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

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Homework Help Overview

The problem involves a 2.5 kg block on a horizontal surface subjected to a horizontal force and a vertical force. The objective is to determine the frictional force acting on the block for different magnitudes of the vertical force P (8.0 N, 10.0 N, and 12.0 N), given the coefficients of static and kinetic friction.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the calculation of the normal force and the frictional force using the equations provided. There is an exploration of whether the friction is static or kinetic, and how the forces interact to determine if the block moves.

Discussion Status

Participants are actively questioning the assumptions about the movement of the block and the nature of friction involved. Some have provided calculations, while others are seeking clarification on the conditions under which static friction applies.

Contextual Notes

There is an ongoing discussion about the maximum static friction and its relationship to the applied forces, with participants noting the importance of understanding the conditions for movement versus static equilibrium.

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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