region in which the magnitude of f is f sub k which is also equal to mu sub k times N. — **Figure 6.11** (a) The force of friction
$\overset{\to }{f}$

between the block and the rough surface opposes the direction of the applied forcefulness

$\overset{\to }{F}.$

The magnitude of the static friction balances that of the applied force. This is shown in the left side of the graph in (c). (b) At some bespeak, the magnitude of the applied force is greater than the forcefulness of kinetic friction, and the block moves to the right. This is shown in the right side of the graph. (c) The graph of the frictional force versus the practical force; note that

${f}_{\text{s}}(\text{max})>{f}_{\text{k}}.$

This means that

${\mu }_{\text{s}}>{\mu }_{\text{k}}.$

Equally you can see in (Figure), the coefficients of kinetic friction are less than their static counterparts. The approximate values of

$\mu$

are stated to but 1 or two digits to indicate the approximate clarification of friction given past the preceding two equations.

Approximate Coefficients of Static and Kinetic Friction
System	Static Friction ${\mu }_{\text{s}}$	Kinetic Friction ${\mu }_{\text{k}}$
Rubber on dry concrete	one.0	0.7
Rubber on wet concrete	0.5-0.vii	0.3-0.5
Wood on wood	0.v	0.3
Waxed woods on wet snow	0.xiv	0.1
Metallic on wood	0.5	0.three
Steel on steel (dry)	0.half-dozen	0.iii
Steel on steel (oiled)	0.05	0.03
Teflon on steel	0.04	0.04
Bone lubricated by synovial fluid	0.016	0.015
Shoes on wood	0.9	0.7
Shoes on water ice	0.i	0.05
Water ice on ice	0.one	0.03
Steel on ice	0.iv	0.02

(Figure) and (Figure) include the dependence of friction on materials and the normal force. The direction of friction is always opposite that of motion, parallel to the surface between objects, and perpendicular to the normal force. For example, if the crate you try to push (with a force parallel to the floor) has a mass of 100 kg, then the normal force is equal to its weight,

$w=mg=(100\,\text{kg})(9.80\,{\text{m/s}}^{2})=980\,\text{N,}$

perpendicular to the floor. If the coefficient of static friction is 0.45, you would have to exert a force parallel to the floor greater than

${f}_{\text{s}}(\text{max})={\mu }_{\text{s}}N=(0.45)(980\,\text{N})=440\,\text{N}$

to move the crate. Once in that location is motion, friction is less and the coefficient of kinetic friction might be 0.30, so that a force of only

${f}_{\text{k}}={\mu }_{\text{k}}N=(0.30)(980\,\text{N})=290\,\text{N}$

keeps it moving at a constant speed. If the floor is lubricated, both coefficients are considerably less than they would be without lubrication. Coefficient of friction is a unitless quantity with a magnitude usually between 0 and i.0. The bodily value depends on the two surfaces that are in contact.

Many people take experienced the slipperiness of walking on ice. However, many parts of the body, especially the joints, accept much smaller coefficients of friction—often three or 4 times less than ice. A joint is formed by the ends of two basic, which are connected by thick tissues. The knee joint is formed past the lower leg bone (the tibia) and the thighbone (the femur). The hip is a ball (at the cease of the femur) and socket (office of the pelvis) joint. The ends of the bones in the joint are covered by cartilage, which provides a smooth, most-glassy surface. The joints also produce a fluid (synovial fluid) that reduces friction and wear. A damaged or arthritic articulation tin be replaced past an artificial joint ((Effigy)). These replacements tin can exist made of metals (stainless steel or titanium) or plastic (polyethylene), as well with very modest coefficients of friction.

Two X ray photos of artificial knee replacements. — **Figure 6.12** Artificial knee replacement is a procedure that has been performed for more than 20 years. These post-operative X-rays testify a right knee joint replacement. (credit: Mike Baird)

Natural lubricants include saliva produced in our mouths to help in the swallowing procedure, and the glace mucus found between organs in the body, assuasive them to motility freely past each other during heartbeats, during breathing, and when a person moves. Hospitals and doctor's clinics commonly utilize bogus lubricants, such equally gels, to reduce friction.

The equations given for static and kinetic friction are empirical laws that draw the beliefs of the forces of friction. While these formulas are very useful for practical purposes, they do not accept the condition of mathematical statements that represent full general principles (e.1000., Newton's 2d constabulary). In fact, at that place are cases for which these equations are not even skilful approximations. For instance, neither formula is authentic for lubricated surfaces or for two surfaces siding beyond each other at high speeds. Unless specified, nosotros will not be concerned with these exceptions.

Example

Static and Kinetic Friction

A 20.0-kg crate is at rest on a floor as shown in (Figure). The coefficient of static friction between the crate and floor is 0.700 and the coefficient of kinetic friction is 0.600. A horizontal forcefulness

$\overset{\to }{P}$

is applied to the crate. Find the force of friction if (a)

$\overset{\to }{P}=20.0\,\text{N,}$

(b)

$\overset{\to }{P}=30.0\,\text{N,}$

(c)

$\overset{\to }{P}=120.0\,\text{N,}$

and (d)

$\overset{\to }{P}=180.0\,\text{N}\text{.}$

Here, may represent either the static or the kinetic frictional force. (a) An illustration of a man pushing a crate on a horizontal floor, exerting a force P directed horizontally to the right. (b) A free body diagram of the crate showing force P directed horizontally to the right, force f directed horizontally to the left, force N directed vertically up, and force w directed vertically down. An x y coordinate system is shown with positive x to the right and positive y up. — **Figure half dozen.13** (a) A crate on a horizontal surface is pushed with a force
$\overset{\to }{P}.$

(b) The forces on the crate. Here,

$\overset{\to }{f}$

may represent either the static or the kinetic frictional force.

Strategy

The costless-body diagram of the crate is shown in (Figure)(b). Nosotros employ Newton'due south second law in the horizontal and vertical directions, including the friction forcefulness in opposition to the direction of motion of the box.

Solution

Newton's second law gives

$\begin{array}{cccc}\sum {F}_{x}=m{a}_{x}\hfill & & & \sum {F}_{y}=m{a}_{y}\hfill \\ P-f=m{a}_{x}\hfill & & & N-w=0.\hfill \end{array}$

Here we are using the symbol f to represent the frictional force since we have not even so adamant whether the crate is subject to station friction or kinetic friction. We do this whenever we are unsure what type of friction is acting. At present the weight of the crate is

$w=(20.0\,\text{kg})(9.80\,{\text{m/s}}^{2})=196\,\text{N,}$

which is too equal to N. The maximum force of static friction is therefore

$(0.700)(196\,\text{N})=137\,\text{N}\text{.}$

Every bit long as

$\overset{\to }{P}$

is less than 137 Northward, the force of static friction keeps the crate stationary and

${f}_{\text{s}}=\overset{\to }{P}.$

Thus, (a)

${f}_{s}=20.0\,\text{N,}$

(b)

${f}_{s}=30.0\,\text{N,}$

and (c)

${f}_{s}=120.0\,\text{N}\text{.}$

(d) If

$\overset{\to }{P}=180.0\,\text{N,}$

the practical force is greater than the maximum force of static friction (137 N), so the crate tin can no longer remain at residue. Once the crate is in motility, kinetic friction acts. And then

${f}_{\text{k}}={\mu }_{\text{k}}N=(0.600)(196\,\text{N})=118\,\text{N,}$

and the acceleration is

${a}_{x}=\frac{\overset{\to }{P}-{f}_{\text{k}}}{m}=\frac{180.0\,\text{N}-118\,\text{N}}{20.0\,\text{kg}}=3.10\,{\text{m/s}}^{2}\text{.}$

Significance

This instance illustrates how we consider friction in a dynamics problem. Observe that static friction has a value that matches the applied force, until we achieve the maximum value of static friction. Also, no motility tin occur until the applied strength equals the forcefulness of static friction, but the force of kinetic friction will so become smaller.

Check Your Understanding

A cake of mass 1.0 kg rests on a horizontal surface. The frictional coefficients for the block and surface are

${\mu }_{s}=0.50$

and

${\mu }_{k}=0.40.$

(a) What is the minimum horizontal forcefulness required to move the block? (b) What is the block'southward acceleration when this force is applied?

[reveal-reply q="fs-id1165036788230″]Show Solution[/reveal-reply]

[hidden-answer a="fs-id1165036788230″]

a. iv.9 Northward; b. 0.98 m/s²

[/hidden-answer]

Friction and the Inclined Plane

I state of affairs where friction plays an obvious role is that of an object on a gradient. It might be a crate beingness pushed upwardly a ramp to a loading dock or a skateboarder benumbed downward a mountain, but the basic physics is the same. We usually generalize the sloping surface and phone call it an inclined airplane but so pretend that the surface is apartment. Let's look at an case of analyzing motion on an inclined aeroplane with friction.

Example

Downhill Skier

A skier with a mass of 62 kg is sliding down a snowy gradient at a constant velocity. Find the coefficient of kinetic friction for the skier if friction is known to exist 45.0 Due north.

Strategy

The magnitude of kinetic friction is given as 45.0 Northward. Kinetic friction is related to the normal force

$N$

past

${f}_{\text{k}}={\mu }_{\text{k}}N$

; thus, we can find the coefficient of kinetic friction if nosotros tin notice the normal force on the skier. The normal force is always perpendicular to the surface, and since there is no motility perpendicular to the surface, the normal force should equal the component of the skier's weight perpendicular to the slope. (See (Figure), which repeats a effigy from the affiliate on Newton'south laws of motion.)

The figure shows a skier going down a slope that forms an angle of 25 degrees with the horizontal. An x y coordinate system is shown, tilted so that the positive x direction is parallel to the slop, pointing up the slope, and the positive y direction is out of the slope, perpendicular to it. The weight of the skier, labeled w, is represented by a red arrow pointing vertically downward. This weight is divided into two components, w sub y is perpendicular to the slope pointing in the minus y direction, and w sub x is parallel to the slope, pointing in the minus x direction. The normal force, labeled N, is also perpendicular to the slope, equal in magnitude but pointing out, opposite in direction to w sub y. The friction, f, is represented by a red arrow pointing upslope. In addition, the figure shows a free body diagram that shows the relative magnitudes and directions of f, N, w, and the components w sub x and w sub y of w. In both diagrams, the w vector is scribbled out, as it is replaced by its components. — **Figure 6.14** The motion of the skier and friction are parallel to the slope, so it is about user-friendly to project all forces onto a coordinate system where i centrality is parallel to the slope and the other is perpendicular (axes shown to left of skier). The normal forcefulness
$\overset{\to }{N}$

is perpendicular to the slope, and friction

$\overset{\to }{f}$

is parallel to the slope, but the skier's weight

$\overset{\to }{w}$

has components along both axes, namely

${\overset{\to }{w}}_{y}$

and

${\overset{\to }{w}}_{x}.$

The normal force

$\overset{\to }{N}$

is equal in magnitude to

${\overset{\to }{w}}_{y},$

so there is no motion perpendicular to the slope. However,

$\overset{\to }{f}$

is less than

${\overset{\to }{w}}_{x}$

in magnitude, then at that place is acceleration down the gradient (along the x-axis).

We take

$N={w}_{y}=w\,\text{cos}\,25\text{°}=mg\,\text{cos}\,25\text{°}.$

Substituting this into our expression for kinetic friction, we obtain

${f}_{\text{k}}={\mu }_{\text{k}}mg\,\text{cos}\,25\text{°},$

which can now be solved for the coefficient of kinetic friction

${\mu }_{\text{k}}.$

Solution

Solving for

${\mu }_{\text{k}}$

gives

${\mu }_{\text{k}}=\frac{{f}_{\text{k}}}{N}=\frac{{f}_{\text{k}}}{w\,\text{cos}\,25\text{°}}=\frac{{f}_{\text{k}}}{mg\,\text{cos}\,25\text{°}}.$

Substituting known values on the right-hand side of the equation,

${\mu }_{\text{k}}=\frac{45.0\,\text{N}}{(62\,\text{kg})(9.80\,{\text{m/s}}^{2})(0.906)}=0.082.$

Significance

This result is a picayune smaller than the coefficient listed in (Figure) for waxed wood on snow, but it is still reasonable since values of the coefficients of friction can vary profoundly. In situations similar this, where an object of mass thousand slides downwards a gradient that makes an angle

$\theta$

with the horizontal, friction is given by

${f}_{\text{k}}={\mu }_{\text{k}}mg\,\text{cos}\,\theta .$

All objects slide down a slope with constant dispatch nether these circumstances.

Nosotros take discussed that when an object rests on a horizontal surface, the normal force supporting information technology is equal in magnitude to its weight. Furthermore, simple friction is always proportional to the normal force. When an object is non on a horizontal surface, as with the inclined airplane, we must find the force interim on the object that is directed perpendicular to the surface; information technology is a component of the weight.

Nosotros now derive a useful human relationship for calculating coefficient of friction on an inclined aeroplane. Detect that the issue applies only for situations in which the object slides at constant speed down the ramp.

An object slides downward an inclined airplane at a constant velocity if the internet forcefulness on the object is nix. We can use this fact to measure the coefficient of kinetic friction between ii objects. As shown in (Figure), the kinetic friction on a slope is

${f}_{k}={\mu }_{k}mg\,\text{cos}\,\theta$

. The component of the weight down the slope is equal to

$mg\,\text{sin}\,\theta$

(see the free-body diagram in (Figure)). These forces act in opposite directions, and then when they have equal magnitude, the acceleration is zero. Writing these out,

${\mu }_{\text{k}}mg\,\text{cos}\,\theta =mg\,\text{sin}\,\theta .$

Solving for

${\mu }_{\text{k}},$

nosotros find that

${\mu }_{\text{k}}=\frac{mg\,\text{sin}\,\theta }{mg\,\text{cos}\,\theta }=\text{tan}\,\theta .$

Put a coin on a book and tilt it until the money slides at a constant velocity downwardly the book. You might need to tap the book lightly to become the coin to move. Mensurate the angle of tilt relative to the horizontal and discover

${\mu }_{\text{k}}.$

Note that the money does not start to slide at all until an angle greater than

$\theta$

is attained, since the coefficient of static friction is larger than the coefficient of kinetic friction. Call up about how this may touch on the value for

${\mu }_{\text{k}}$

and its uncertainty.

Atomic-Scale Explanations of Friction

The simpler aspects of friction dealt with so far are its macroscopic (large-calibration) characteristics. Not bad strides accept been fabricated in the atomic-scale explanation of friction during the past several decades. Researchers are finding that the atomic nature of friction seems to accept several key characteristics. These characteristics not only explain some of the simpler aspects of friction—they also hold the potential for the development of nearly friction-free environments that could relieve hundreds of billions of dollars in energy which is currently being converted (unnecessarily) into heat.

(Figure) illustrates one macroscopic feature of friction that is explained past microscopic (pocket-sized-calibration) research. We have noted that friction is proportional to the normal strength, but non to the corporeality of area in contact, a somewhat counterintuitive notion. When 2 rough surfaces are in contact, the bodily contact area is a tiny fraction of the total area considering only high spots touch. When a greater normal force is exerted, the actual contact expanse increases, and we find that the friction is proportional to this surface area.

This figure has two parts, each of which shows two rough parallel surfaces in close proximity to each other. Because the surfaces are irregular, the two surfaces contact each other only at certain points, leaving gaps in between. In the first part, the normal force is small, so that the surfaces are farther apart and area of contact between the two surfaces is much smaller than their total area. In the second part, the normal force is large, so that the two surfaces are very close to each other and area of contact between the two surfaces has increased. — **Figure 6.15** 2 rough surfaces in contact have a much smaller area of bodily contact than their total expanse. When the normal force is larger as a result of a larger applied strength, the area of actual contact increases, as does friction.

Still, the atomic-scale view promises to explain far more than the simpler features of friction. The mechanism for how heat is generated is now being determined. In other words, why do surfaces become warmer when rubbed? Essentially, atoms are linked with one another to form lattices. When surfaces rub, the surface atoms adhere and crusade atomic lattices to vibrate—essentially creating audio waves that penetrate the material. The sound waves diminish with distance, and their energy is converted into heat. Chemical reactions that are related to frictional habiliment can as well occur between atoms and molecules on the surfaces. (Figure) shows how the tip of a probe drawn across another material is plain-featured past atomic-scale friction. The force needed to drag the tip can be measured and is found to be related to shear stress , which is discussed in Static Equilibrium and Elasticity. The variation in shear stress is remarkable (more than a factor of

${10}^{12}$

) and difficult to predict theoretically, but shear stress is yielding a fundamental understanding of a large-scale phenomenon known since ancient times—friction.

This figure shows a molecular model of a probe that is dragged over the surface of a substrate. The substrate is represented by a rectangular grid of small spheres, each sphere representing an atom. The probe, made up of a different grid of small spheres, is in the form of an inverted pyramid with a flattened peak and horizontal layers of atoms. The pyramid is somewhat distorted because of friction. The atomic and molecular interactions occur at the interface between the probe and the substrate. The friction, f, is parallel to the surface and in the opposite direction of the motion of the probe. — **Figure half dozen.16** The tip of a probe is plain-featured sideways past frictional force as the probe is dragged across a surface. Measurements of how the forcefulness varies for unlike materials are yielding key insights into the diminutive nature of friction.

Describe a model for friction on a molecular level. Depict matter in terms of molecular motion. The description should include diagrams to support the clarification; how the temperature affects the image; what are the differences and similarities between solid, liquid, and gas particle movement; and how the size and speed of gas molecules chronicle to everyday objects.

Example

Sliding Blocks

The two blocks of (Figure) are attached to each other by a massless string that is wrapped around a frictionless pulley. When the lesser 4.00-kg block is pulled to the left past the constant force

$\overset{\to }{P},$

the top 2.00-kg block slides across it to the right. Discover the magnitude of the force necessary to motility the blocks at constant speed. Assume that the coefficient of kinetic friction betwixt all surfaces is 0.400.

Figure (a) shows an illustration of a 4.0 kilogram block on a horizontal surface and a 2.0 kilogram block resting on top of it. A pulley is connected horizontally to a wall to the right of the blocks. The blocks are connected by a string that passes from one block, over the pulley, and to the other block so that the string is horizontal and to the right of each block. A force P pulls the lower block to the left. An x y coordinate system is shown, with positive x to the right and positive y up. Figure (b) shows the free body diagrams of the blocks. The upper block has forces mu times vector N sub 1 to the left, vector T to the right, 19.6 N vertically down, and vector N sub 1 up. The lower block has forces mu times vector N sub 1 to the right, mu times vector N sub 2 to the right, Vector P to the left, vector T sub i to the right, Vector N sub 1 vertically down, weight w down, and vector N sub 2 up. — **Figure half dozen.17** (a) Each block moves at constant velocity. (b) Costless-body diagrams for the blocks.

Strategy

We clarify the motions of the two blocks separately. The elevation cake is subjected to a contact forcefulness exerted by the bottom cake. The components of this forcefulness are the normal forcefulness

${N}_{1}$

and the frictional strength

$-0.400{N}_{1}.$

Other forces on the acme block are the tension

$T\text{i}$

in the string and the weight of the meridian block itself, 19.6 N. The bottom block is subjected to contact forces due to the top block and due to the floor. The kickoff contact force has components

$\text{−}{N}_{1}$

and

$0.400{N}_{1},$

which are simply reaction forces to the contact forces that the bottom block exerts on the top block. The components of the contact force of the floor are

${N}_{2}$

and

$0.400{N}_{2}.$

Other forces on this block are

$\text{−}P,$

the tension

$T\text{i},$

and the weight –39.2 N.

Solution

Since the tiptop block is moving horizontally to the right at constant velocity, its acceleration is zero in both the horizontal and the vertical directions. From Newton's second law,

$\begin{array}{cccccccc}\hfill \sum {F}_{x}& =\hfill & {m}_{1}{a}_{x}\hfill & & & \hfill \sum {F}_{y}& =\hfill & {m}_{1}{a}_{y}\hfill \\ \hfill T-0.400{N}_{1}& =\hfill & 0\hfill & & & \hfill {N}_{1}-19.6\,\text{N}& =\hfill & 0.\hfill \end{array}$

Solving for the two unknowns, we obtain

${N}_{1}=19.6\,\text{N}$

and

$T=0.40{N}_{1}=7.84\,\text{N}\text{.}$

The bottom cake is as well non accelerating, so the application of Newton'south 2d police to this cake gives

$\begin{array}{cccc}\sum {F}_{x}={m}_{2}{a}_{x}\hfill & & & \sum {F}_{y}={m}_{2}{a}_{y}\hfill \\ T-P+0.400\,{N}_{1}+0.400\,{N}_{2}=0\hfill & & & {N}_{2}-39.2\,\text{N}-{N}_{1}=0.\hfill \end{array}$

The values of

${N}_{1}$

and T were establish with the first set of equations. When these values are substituted into the 2nd set of equations, we can determine

${N}_{2}$

and P. They are

${N}_{2}=58.8\,\text{N}\enspace\text{and}\enspaceP=39.2\,\text{N}\text{.}$

Significance

Understanding what direction in which to draw the friction force is often troublesome. Notice that each friction force labeled in (Figure) acts in the management reverse the motion of its corresponding block.

Case

A Crate on an Accelerating Truck

A 50.0-kg crate rests on the bed of a truck as shown in (Figure). The coefficients of friction betwixt the surfaces are

${\mu }_{\text{k}}=0.300$

and

${\mu }_{\text{s}}=0.400.$

Observe the frictional force on the crate when the truck is accelerating forward relative to the ground at (a) 2.00 chiliad/s², and (b) 5.00 one thousand/due southⁱⁱ.

Figure (a) shows an illustration of a 50 kilogram crate on the bed of a truck. A horizontal arrow indicates an acceleration, a, to the right. An x y coordinate system is shown, with positive x to the right and positive y up. Figure (b) shows the free body diagram of the crate. The forces are 490 Newtons vertically down, vector N vertically up, and vector f horizontally to the right. — **Effigy 6.18** (a) A crate rests on the bed of the truck that is accelerating frontward. (b) The free-torso diagram of the crate.

Strategy

The forces on the crate are its weight and the normal and frictional forces due to contact with the truck bed. We start past assuming that the crate is not slipping. In this case, the static frictional force

${f}_{\text{s}}$

acts on the crate. Furthermore, the accelerations of the crate and the truck are equal.

Solution

Awarding of Newton'south second police force to the crate, using the reference frame attached to the basis, yields
$\begin{array}{cccccccc}\hfill \sum {F}_{x}& =\hfill & m{a}_{x}\hfill & & & \hfill \sum {F}_{y}& =\hfill & m{a}_{y}\hfill \\ \hfill {f}_{\text{s}}& =\hfill & (50.0\,\text{kg})(2.00\,{\text{m/s}}^{2})\hfill & & & \hfill N-4.90\,×\,{10}^{2}\,\text{N}& =\hfill & (50.0\,\text{kg})(0)\hfill \\ & =\hfill & 1.00\,×\,{10}^{2}\,\text{N}\hfill & & & \hfill N& =\hfill & 4.90\,×\,{10}^{2}\,\text{N}\text{.}\hfill \end{array}$

We can now check the validity of our no-slip assumption. The maximum value of the force of static friction is

${\mu }_{\text{s}}N=(0.400)(4.90\,×\,{10}^{2}\,\text{N})=196\,\text{N,}$

whereas the actual force of static friction that acts when the truck accelerates forward at

$2.00\,{\text{m/s}}^{2}$

is only

$1.00\,×\,{10}^{2}\,\text{N}\text{.}$

Thus, the assumption of no slipping is valid.
If the crate is to motility with the truck when it accelerates at
$5.0\,{\text{m/s}}^{2},$

the forcefulness of static friction must exist

${f}_{\text{s}}=m{a}_{x}=(50.0\,\text{kg})(5.00\,{\text{m/s}}^{2})=250\,\text{N}\text{.}$

Since this exceeds the maximum of 196 N, the crate must skid. The frictional force is therefore kinetic and is

${f}_{k}={\mu }_{k}N=(0.300)(4.90\,×\,{10}^{2}\,\text{N})=147\,\text{N}\text{.}$

The horizontal acceleration of the crate relative to the ground is now plant from

$\begin{array}{ccc}\hfill \sum {F}_{x}& =\hfill & m{a}_{x}\hfill \\ \hfill 147\,\text{N}& =\hfill & (50.0\,\text{kg}){a}_{x},\hfill \\ \hfill \text{so}\,{a}_{x}& =\hfill & 2.94\,{\text{m/s}}^{2}.\hfill \end{array}$

Significance

Relative to the ground, the truck is accelerating forward at

$5.0\,{\text{m/s}}^{2}$

and the crate is accelerating forward at

$2.94\,{\text{m/s}}^{2}$

. Hence the crate is sliding backward relative to the bed of the truck with an acceleration

$2.94\,{\text{m/s}}^{2}-5.00\,{\text{m/s}}^{2}=-2.06\,{\text{m/s}}^{2}\text{.}$

Example

Snowboarding

Earlier, we analyzed the situation of a downhill skier moving at constant velocity to determine the coefficient of kinetic friction. Now let's practise a similar analysis to determine acceleration. The snowboarder of (Figure) glides downwardly a slope that is inclined at

$\theta ={13}^{0}$

to the horizontal. The coefficient of kinetic friction between the board and the snow is

${\mu }_{\text{k}}=0.20.$

What is the dispatch of the snowboarder?

Figure (a) shows an illustration of a snowboarder on a slope inclined at 13 degrees above the horizontal. An arrow indicates an acceleration, a, downslope. Figure (b) shows the free body diagram of the snowboarder. The forces are m g cosine 13 degrees into the slope, perpendicular to the surface, N, out of the slope, perpendicular to the surface, m g sine 13 degrees downslope parallel to the surface and mu sub k times N, upslope parallel to the surface. — **Figure half-dozen.19** (a) A snowboarder glides downward a slope inclined at 13° to the horizontal. (b) The gratuitous-body diagram of the snowboarder.

Strategy

The forces acting on the snowboarder are her weight and the contact forcefulness of the slope, which has a component normal to the incline and a component forth the incline (force of kinetic friction). Because she moves forth the slope, the most convenient reference frame for analyzing her motion is one with the 10-centrality along and the y-axis perpendicular to the incline. In this frame, both the normal and the frictional forces prevarication along coordinate axes, the components of the weight are

$mg\,\text{sin}\,\theta \,\text{along the slope and}\,mg\,\text{cos}\,\theta \,\text{at right angles into the slope}$

, and the only acceleration is along the ten-axis

$({a}_{y}=0).$

Solution

We can at present employ Newton's second police to the snowboarder:

$\begin{array}{cccccc}\hfill \sum {F}_{x}& =\hfill & m{a}_{x}\hfill & & & \sum {F}_{y}=m{a}_{y}\hfill \\ \hfill mg\,\text{sin}\,\theta -{\mu }_{k}N& =\hfill & m{a}_{x}\hfill & & & \hfill N-mg\,\text{cos}\,\theta =m(0)\text{.}\end{array}$

From the second equation,

$N=mg\,\text{cos}\,\theta .$

Upon substituting this into the first equation, we detect

$\begin{array}{cc}\hfill {a}_{x}& =g(\text{sin}\,\theta -{\mu }_{\text{k}}\,\text{cos}\,\theta )\hfill \\ & =g(\text{sin}\,13\text{°}-0.20\,\text{cos}\,13\text{°})=0.29\,{\text{m/s}}^{2}\text{.}\hfill \end{array}$

Significance

Notice from this equation that if

$\theta$

is small enough or

${\mu }_{\text{k}}$

is large enough,

${a}_{x}$

is negative, that is, the snowboarder slows down.

Bank check Your Understanding

The snowboarder is now moving down a colina with incline

$10.0\text{°}$

. What is the skier'due south acceleration?

[reveal-answer q="fs-id1165037850930″]Show Solution[/reveal-answer]

[hidden-answer a="fs-id1165037850930″]

$\text{−0.23}\,{\text{m/s}}^{2}$

; the negative sign indicates that the snowboarder is slowing downwards.
[/hidden-answer]

Summary

Friction is a contact force that opposes the motion or attempted motility betwixt two systems. Unproblematic friction is proportional to the normal forcefulness Due north supporting the two systems.
The magnitude of static friction force between 2 materials stationary relative to each other is determined using the coefficient of static friction, which depends on both materials.
The kinetic friction strength between two materials moving relative to each other is determined using the coefficient of kinetic friction, which too depends on both materials and is always less than the coefficient of static friction.

Conceptual Questions

The glue on a piece of tape can exert forces. Can these forces be a type of simple friction? Explain, because peculiarly that tape can stick to vertical walls and fifty-fifty to ceilings.

When y'all learn to bulldoze, you discover that you need to let up slightly on the restriction pedal as you come to a stop or the machine will stop with a jerk. Explain this in terms of the human relationship between static and kinetic friction.

[reveal-answer q="fs-id1165038273214″]Show Solution[/reveal-respond]

[subconscious-answer a="fs-id1165038273214″]

If you exercise non allow up on the restriction pedal, the machine's wheels will lock then that they are not rolling; sliding friction is at present involved and the sudden modify (due to the larger forcefulness of static friction) causes the jerk.

[/hidden-answer]

When you button a slice of chalk across a chalkboard, it sometimes screeches because it rapidly alternates betwixt slipping and sticking to the board. Draw this process in more than detail, in particular, explaining how it is related to the fact that kinetic friction is less than static friction. (The same sideslip-grab process occurs when tires screech on pavement.)

A physics major is cooking breakfast when she notices that the frictional force between her steel spatula and Teflon frying pan is only 0.200 N. Knowing the coefficient of kinetic friction betwixt the two materials, she quickly calculates the normal force. What is it?

[reveal-answer q="fs-id1165036988460″]Show Solution[/reveal-answer]

[subconscious-answer a="fs-id1165036988460″]

5.00 N

[/hidden-answer]

Problems

(a) When rebuilding his machine's engine, a physics major must exert

$3.00\,×\,{10}^{2}$

N of force to insert a dry steel piston into a steel cylinder. What is the normal strength between the piston and cylinder? (b) What force would he have to exert if the steel parts were oiled?

(a) What is the maximum frictional force in the knee of a person who supports 66.0 kg of her mass on that knee joint? (b) During strenuous do, information technology is possible to exert forces to the joints that are hands 10 times greater than the weight existence supported. What is the maximum force of friction under such conditions? The frictional forces in joints are relatively minor in all circumstances except when the joints deteriorate, such equally from injury or arthritis. Increased frictional forces can cause further damage and hurting.

[reveal-answer q="fs-id1165037987870″]Show Solution[/reveal-answer]

[hidden-answer a="fs-id1165037987870″]

a. ten.0 Due north; b. 97.0 N

[/hidden-respond]

Suppose you take a 120-kg wooden crate resting on a wood floor, with coefficient of static friction 0.500 between these wood surfaces. (a) What maximum force can you lot exert horizontally on the crate without moving it? (b) If y'all continue to exert this force once the crate starts to skid, what will its acceleration so exist? The coefficient of sliding friction is known to exist 0.300 for this situation.

(a) If half of the weight of a modest

$1.00\,×\,{10}^{3}\text{-kg}$

utility truck is supported by its two bulldoze wheels, what is the maximum acceleration information technology tin can accomplish on dry out physical? (b) Volition a metallic chiffonier lying on the wooden bed of the truck slip if it accelerates at this rate? (c) Solve both bug assuming the truck has four-wheel bulldoze.

[reveal-respond q="fs-id1165036766851″]Evidence Solution[/reveal-respond]

[hidden-answer a="fs-id1165036766851″]

$4.9\,{\text{m/s}}^{2}$

; b. The chiffonier will non skid. c. The cabinet volition sideslip.
[/hidden-answer]

A squad of viii dogs pulls a sled with waxed wood runners on moisture snow (mush!). The dogs have average masses of 19.0 kg, and the loaded sled with its passenger has a mass of 210 kg. (a) Calculate the acceleration of the dogs starting from rest if each dog exerts an average forcefulness of 185 Due north backward on the snowfall. (b) Calculate the force in the coupling between the dogs and the sled.

Consider the 65.0-kg water ice skater existence pushed by two others shown below. (a) Find the management and magnitude of

${F}_{\text{tot}},$

the total strength exerted on her by the others, given that the magnitudes

${F}_{1}$

and

${F}_{2}$

are 26.4 Due north and xviii.half dozen N, respectively. (b) What is her initial acceleration if she is initially stationary and wearing steel-bladed skates that point in the direction of

${F}_{\text{tot}}?$

(Remember that friction e'er acts in the direction contrary that of motion or attempted motility between surfaces in contact.)

[reveal-answer q="659751″]Evidence Solution[/reveal-answer]
[hidden-answer a="659751″]a. 32.iii N,

$35.2\text{°};$

b. 0; c.

$0.301\,{\text{m/s}}^{2}$

in the direction of

${\overset{\to }{F}}_{\text{tot}}$

[/hidden-answer]

Show that the acceleration of whatever object down a frictionless incline that makes an angle

$\theta$

with the horizontal is

$a=g\,\text{sin}\,\theta$

. (Note that this acceleration is independent of mass.)

Show that the acceleration of whatever object down an incline where friction behaves just (that is, where

${f}_{\text{k}}={\mu }_{\text{k}}N)$

$a=g(\text{sin}\,\theta -{\mu }_{\text{k}}\,\text{cos}\,\theta ).$

Note that the acceleration is independent of mass and reduces to the expression constitute in the previous problem when friction becomes negligibly modest

$({\mu }_{\text{k}}=0).$

[reveal-answer q="607581″]Show Solution[/reveal-reply]
[hidden-answer a="607581″]

$\begin{array}{ccc}\hfill \text{net}\,{F}_{y}& =\hfill & 0⇒N=mg\,\text{cos}\,\theta \hfill \\ \hfill \text{net}\,{F}_{x}& =\hfill & ma\hfill \\ \hfill a& =\hfill & g(\text{sin}\,\theta -{\mu }_{\text{k}}\,\text{cos}\,\theta )\hfill \end{array}$

[/subconscious-answer]

Calculate the deceleration of a snowfall boarder going up a

$5.00\text{°}$

slope, assuming the coefficient of friction for waxed woods on wet snow. The effect of the preceding problem may exist useful, but exist careful to consider the fact that the snow boarder is going uphill.

A auto at a post part sends packages out a chute and downwards a ramp to be loaded into delivery vehicles. (a) Calculate the dispatch of a box heading downward a

$10.0\text{°}$

slope, bold the coefficient of friction for a parcel on waxed wood is 0.100. (b) Find the angle of the slope down which this box could movement at a abiding velocity. You tin neglect air resistance in both parts.

[reveal-respond q="fs-id1165037974281″]Show Solution[/reveal-answer]

[hidden-answer a="fs-id1165037974281″]

$1.69\,{\text{m/s}}^{2};$

$5.71\text{°}$

[/hidden-answer]

If an object is to rest on an incline without slipping, so friction must equal the component of the weight of the object parallel to the incline. This requires greater and greater friction for steeper slopes. Bear witness that the maximum angle of an incline above the horizontal for which an object volition not slide down is

$\theta ={\text{tan}}^{-1}\,{\mu }_{\text{s}}.$

Y'all may use the effect of the previous problem. Assume that

$a=0$

and that static friction has reached its maximum value.
An illustration of a block mass m on a slope. The slope angles up and to the right at an angle of theta degrees to the horizontal. The mass feels force w sub parallel in a direction parallel to the slope toward its bottom, and f in a direction parallel to the slope toward its top.

Calculate the maximum acceleration of a auto that is heading downwards a

$6.00\text{°}$

slope (i that makes an bending of

$6.00\text{°}$

with the horizontal) nether the post-obit road atmospheric condition. You may presume that the weight of the automobile is evenly distributed on all four tires and that the coefficient of static friction is involved—that is, the tires are non allowed to slip during the deceleration. (Ignore rolling.) Calculate for a car: (a) On dry physical. (b) On wet concrete. (c) On ice, bold that

${\mu }_{\text{s}}=0.100$

, the same equally for shoes on ice.

[reveal-answer q="fs-id1165036838774″]Evidence Solution[/reveal-answer]

[subconscious-answer a="fs-id1165036838774″]

$10.8\,{\text{m/s}}^{2};$

$7.85\,{\text{m/s}}^{2};$

c.
[/hidden-answer]

Calculate the maximum acceleration of a car that is heading up a

$4.00\text{°}$

slope (1 that makes an angle of

$4.00\text{°}$

with the horizontal) under the following road conditions. Assume that simply half the weight of the car is supported by the two bulldoze wheels and that the coefficient of static friction is involved—that is, the tires are not allowed to slip during the acceleration. (Ignore rolling.) (a) On dry physical. (b) On wet concrete. (c) On water ice, assuming that

${\mu }_{\text{s}}=0.100$

, the aforementioned as for shoes on ice.

Repeat the preceding trouble for a auto with 4-cycle drive.

[reveal-answer q="fs-id1165037157329″]Evidence Solution[/reveal-respond]

[hidden-answer a="fs-id1165037157329″]

$9.09\,{\text{m/s}}^{2};$

$6.16\,{\text{m/s}}^{2};$

$0.294\,{\text{m/s}}^{2}$

[/hidden-answer]

A freight railroad train consists of two

$8.00\,×\,{10}^{5}\text{-kg}$

engines and 45 cars with average masses of

$5.50\,×\,{10}^{5}\,\text{kg}\text{.}$

(a) What strength must each engine exert backward on the track to accelerate the train at a rate of

$5.00\,×\,{10}^{-2}\text{m}\text{/}{\text{s}}^{2}$

if the strength of friction is

$7.50\,×\,{10}^{5}\text{N}$

, assuming the engines exert identical forces? This is not a large frictional force for such a massive system. Rolling friction for trains is small, and consequently, trains are very energy-efficient transportation systems. (b) What is the force in the coupling between the 37th and 38th cars (this is the force each exerts on the other), assuming all cars have the same mass and that friction is evenly distributed amid all of the cars and engines?

Consider the 52.0-kg mountain climber shown below. (a) Find the tension in the rope and the strength that the mount climber must exert with her feet on the vertical stone face to remain stationary. Assume that the forcefulness is exerted parallel to her legs. Also, assume negligible force exerted past her arms. (b) What is the minimum coefficient of friction between her shoes and the cliff?

[reveal-reply q="129819″]Testify Solution[/reveal-answer]
[hidden-answer a="129819″]a. 272 N, 512 N; b. 0.268[/hidden-reply]

A contestant in a wintertime sporting event pushes a 45.0-kg block of ice across a frozen lake as shown beneath. (a) Calculate the minimum force F he must exert to get the block moving. (b) What is its acceleration once it starts to motility, if that force is maintained?

A block of ice is being pushed with a force F that is directed at an angle of twenty five degrees below the horizontal.

The contestant now pulls the block of ice with a rope over his shoulder at the same angle above the horizontal as shown below. Calculate the minimum forcefulness F he must exert to get the cake moving. (b) What is its dispatch one time it starts to move, if that force is maintained?

A block of ice is being pulled with a force F that is directed at an angle of twenty five degrees above the horizontal.
[reveal-respond q="665628″]Show Solution[/reveal-answer]
[hidden-answer a="665628″]a. 46.five North; b. $0.629\,{\text{m/s}}^{2}$ [/hidden-respond]

At a post role, a bundle that is a xx.0-kg box slides downward a ramp inclined at

$30.0\text{°}$

with the horizontal. The coefficient of kinetic friction betwixt the box and aeroplane is 0.0300. (a) Notice the dispatch of the box. (b) Discover the velocity of the box equally it reaches the end of the plane, if the length of the plane is 2 m and the box starts at residue.

Glossary

friction: strength that opposes relative motion or attempts at motion between systems in contact

kinetic friction: strength that opposes the motion of two systems that are in contact and moving relative to each other

static friction: force that opposes the motion of two systems that are in contact and are not moving relative to each other

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Source: https://opentextbc.ca/universityphysicsv1openstax/chapter/6-2-friction/

when the crate falls off in what direction is it moving relative to the ground

half dozen.ii Friction

Learning Objectives

Static and Kinetic Friction

Friction

Static and Kinetic Friction

Magnitude of Static Friction

Magnitude of Kinetic Friction

Example

Static and Kinetic Friction

Strategy

Solution

Significance

Check Your Understanding

Friction and the Inclined Plane

Example

Downhill Skier

Strategy

Solution

Significance

Atomic-Scale Explanations of Friction

Example

Sliding Blocks

Strategy

Solution

Significance

Case

A Crate on an Accelerating Truck

Strategy

Solution

Significance

Example

Snowboarding

Strategy

Solution

Significance

Bank check Your Understanding

Summary

Conceptual Questions

Problems

Glossary

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