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Chapter 6: Q85P (page 147)

In the early afternoon, a car is parked on a street that runs down a steep hill, at an angle of35.0掳relative to the horizontal. Just then the coefficient of static friction between the tires and the street surface is 0.725. Later, after nightfall, a sleet storm hits the area, and the coefficient decreases due to both the ice and a chemical change in the road surface because of the temperature decrease. By what percentage must the coefficient decrease if the car is to be in danger of sliding down the street?

Short Answer

Expert verified

At 3.4% the coefficient decreases from the given 0.725 value of coefficient of static friction.

Step by step solution

01

Given

Original coefficient of static friction:渭s=0.725

02

Understanding the concept

The problem deals with the Newton鈥檚 second law of motion which states that the acceleration of an object is dependent upon the net force acting upon the object and the mass of the object. Draw the free body diagram and then use Newton鈥檚 second law.

03

Draw the free body diagram

Free body diagram:

04

Calculate the percentage decrease in coefficient if the car is to be in danger of sliding down the street

The car is in 鈥渄anger of sliding鈥� down when,

mgsin35掳=fs=fs.maxmgsin35掳=渭s脳mgcos35掳sin35掳=渭s脳cos35掳

鈬�tan35掳=渭s鈬�渭s=0.700

Comparison with original value:鈬�1-0.7000.725=0.034鈬�3.4%less

This value represents a 3.45% decrease from the given 0.725 value of coefficient of static friction.

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Most popular questions from this chapter

In Fig. 6-13, horizontal forceF鈫�1of magnitude 10N is applied to a box on a floor, but the box does not slide. Then, as the magnitude of vertical forceF鈫�2is increased from zero, do the following quantities increase, decrease, or stay the same: (a) the magnitude of the frictional forcef鈫�son the box; (b) the magnitude of the normal force on the box from the floor; (c) the maximum valuef鈫�s.maxof the magnitude of the static frictional force on the box? (d) Does the box eventually slide?

A baseball player with massm=79kg, sliding into second base, is retarded by a frictional force of magnitude470N.What is the coefficient of kinetic friction渭kbetween the player and the ground?

A slide-loving pig slides down a certain35鈭�slide in twice the time it would take to slide down a frictionless35鈭�slide.What is the coefficient of kinetic friction between the pig and the slide?

In Fig. 6-37, a slab of mass m1=40kgrests on a frictionless floor, and a block of mas m2=10kgrests on top of the slab. Between block and slab, the coefficient of static friction is 0.60, and the coefficient of kinetic friction is 0.40. A horizontal force of magnitude 100Nbegins to pull directly on the block, as shown. In unit-vector notation, what are the resulting accelerations of (a) the block and (b) the slab?

In Fig. 6-34, blocks A and B have weights of 44Nand 22N, respectively. (a) Determine the minimum weight of block C to keep A from sliding if 渭kbetween A and the table is 0.20. (b) Block C suddenly is lifted off A. What is the acceleration of block A if 渭kbetween A and the table is 0.15?
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