Statics: Crate Problem - Solving for N = 132N

[sandmanvgc]

Homework Statement

Answer key say answer is d. I took combined weight and vertical component of 200 force to calculate frictional force would be 132 N. Am I missing something?

Relevant Equations

$F_f = uF_n$

(4/5)200 + 500 = 660
660(0.2) = 132 N

 

[PhanthomJay]

But what’s the applied force in the horizontal direction

 

[sandmanvgc]

But what’s the applied force in the horizontal direction

120. Is that what they were asking for? Is frictional force not what I calculated above?

EDIT: 120 is the answer but I thought that would've been the applied force and not the Frictional force.

Or is the frictional force just equal to whatever the applied perpendicular force is until max possible value is reached?

 

[PhanthomJay]

Yes, for static friction, the friction force is less than or equal to uN, from equilibrium equation Newton’s first law. If the box is moving , then kinetic friction applies and the friction force is equal to uN,
Suppose the box was just resting there and there was no applied force at all. Using u =0.2, what would be the friction force for that situation?

 

[sandmanvgc]

Yes, for static friction, the friction force is less than or equal to uN, from equilibrium equation Newton’s first law. If the box is moving , then kinetic friction applies and the friction force is equal to uN,
Suppose the box was just resting there and there was no applied force at all. Using u =0.2, what would be the friction force for that situation?

So it would be 0

 

[PhanthomJay]

Ok you got it!