What is the Magnitude of Frictional Force with Varying Applied Forces?

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Homework Help Overview

The problem involves a 2.5 kg block on a horizontal surface subjected to both horizontal and vertical forces. The task is to determine the magnitude of the frictional force for varying vertical forces while considering the coefficients of static and kinetic friction.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • The original poster attempts to calculate the normal force and the frictional force for different vertical forces applied to the block. Some participants question the assumption of motion and clarify the relationship between static friction and applied forces.

Discussion Status

Participants have engaged in a productive discussion regarding the calculations of normal force and static friction. Guidance has been offered on how to approach the problem for different cases, particularly emphasizing the conditions under which the block remains at rest.

Contextual Notes

There is a focus on the maximum static frictional force and its relationship to the applied horizontal force. The discussion highlights the importance of understanding the forces acting on the block before determining the frictional force.

hansel13
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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 coefficients of friction for the block and surface are us = 0.40 and uk = 0.25. Determine the magnitude of the frictional force acting on the block if the magnitude of P is
(a) 8 N
(b) 10 N
(c) 12 N

Homework Equations


fk = UkN
N = mg

The Attempt at a Solution



part (a)
N = 2.5kg*9.8m/s2 = 24.5 N

So, fk = 0.25*24.5 N = 6.125

But the answer to part (a) was 6.0, not 6.125... It feels like I'm quite a bit off track. Could someone give me a little step forward on part (a)?
 
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You are assuming that the block is actually moving when the horizontal and vertical forces are applied.

Remember, the static frictional force increases as the applied force on your object increases (but in the opposite direction) until it reaches a maximum [tex]f_{smax}=\mu_s N[/tex].

In other words, until the applied force reaches a value equal to that of [tex]f_{smax}[/tex] there will be no motion AND the magnitude of the static frictional force will be equal to the magnitude of the applied force.

For (a) in your example (taking the negative direction of an imaginary y-axis as positive) when calculating the resultant "normal" force [tex]N_{res}[/tex] on your block (y-axis only) we get

[tex]N_{res} = F_{mg} + P = (2.5kg \times 9.8 m/s^2) \\- \\8N = 16.5N[/tex]

You know the coefficient of static friction is [tex]\mu_s=0.40[/tex] and therefore

[tex]f_{smax} = 16.5N \times 0.40 = 6.6N[/tex]

This is clearly larger than the horizontal force of 6.0N so we can conclude that there is no horizontal motion when a vertical force of 8N is applied (since [tex]f_{smax}[/tex] opposes the horizontal force until the horizontal force reaches this maximum value).

Therefore the magnitude of the frictional force at this point must still equal to that of the horizontal force, i.e. frictional force = 6.0N.

Now you should just apply the same principles to the remaining parts of your problem and you should be ok.

Hope that helps :smile:
 
Got it, thanks.
 
Good stuff! :biggrin:
 

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