A 10.0 kg block is on an incline plane of variable angle. The coefficients of st

Question

A 10.0 kg block is on an incline plane of variable angle. The coefficients of static and kinetic friction are mu s=0.31 and mu k=0.22 (a) What is the minimum angle such that the block slides down the plane? (b) What is the frictional force at this angle? N (c) If the angle of the incline plane is half of that in part (a) what are the force of friction and acceleration of the block down the plane if it released from rest? (d) If the angle of the incline plane is twice that in part (a) what are the force of friction and acceleration of the block down the plane if it released from rest? (e) Suppose the block is given an initial velocity up the incline plane when the angle of the plane is half of that in part a. What is the force of friction and acceleration of the block as it moves up the plane? What is the force of friction and acceleration of the block after it reaches its maximum height on the plane? (f) Suppose the block is given an initial velocity up the incline plane when the angle of the plane is twice that in part a. What is the force of friction and acceleration of the block as it moves up the plane? What is the force of friction and acceleration of the block after it reaches its maximum height on the plane? L N

Explanation / Answer

1)We can easily find, through geometry of angles, that the component of gravity acting in the direction parallel to the inclined plane is simply gsin(thetha) and the component of gravity acting in the direction perpendicular to plane is gcos(thetha).
From this, we deduce that the force of gravity on the object that can contribute to motion is mgsin(30), and the normal force is mgcos(30).
Since Ff = u(N) = u(mgcos(thetha)), and this maximum static frictional force must be able to sustain the object's parallel force of mgsin(thetha), we equate them:
us(mgcos(thetha)) = ukmgsin(thetha), and ucos(thetha) = sin(thetha), which leads to u = tan(thetha).

tan(thetha)=0.31/0.22

thetha=51.5 degree

b) The force pulling the crate down slope is
sin(51.5)*9.81*10
92.5 N
The frictional force must be equal to this since the crate is static.

Let's check to see if 92.5N is the right perpendicular force to achieve it

the normal force is
10*9.81*cos(51.5)+92.5
124.8N

The frictional force is
124.8*0.31
38.6 N

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