A. Physics Frictional force

[narcoyam]

A block is at rest on the incline shown in the figure. The coefficients of static and kinetic friction are µs= 0.67 and µk = 0.57, respectively.
The acceleration of gravity is 9.8m/s^2


1. -> What is the frictional force action on the 42kg mass?
Answer in units of N.

N = mg cos x
(42) (9.8) cos (32) = 349.0565964
then 0.67 = Fs/ 349.0565964
so the answer is 233.8679196N

2. -> What is the largest angle which the incline can have so that the mass does not slide down the incline?
Answer in unit of degree.

0.67 = tan x
x= 33.82208522 degree


3. -> What is the acceleration of the block down the incline if the angle of the incline is 37 degree?
Answer in units of m/s^2

a=-ukg
so (-0.57)(9.8) = 5.586m/s^2
But this is wrong




I'm not sure if my answers are right but 3rd one is definitely wrong
can you check if this is right? Thank u

 

[kuruman]

Part (1) is incorrect. You say that the force f static friction has reached its maximum value when the angle is 32 degrees. That is not correct. As you calculate in part (2), the force of static friction reaches its maximum value when the angle is 33.82 degrees.

In part (1) the force of static friction is whatever is necessary to keep the mass at rest. Draw a free body diagram to find what that is.

 

[kerimek]

For the third question, the acceleration of the block down the incline can be calculated using Newton's Second Law: F=ma. In this case, the only force acting on the block is the component of its weight parallel to the incline, which is given by mg sin x, where x is the angle of the incline. So we have:

F=mg sin x = ma

Rearranging for a, we get:

a = g sin x

Substituting the given values, we get:

a = (9.8) sin (37) = 5.886 m/s^2

So the acceleration of the block down the incline with an angle of 37 degrees is 5.886 m/s^2.