Find Friction Force for P= 8.0, 10.0 and 12.0 N

AI Thread Summary
The discussion revolves around calculating the friction force acting on a 2.5 kg block subjected to both horizontal and vertical forces. The applied vertical force P varies at 8.0 N, 10.0 N, and 12.0 N, while a constant horizontal force of 6.0 N is also applied. The coefficients of static and kinetic friction are given as Ms = 0.40 and Mk = 0.25. It is established that the block does not move when the static friction force exceeds the applied force, with static friction needing to match the applied force of 6.0 N to prevent slipping. The participants emphasize understanding the difference between static and kinetic friction in determining the block's movement.
Shatzkinator
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Homework Statement


A 2.5 kg block is initially at rest on a horizontal surface. A horizontal force F of magnitude 6.0 N and a vertical force P are then applied to the block (the vertical force is upwards). The coefficients of friction for the block and surface are Ms = 0.40 and Mk = 0.25. Determine th emagnitude of the frictional force acting on the block if the magnitude of P is a) 8.0 N, b) 10 N, and c) 12 N


Homework Equations


fs = Mk * Fn
Fnet = ma
Fg - Fn - P = 0

The Attempt at a Solution


I tried using the third equation to get Fn and plug this value into the first equation to get force of friction. I didn't get the right answer... for P = 8.0 N the answer should be 6.0 N. How?
 
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Show exactly what you did. Hint: First you must determine whether the relevant friction is static or kinetic.
 
Doc Al said:
Show exactly what you did. Hint: First you must determine whether the relevant friction is static or kinetic.

Fg - Fn - P = 0
24.5 = Fn + 8
Fn = 16.5

fs = 16.5 * 0.25 = 4.1 N
 
Shatzkinator said:
Fg - Fn - P = 0
24.5 = Fn + 8
Fn = 16.5
Good.

fs = 16.5 * 0.25 = 4.1 N
This assumes that the block moves. Does it?
 
Doc Al said:
Good.


This assumes that the block moves. Does it?

well Yea if the force applied is 6.0 N and the friction is 4.1 then yes it should move.
 
Shatzkinator said:
well Yea if the force applied is 6.0 N and the friction is 4.1 then yes it should move.

Oh wait no it doesn't because Fn * Ms = 8.25 which is greater than 6 N so it doesn't move. Now what?
 
Shatzkinator said:
Oh wait no it doesn't because Fn * Ms = 8.25 which is greater than 6 N so it doesn't move. Now what?
How does static friction work? Note that for static friction, μN is the maximum possible static friction that the surfaces can generate before being forced to move. But static friction will be only what it needs to be to prevent slipping (up to the maximum). How much does static friction need to be in this case to prevent slipping?
 
Doc Al said:
How does static friction work? Note that for static friction, μN is the maximum possible static friction that the surfaces can generate before being forced to move. But static friction will be only what it needs to be to prevent slipping (up to the maximum). How much does static friction need to be in this case to prevent slipping?

6 to prevent slipping?
 
Shatzkinator said:
6 to prevent slipping?
Right!
 
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