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In Fig. 6-51, a crate slides down an inclined right-angled trough. The coefficient of kinetic friction between the crate and the trough isk. What is the acceleration of the crate in terms of k,, and g?

Short Answer

Expert verified

a=gsin()-2kcos()

Step by step solution

01

Given data

  • The tilted angle is.
  • The kinetic friction coefficient between the crate and the inclined surface is k.
02

To understand the concept

The problem deals with Newton's second law of motion, which states that the acceleration of an object can be expressed in terms of the net force acting upon the object and the mass of the object.First, calculate the resultant normal force, then apply Newton's second law to solve the question.

Formula:

The frictional force is given by,

f=2kFN

03

Calculate the acceleration of the crate in terms of μk, θ, and g

The free body diagram for the front view is shown below:


As both normal forces make45with the center line, the resultant normal force is:

FNr=2FNcos45=2FN

The free body diagram for the side view is shown below:

Newton's second law of motion for the create (+x direction is down the plane, +y direction is in the direction of FNr):

x:mgsin()-f=ma

y:FNr-mgcos()=0

The frictional force on the crate is given by:

f=2kFN=2kFNr2=2kFNr

Substituting frictional force in x component and combining it with y component:

mgsin-2kmgcos=maa=gsin-2kcos

Thus, the acceleration of the crate is a=gsin-2kcos.

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Most popular questions from this chapter

Figure 6-20 shows an initially stationary block of masson a floor. A force of magnitudeis then applied at upward angle=20.What is the magnitude of the acceleration of the block across the floor if the friction coefficients are (a)s=0.600andk=0.500and (b)s=0.400andk=0.300?

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