Frictional Forces on a Wooden Crate with External Forces

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SUMMARY

The discussion focuses on calculating the frictional forces acting on a wooden crate with a mass of 60 kg, subjected to an external force of 220 N parallel to the surface and an additional perpendicular force of 40 N. The coefficient of static friction is 0.5. For part A, the frictional force can be determined using the equation Fs ≤ U * Fn, where Fn is the normal force. For part B, the smallest force necessary to keep the block stationary is derived from the equation Ff ≤ U * mg + f, leading to the conclusion that the frictional force must balance the applied forces.

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having major issue with this question.

Homework Statement


a wooden crate of mass m= 60 kg sits on the horizontal surface. the coefficient of static friction between the block and the surface is u=0.5 . an external force P = 220 N, is also applied to the block in a direction parallel to the surface. Also an external force F is applied to the block in a direction perpendicular downwards to the surface.
A) if the force F = 40 N, what is the magnitude and direction of the frictional force that the surface exerts on the block?
B) what is the smallest force F, necessary to hold the block stationary on the surface?
upload_2017-6-19_9-35-5.png


variables
m = 60kg
P = 220N
U = 0.5

Homework Equations


equations
Fs<= U*Fn
F = ma
Fn = mg

The Attempt at a Solution


not sure on where to start for part a in this question. but was thinking for part b you let Ff = P so that
Ff<= U*Fn+f
Ff<= U*mg + f
220<= U*mg + f
and then rearrange to solve for f
am i on the right track for part b at least?
 
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braydon said:
Fn = mg
Not always, and not in this case.
braydon said:
Ff<= U*mg + f
Is f the same as the given F?
What you have written means Ff<=(U*mg)+f. Is that what you intended?
 

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