Why When U Rub Something Over and Over Again the Friction Changes

Learning Objectives

By the finish of this section, you will be able to:

Discuss the general characteristics of friction.
Depict the diverse types of friction.
Calculate the magnitude of static and kinetic friction.

Friction is a force that is around us all the time that opposes relative move between systems in contact simply also allows us to move (which you lot have discovered if you take ever tried to walk on water ice). While a common force, the behavior of friction is actually very complicated and is nevertheless non completely understood. We take to rely heavily on observations for any understandings we can gain. Nonetheless, we can nonetheless deal with its more than elementary general characteristics and sympathise the circumstances in which information technology behaves.

Friction

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

One of the simpler characteristics of friction is that it is parallel to the contact surface between systems and always in a direction that opposes movement or attempted motion of the systems relative to each other. If two systems are in contact and moving relative to one another, and so the friction between them is called kinetic friction. For example, friction slows a hockey puck sliding on water ice. But when objects are stationary, static friction tin act between them; the static friction is usually greater than the kinetic friction betwixt the objects.

Kinetic Friction

If two systems are in contact and moving relative to ane another, so the friction between them is chosen kinetic friction.

Imagine, for instance, trying to slide a heavy crate across a concrete floor—y'all may push harder and harder on the crate and non motility it at all. This means that the static friction responds to what y'all do—it increases to be equal to and in the opposite direction of your push. Simply if you finally push hard enough, the crate seems to slip suddenly and starts to move. In one case in motion it is easier to keep information technology in motility than information technology was to get it started, indicating that the kinetic friction force is less than the static friction force. If you lot add mass to the crate, say by placing a box on top of it, you demand to push button even harder to get it started and also to keep information technology moving. Furthermore, if you oiled the physical you lot would find it to be easier to go the crate started and keep it going (equally you might await).

Figure i is a crude pictorial representation of how friction occurs at the interface betwixt ii objects. Close-up inspection of these surfaces shows them to be crude. Then when you push to get an object moving (in this example, a crate), y'all must enhance the object until it can skip forth with only the tips of the surface hit, break off the points, or practice both. A considerable force can be resisted by friction with no apparent move. The harder the surfaces are pushed together (such as if another box is placed on the crate), the more force is needed to move them. Part of the friction is due to adhesive forces between the surface molecules of the two objects, which explain the dependence of friction on the nature of the substances. Adhesion varies with substances in contact and is a complicated aspect of surface physics. One time an object is moving, there are fewer points of contact (fewer molecules adhering), so less force is required to keep the object moving. At small just nonzero speeds, friction is well-nigh independent of speed.

Figure 1.

Frictional forces, such as f, always oppose movement or attempted motion between objects in contact. Friction arises in function because of the roughness of the surfaces in contact, as seen in the expanded view. In order for the object to motion, it must ascent to where the peaks can skip forth the bottom surface. Thus a force is required merely to gear up the object in movement. Some of the peaks will be broken off, also requiring a strength to maintain motion. Much of the friction is actually due to attractive forces between molecules making up the ii objects, and then that fifty-fifty perfectly smooth surfaces are non friction-gratis. Such adhesive forces as well depend on the substances the surfaces are made of, explaining, for example, why rubber-soled shoes skid less than those with leather soles.

The magnitude of the frictional force has ii forms: one for static situations (static friction), the other for when there is motion (kinetic friction).

When there is no move between the objects, the magnitude of static friction f _s is f _s ≤μ _s N, where μ _s is the coefficient of static friction and N is the magnitude of the normal force (the strength perpendicular to the surface).

Magnitude of Static Friction

Magnitude of static friction f _s is f _southward ≤μ _{due south} N, where μ _s is the coefficient of static friction and Northward is the magnitude of the normal force.

The symbol ≤ ways less than or equal to, implying that static friction can have a minimum and a maximum value of μ _s N. Static friction is a responsive forcefulness that increases to exist equal and opposite to any force is exerted, upwardly to its maximum limit. One time the applied forcefulness exceeds f _s(max), the object will move. Thus f _s(max) = μ _s Northward.

Once an object is moving, the magnitude of kinetic friction f _k is given by f _k = μ _m N, where μ _k is the coefficient of kinetic friction. A organisation in whichf _k = μ _k N is described as a system in which friction behaves simply.

Magnitude of Kinetic Friction

The magnitude of kinetic friction f ₁₀₀₀ is given by f ₁₀₀₀=μ _m N, where μ _k is the coefficient of kinetic friction.

As seen in Tabular array 1, the coefficients of kinetic friction are less than their static counterparts. That values of μ in Table ane are stated to only one or, at most, two digits is an indication of the approximate description of friction given by the above two equations.

Table 1. Coefficients of Static and Kinetic Friction
Arrangement	Static frictionμ _south	Kinetic frictionμ _k
Prophylactic on dry out concrete	1.0	0.7
Prophylactic on wet concrete	0.7	0.5
Wood on wood	0.5	0.3
Waxed wood on wet snow	0.14	0.i
Metal on wood	0.v	0.3
Steel on steel (dry out)	0.6	0.3
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.vii
Shoes on water ice	0.1	0.05
Ice on ice	0.1	0.03
Steel on ice	0.four	0.02

The equations given earlier include the dependence of friction on materials and the normal force. The direction of friction is e'er opposite that of motion, parallel to the surface between objects, and perpendicular to the normal force. For example, if the crate yous try to push button (with a strength parallel to the flooring) has a mass of 100 kg, and so the normal force would be equal to its weight, W =mg = (100 kg)(9.80 g/sⁱⁱ) = 980 Northward, perpendicular to the floor. If the coefficient of static friction is 0.45, you lot would have to exert a force parallel to the floor greater thanf_s _(max) =μ_{due south}N = (0.45)(980)N = 440N to motility the crate. In one case there is movement, friction is less and the coefficient of kinetic friction might be 0.30, then that a force of only 290 N f _thousand =μ _k N = (0.thirty)(980N) = 290N would keep it moving at a constant speed. If the floor is lubricated, both coefficients are considerably less than they would exist without lubrication. Coefficient of friction is a unit of measurement less quantity with a magnitude usually betwixt 0 and i.0. The coefficient of the friction depends on the 2 surfaces that are in contact.

Accept-Home Experiment

Find a small-scale plastic object (such equally a food container) and slide it on a kitchen table by giving it a gentle tap. Now spray h2o on the table, simulating a low-cal shower of rain. What happens now when y'all give the object the same-sized tap? Now add a few drops of (vegetable or olive) oil on the surface of the h2o and requite the same tap. What happens now? This latter situation is specially important for drivers to notation, especially later a light rain shower. Why?

Many people accept experienced the slipperiness of walking on water ice. Even so, many parts of the body, peculiarly the joints, have much smaller coefficients of friction—frequently three or 4 times less than ice. A joint is formed past the ends of two basic, which are connected by thick tissues. The knee articulation is formed by the lower leg bone (the tibia) and the thighbone (the femur). The hip is a ball (at the end of the femur) and socket (part of the pelvis) joint. The ends of the bones in the joint are covered by cartilage, which provides a smooth, nigh burnished surface. The joints besides produce a fluid (synovial fluid) that reduces friction and wear. A damaged or arthritic joint tin can be replaced past an artificial joint (Figure 2). These replacements can be made of metals (stainless steel or titanium) or plastic (polyethylene), besides with very small coefficients of friction.

Two x rays of an artificial knee replacement are shown.

Effigy 2. Artificial knee replacement is a process that has been performed for more than 20 years. In this figure, we run into the post-op 10 rays of the correct human knee joint replacement. (credit: Mike Baird, Flickr)

Other natural lubricants include saliva produced in our mouths to assist in the swallowing process, and the slippery fungus constitute between organs in the trunk, allowing them to move freely by each other during heartbeats, during breathing, and when a person moves. Artificial lubricants are also common in hospitals and doctor's clinics. For instance, when ultrasonic imaging is carried out, the gel that couples the transducer to the skin likewise serves to to lubricate the surface between the transducer and the peel—thereby reducing the coefficient of friction betwixt the two surfaces. This allows the transducer to mover freely over the skin.

Case 1. Skiing Exercise

A skier with a mass of 62 kg is sliding down a snowy slope. Find the coefficient of kinetic friction for the skier if friction is known to be 45.0 N.

Strategy

The magnitude of kinetic friction was given in to be 45.0 N. Kinetic friction is related to the normal strength N as f _k =μ _k Northward; thus, the coefficient of kinetic friction can exist institute if we can find the normal force of the skier on a slope. The normal strength is always perpendicular to the surface, and since there is no motion perpendicular to the surface, the normal force should equal the component of the skier's weight perpendicular to the gradient. (See the skier and costless-body diagram in Figure three.)

Figure iii.

The movement of the skier and friction are parallel to the slope and and so it is most convenient to project all forces onto a coordinate system where one axis is parallel to the gradient and the other is perpendicular (axes shown to left of skier). N (the normal force) is perpendicular to the gradient, and f (the friction) is parallel to the slope, just westward (the skier's weight) has components along both axes, namely w_⊥ and W_//. N is equal in magnitude to w_⊥, so in that location is no motion perpendicular to the slope. Still, f is less than Due west_// in magnitude, so there is acceleration downward the slope (along the ten-axis).

That is,N =due west _⊥ =w cos 25º =mg cos 25º.

Substituting this into our expression for kinetic friction, we gof _{one thousand} =μ _k mgcos 25º, which tin can at present be solved for the coefficient of kinetic friction μ _k.

Solution

Solving for μ _yard gives [latex]\displaystyle\mu_k=\frac{f_k}{N}=\frac{f_k}{w\text{ cos }25^{\circ}}=\frac{f_k}{mg\text{ cos }25^{\circ}}\\[/latex]

Substituting known values on the right-hand side of the equation, [latex]\displaystyle\mu_k=\frac{45.0}{(62\text{ kg})(9.80\text{ one thousand/southward}^two)(0.906)}=0.082\\[/latex].

Discussion

This result is a piffling smaller than the coefficient listed in Table 5.i for waxed wood on snow, but it is still reasonable since values of the coefficients of friction can vary profoundly. In situations like this, where an object of mass 1000 slides downwardly a slope that makes an angle θ with the horizontal, friction is given byf _k =μ _{one thousand}mg cos θ. All objects will slide downwardly a gradient with abiding acceleration under these circumstances. Proof of this is left for this chapter's Issues and Exercises.

Take-Dwelling house Experiment

An object will slide down an inclined airplane at a constant velocity if the net strength on the object is zero. We can use this fact to measure the coefficient of kinetic friction betwixt two objects. As shown in Example 1, the kinetic friction on a slope f _k =μ _kmg cos θ. The component of the weight down the slope is equal to mg sin θ (encounter the gratis-body diagram in Figure 3). These forces act in opposite directions, so when they have equal magnitude, the acceleration is zero. Writing these out:

f _k =Fg _ten

μ _grandmg cos θ =mg sin θ.

Solving for μ _k, we notice that

[latex]\displaystyle\mu_{\text{chiliad}}=\frac{mg\sin\theta}{mg\cos\theta}=\tan\theta\\[/latex]

Put a coin on a book and tilt it until the coin slides at a abiding velocity downward the book. You might need to tap the volume lightly to get the coin to motility. Measure the bending of tilt relative to the horizontal and discover μ _k. Note that the coin volition not start to slide at all until an angle greater than θ is attained, since the coefficient of static friction is larger than the coefficient of kinetic friction. Discuss how this may impact the value for μ _yard and its doubt.

Nosotros have discussed that when an object rests on a horizontal surface, there is a normal forcefulness supporting information technology equal in magnitude to its weight. Furthermore, simple friction is always proportional to the normal forcefulness.

Making Connections: Submicroscopic Explanations of Friction

The simpler aspects of friction dealt with and then far are its macroscopic (large-scale) characteristics. Great strides have been made in the atomic-scale explanation of friction during the past several decades. Researchers are finding that the diminutive nature of friction seems to have several fundamental characteristics. These characteristics not simply explain some of the simpler aspects of friction—they also hold the potential for the development of nearly friction-complimentary environments that could save hundreds of billions of dollars in energy which is currently being converted (unnecessarily) to heat.

Figure iv illustrates one macroscopic feature of friction that is explained past microscopic (small-scale) inquiry. Nosotros take noted that friction is proportional to the normal force, just not to the expanse in contact, a somewhat counterintuitive notion. When ii crude surfaces are in contact, the bodily contact area is a tiny fraction of the full expanse since just high spots touch. When a greater normal strength is exerted, the actual contact area increases, and it is plant that the friction is proportional to this area.

Figure 4. Two crude surfaces in contact accept a much smaller area of bodily contact than their total surface area. When there is a greater normal force as a effect of a greater practical force, the area of actual contact increases equally does friction.

Merely the atomic-calibration view promises to explain far more than the simpler features of friction. The mechanism for how heat is generated is now existence determined. In other words, why practice surfaces go warmer when rubbed? Essentially, atoms are linked with one another to form lattices. When surfaces rub, the surface atoms attach and crusade atomic lattices to vibrate—essentially creating sound waves that penetrate the textile. The sound waves diminish with distance and their energy is converted into heat. Chemical reactions that are related to frictional wearable can likewise occur between atoms and molecules on the surfaces. Figure v shows how the tip of a probe drawn across some other material is plain-featured by diminutive-calibration friction. The force needed to drag the tip can be measured and is found to be related to shear stress, which will exist discussed afterward in this chapter. The variation in shear stress is remarkable (more than a factor of 10¹² ) 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 prism, made up of a 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. The pyramid is somewhat distorted because of friction.

Figure five. The tip of a probe is plain-featured sideways past frictional forcefulness every bit the probe is dragged across a surface. Measurements of how the force varies for unlike materials are yielding fundamental insights into the atomic nature of friction.

PhET Explorations: Forces and Move

Explore the forces at piece of work when you try to push a filing cabinet. Create an applied force and see the resulting friction force and full force acting on the cabinet. Charts testify the forces, position, velocity, and acceleration vs. time. Draw a free-body diagram of all the forces (including gravitational and normal forces).

Forces and Motion

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Section Summary

Friction is a contact force between systems that opposes the motility or attempted motion between them. Simple friction is proportional to the normal force N pushing the systems together. (A normal forcefulness is always perpendicular to the contact surface between systems.) Friction depends on both of the materials involved. The magnitude of static friction [latex]{f}_{\text{south}}\\[/latex] between systems stationary relative to one some other is given by [latex]{f}_{\text{due south}}\le {\mu }_{\text{s}}N\\[/latex], where [latex]{\mu }_{\text{south}}\\[/latex] is the coefficient of static friction, which depends on both of the materials.
The kinetic friction force [latex]{f}_{\text{k}}\\[/latex] betwixt systems moving relative to one another is given by [latex]{f}_{\text{k}}={\mu }_{\text{yard}}N\\[/latex], where [latex]{\mu }_{\text{k}}\\[/latex] is the coefficient of kinetic friction, which besides depends on both materials.

Conceptual Questions

Define normal force. What is its relationship to friction when friction behaves simply?
The gum on a piece of tape can exert forces. Can these forces be a blazon of simple friction? Explain, because especially that tape tin stick to vertical walls and even to ceilings.
When y'all learn to drive, you discover that you need to let up slightly on the brake pedal equally you come to a finish or the motorcar will stop with a wiggle. Explain this in terms of the human relationship between static and kinetic friction.
When you button a piece of chalk beyond a chalkboard, information technology sometimes screeches considering it chop-chop alternates between slipping and sticking to the board. Describe this process in more item, in particular explaining how it is related to the fact that kinetic friction is less than static friction. (The aforementioned slip-grab process occurs when tires screech on pavement.)

Problems & Exercises

Limited your answers to problems in this section to the right number of significant figures and proper units.

A physics major is cooking breakfast when he notices that the frictional forcefulness between his steel spatula and his Teflon frying pan is only 0.200 Northward. Knowing the coefficient of kinetic friction betwixt the two materials, he quickly calculates the normal force. What is information technology?
When rebuilding her motorcar's engine, a physics major must exert 300 N of force to insert a dry steel piston into a steel cylinder. (a) What is the magnitude of the normal force between the piston and cylinder? (b) What is the magnitude of the force would she take to exert if the steel parts were oiled?
(a) What is the maximum frictional force in the knee joint of a person who supports 66.0 kg of her mass on that knee? (b) During strenuous practice information technology is possible to exert forces to the joints that are easily 10 times greater than the weight being supported. What is the maximum strength of friction under such conditions? The frictional forces in joints are relatively small in all circumstances except when the joints deteriorate, such every bit from injury or arthritis. Increased frictional forces tin cause further damage and pain.
Suppose you have a 120-kg wooden crate resting on a wood floor. (a) What maximum strength can you exert horizontally on the crate without moving it? (b) If you continue to exert this forcefulness one time the crate starts to slip, what volition the magnitude of its acceleration then be?
(a) If one-half of the weight of a small 1.00 × 10³ kg utility truck is supported by its 2 bulldoze wheels, what is the magnitude of the maximum acceleration it can accomplish on dry concrete? (b) Will a metal cabinet lying on the wooden bed of the truck slip if it accelerates at this rate? (c) Solve both bug assuming the truck has four-bike bulldoze.
A team of eight dogs pulls a sled with waxed wood runners on moisture snowfall (mush!). The dogs have average masses of 19.0 kg, and the loaded sled with its rider has a mass of 210 kg. (a) Summate the magnitude of the dispatch starting from rest if each canis familiaris exerts an average force of 185 N backward on the snow. (b) What is the magnitude of the dispatch once the sled starts to move? (c) For both situations, calculate the magnitude of the strength in the coupling betwixt the dogs and the sled.
Consider the 65.0-kg water ice skater being pushed by 2 others shown in Effigy 6. (a) Find the direction and magnitude of [latex]{\mathbf{F}}_{\text{tot}}\\[/latex], the total force exerted on her by the others, given that the magnitudes [latex]{F}_{1}\\[/latex] and [latex]{F}_{2}\\[/latex] are 26.4 N and 18.6 Due north, respectively; (b) What is her initial dispatch if she is initially stationary and wearing steel-bladed skates that point in the direction of [latex]{\mathbf{F}}_{\text{tot}}\\[/latex]? (c) What is her acceleration assuming she is already moving in the direction of [latex]{\mathbf{F}}_{\text{tot}}\\[/latex]? (Remember that friction ever acts in the direction opposite that of move or attempted motion betwixt surfaces in contact.)

Figure half dozen.
Show that the acceleration of any object downwardly a frictionless incline that makes an angle θ with the horizontal is a= msin θ. (Note that this acceleration is contained of mass.)
Show that the acceleration of whatsoever object downward an incline where friction behaves but (that is, where f _grand= μ _{one thousand} Northward) is a=g(sin θ− μ ₁₀₀₀cos θ). Annotation that the acceleration is independent of mass and reduces to the expression found in the previous trouble when friction becomes negligibly small (μ _thou=0).
Calculate the deceleration of a snow boarder going up a 5.0º, slope assuming the coefficient of friction for waxed woods on moisture snow. The issue of question 9 may be useful, just be careful to consider the fact that the snowfall boarder is going uphill. Explicitly show how you follow the steps in Problem-Solving Strategies.
(a) Calculate the dispatch of a skier heading down a 10.0º gradient, assuming the coefficient of friction for waxed wood on wet snow. (b) Discover the angle of the slope down which this skier could coast at a abiding velocity. You tin fail air resistance in both parts, and you lot volition detect the result of question 9 to be useful. Explicitly testify how y'all follow the steps in the Trouble-Solving Strategies.
If an object is to rest on an incline without slipping, and then friction must equal the component of the weight of the object parallel to the incline. This requires greater and greater friction for steeper slopes. Show that the maximum angle of an incline above the horizontal for which an object will not slide down is [latex]\theta=\tan^{-1}\mu _{\text{s}}\\[/latex]. You may use the consequence of the previous problem. Assume that a = 0 and that static friction has reached its maximum value.
Calculate the maximum deceleration of a car that is heading downwardly a 6º slope (one that makes an angle of 6º with the horizontal) under the following route weather. You may assume 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 immune to slip during the deceleration. (Ignore rolling.) Summate for a auto: (a) On dry out physical; (b)On wet concrete; (c) On ice, assuming that [latex]{\mu }_{\text{south}}=0.100\\[/latex] , the same as for shoes on ice.
Calculate the maximum dispatch of a motorcar that is heading up a 4º gradient (1 that makes an angle of 4º with the horizontal) under the following route conditions. Presume that just one-half the weight of the machine is supported by the ii bulldoze wheels and that the coefficient of static friction is involved—that is, the tires are not immune to slip during the dispatch. (Ignore rolling.) (a) On dry out concrete; (b) On wet physical; (c) On water ice, bold that [latex]\mu _{\text{s}}=0.100\\[/latex], the same as for shoes on ice.
Repeat question fourteen for a machine with four-bicycle bulldoze.
A freight railroad train consists of ii [latex]8\text{.}\text{00}\times {\text{10}}^{5}\text{-kg}\\[/latex] engines and 45 cars with boilerplate masses of [latex]5\text{.}\text{50}\times {\text{10}}^{5}\text{kg}\\[/latex]. (a) What force must each engine exert astern on the track to accelerate the train at a charge per unit of [latex]5\text{.}\text{00}\times {\text{ten}}^{-two}m/{south}^{ii}\\[/latex] if the forcefulness of friction is [latex]vii\text{.}\text{50}\times {\text{10}}^{5}N\\[/latex], 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 free energy-efficient transportation systems. (b) What is the magnitude of the forcefulness in the coupling between the 37th and 38th cars (this is the force each exerts on the other), assuming all cars have the aforementioned mass and that friction is evenly distributed among all of the cars and engines?
Consider the 52.0-kg mountain climber in Figure 7. (a) Observe the tension in the rope and the force that the mountain climber must exert with her anxiety on the vertical rock face up to remain stationary. Assume that the force is exerted parallel to her legs. Also, assume negligible forcefulness exerted by her arms; (b) What is the minimum coefficient of friction between her shoes and the cliff?

Effigy 7. Part of the climber'southward weight is supported by her rope and part by friction between her feet and the rock confront.
A contestant in a winter sporting event pushes a 45.0-kg block of ice beyond a frozen lake as shown in Figure 8a. (a) Calculate the minimum forcefulness F he must exert to get the cake moving; (b) What is the magnitude of its acceleration once it starts to movement, if that force is maintained?

Figure viii. Which method of sliding a cake of ice requires less forcefulness—(a) pushing or (b) pulling at the same angle above the horizontal?
Repeat Question 18 with the contestant pulling the block of ice with a rope over his shoulder at the aforementioned angle above the horizontal as shown in Effigy 8b.

Glossary

friction: a force that opposes relative motility or attempts at motion between systems in contact

kinetic friction: a forcefulness that opposes the motility of two systems that are in contact and moving relative to 1 some other

static friction: a force that opposes the movement of two systems that are in contact and are not moving relative to 1 another

magnitude of static friction: [latex]{f}_{\text{south}}\le {\mu }_{\text{s}}North\\[/latex] , where [latex]{\mu }_{\text{s}}\\[/latex] is the coefficient of static friction and N is the magnitude of the normal force

magnitude of kinetic friction: [latex]{f}_{\text{1000}}={\mu }_{\text{k}}N\\[/latex], where [latex]{\mu }_{\text{k}}\\[/latex] is the coefficient of kinetic friction

Selected Solutions to Problems & Exercises

i. v.00 N

4. (a) 588 N; (b) 1.96 m/s²

half dozen. (a) 3.29 m/sⁱⁱ; (b) 3.52 k/s²; (c) 980 Northward, 945 Due north

10. one.83 m/southward²

14. (a) iv.20 m/southⁱⁱ; (b) ii.74 m/southward²; (c) –0.195 m/s^two

16. (a) 1.03 × 106 North; (b) 3.48 × 105 N

eighteen. (a) 51.0 North; (b) 0.720 m/s²

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