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0.7 3.8 friction  (Page 5/10)

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This figure has two parts, each of which shows two rough surfaces in close proximity to each other. In the first part, the normal force is small, so that the 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 area of contact between the two surfaces has increased. As a result, the friction between the two surfaces in the second part is also greater than the friction in the first part.
Two rough surfaces in contact have a much smaller area of actual contact than their total area. When there is a greater normal force as a result of a greater applied force, the area of actual contact increases as does friction.

But 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 get warmer when rubbed? Essentially, atoms are linked with one another to form lattices. When surfaces rub, the surface atoms adhere and cause atomic lattices to vibrate—essentially creating sound 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 wear can also occur between atoms and molecules on the surfaces. [link] shows how the tip of a probe drawn across another material is deformed by atomic-scale friction. The force needed to drag the tip can be measured and is found to be related to shear stress, which will be discussed later in this chapter. The variation in shear stress is remarkable (more than a factor of 10 12 size 12{"10" rSup { size 8{"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 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.
The tip of a probe is deformed sideways by frictional force as the probe is dragged across a surface. Measurements of how the force varies for different materials are yielding fundamental insights into the atomic nature of friction.

Section summary

  • Friction is a contact force between systems that opposes the motion or attempted motion between them. Simple friction is proportional to the normal force N size 12{N} {} pushing the systems together. (A normal force is always perpendicular to the contact surface between systems.) Friction depends on both of the materials involved. The magnitude of static friction f s size 12{f rSub { size 8{s} } } {} between systems stationary relative to one another is given by
    f s ≤ μ s N , size 12{f rSub { size 8{s} }<= μ rSub { size 8{s} } N} {}
    where μ s size 12{μ rSub { size 8{s} } } {} is the coefficient of static friction, which depends on both of the materials.
  • The kinetic friction force f k size 12{f rSub { size 8{k} } } {} between systems moving relative to one another is given by
    f k = μ k N , size 12{f rSub { size 8{k} } =μ rSub { size 8{k} } N} {}
    where μ k size 12{μ rSub { size 8{K} } } {} is the coefficient of kinetic friction, which also depends on both materials.

Conceptual questions

The glue on a piece of tape can exert forces. Can these forces be a type of simple friction? Explain, considering especially that tape can stick to vertical walls and even to ceilings.

Problems&Exercises

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

5.00 N size 12{5 "." "00"`N} {}

Suppose you have a 120-kg wooden crate resting on a wood floor. (a) What maximum force can you exert horizontally on the crate without moving it? (b) If you continue to exert this force once the crate starts to slip, what will the magnitude of its acceleration then be?

(a) 588 N

(b) 1 . 96 m /s 2 size 12{`1 "." "96"`"m/s" rSup { size 8{2} } } {}

Show that the acceleration of any object down an incline where friction behaves simply (that is, where f k = μ k N size 12{f rSub { size 8{k} } =μ rSub { size 8{k} } N} {} ) is a = g ( sin θ − μ k cos θ ). size 12{a=g \( "sin"θ - μ rSub { size 8{k} } "cos"θ \) } {} Note that the acceleration is independent of mass and reduces to the expression found in the previous problem when friction becomes negligibly small ( μ k = 0 ). size 12{ \( μ rSub { size 8{k} } =0 \) "." } {}

Calculate the deceleration of a snow boarder going up a 5.0º size 12{5 "." 0°} {} , slope assuming the coefficient of friction for waxed wood on wet snow. The result of [link] may be useful, but be careful to consider the fact that the snow boarder is going uphill. Explicitly show how you follow the steps in Problem-Solving Strategies .

1 . 83 m/s 2 size 12{ - 1 "." "83"" m/s" rSup { size 8{2} } } {}

(a) Calculate the acceleration of a skier heading down a 10 . 0º size 12{"10" "." 0°} {} slope, assuming the coefficient of friction for waxed wood on wet snow. (b) Find the angle of the slope down which this skier could coast at a constant velocity. You can neglect air resistance in both parts, and you will find the result of [link] to be useful. Explicitly show how you follow the steps in the Problem-Solving Strategies .

Calculate the maximum acceleration of a car that is heading up a 4º slope (one that makes an angle of 4º with the horizontal) under the following road conditions. Assume that only half the weight of the car is supported by the two drive 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 concrete. (b) On wet concrete. (c) On ice, assuming that μ s = 0.100 , the same as for shoes on ice.

(a) 4 . 20 m/s 2 size 12{4 "." "20 m/s" rSup { size 8{2} } } {}

(b) 2 . 74 m/s 2 size 12{2 "." "74 m/s" rSup { size 8{2} } } {}

(c) –0 . 195 m/s 2 size 12{"-0" "." "195 m/s" rSup { size 8{2} } } {}

Repeat [link] for a car with four-wheel drive.

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

(a) 51 . 0 N size 12{"51" "." 0`N} {}

(b) 0 . 720 m /s 2 size 12{0 "." "720"`"m/s" rSup { size 8{2} } } {}

Repeat [link] with the contestant pulling the block of ice with a rope over his shoulder at the same angle above the horizontal as shown in [link] (b).

(a) A block of ice is being pushed by a contestant in a winter sporting event across a frozen lake at an angle of twenty five degrees. (b) A block of ice is being pulled by a contestant in a winter sporting event across a frozen lake at an angle of twenty five degrees.
Which method of sliding a block of ice requires less force—(a) pushing or (b) pulling at the same angle above the horizontal?
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Source:  OpenStax, Newton's laws. OpenStax CNX. Oct 25, 2015 Download for free at https://legacy.cnx.org/content/col11898/1.1
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