Static Friction when pushing a crate on the floor

AI Thread Summary
To determine the minimum force F required to move a crate on a horizontal floor, one must consider the coefficient of static friction μ and the angle θ at which the force is applied. The maximum static friction force is calculated using the equation f = μFn, where Fn is the normal force. In a static scenario, the forces in both horizontal and vertical directions must balance, leading to the equations F cos θ - fs = 0 and -mg + F sin θ + n = 0. When the crate cannot be moved, accelerating the elevator downward can reduce the effective weight of the crate, potentially allowing it to be moved. Understanding these principles is crucial for solving static friction problems effectively.
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Homework Statement



You try to push a crate of weight M with a force F on a horizontal floor. The coefficient of static friction is μ and you exert the force F under an angle θ below the horizontal.

a) Determine the minimum value of F that will move the crate
b) Assume you have not sufficient strength to move the crate. You now do this exercise in an elevator. Which direction should you the elevator accelerate to have a chance to move the box?


Homework Equations



The maximum possible friction force between two surfaces before sliding begins is the product of the coefficient of static friction and the normal force: (f= μ Fn)

The Attempt at a Solution



I have no idea after that...
 
Last edited:
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Welcome to Physics Forums.

Have you drawn a free-body diagram? I always find that they are extremely helpful when solving this type of problem.
 


Yes, i did. Just can't figure out the value F as there is no known data given.
 


White_Light said:
Yes, i did. Just can't figure out the value F as there is no known data given.
Start by summing the forces acting in the vertical and horizontal direction. Since this is a static problem, what do you know about their sums?
 


Horizontal: F cos θ - fs = ma
Vertical : -mg + F sin θ + n = 0

Am i right?
 


White_Light said:
Horizontal: F cos θ - fs = ma
Vertical : -mg + F sin θ + n = 0

Am i right?
Indeed you are. As I said in my previous post, we are dealing with the static case here, so a=0.

Now, can you expression fs in terms of n?
 


Are you referring to the a in the net horizontal force has to be 0 always in static case? Or both for vertical as well?

Horizontal: F cos θ - μn = 0
Vertical : -mg + F sin θ + n = 0

Solving the y equation gives n= F cos θ / μ
 


To get the F value, sub n= F cos θ / μ into x equation. Is this the solution?
 


White_Light said:
To get the F value, sub n= F cos θ / μ into x equation. Is this the solution?
Sounds good to me :approve:
 
  • #10


Thank you so much! ^.^
 
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