Static Friction when pushing a crate on the floor

In summary, the conversation discusses determining the minimum force needed to move a crate on a horizontal floor, taking into account the coefficient of static friction and the angle at which the force is applied. The solution involves drawing a free-body diagram and using the equations for horizontal and vertical forces to find the necessary force value.
  • #1
White_Light
8
0

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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  • #2


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.
 
  • #3


Yes, i did. Just can't figure out the value F as there is no known data given.
 
  • #4


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?
 
  • #5


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

Am i right?
 
  • #6


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?
 
  • #7


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
 
  • #8


To get the F value, sub n= F cos θ / μ into x equation. Is this the solution?
 
  • #9


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! ^.^
 

FAQ: Static Friction when pushing a crate on the floor

What is static friction?

Static friction is the resistance force that occurs when two objects are in contact with each other and one is trying to move or slide past the other, but is not yet in motion.

What factors affect the amount of static friction?

The amount of static friction depends on the types of surfaces in contact, the force pushing the objects together, and any external forces acting on the objects.

How is static friction different from kinetic friction?

Static friction occurs when two objects are at rest, while kinetic friction occurs when the two objects are in motion. Additionally, the amount of force required to overcome static friction is greater than the force required to maintain motion against kinetic friction.

Why is static friction important when pushing a crate on the floor?

Static friction is important when pushing a crate on the floor because it determines the amount of force needed to overcome the resistance and start the crate in motion. If the force applied is not enough to overcome static friction, the crate will not move.

How can static friction be reduced?

Static friction can be reduced by using lubricants between the surfaces, increasing the force applied, or changing the angle of the force applied. Additionally, using smoother or less rough surfaces can also decrease static friction.

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