# Force to get a crate to overcome static friction

1. Mar 15, 2013

### CollegeStudent

1. The problem statement, all variables and given/known data
A crate has a mass of 42.0 kg. The coefficient of static friction between the surface and the crate is 0.422. The coefficient of kinetic friction between the surface and the crate is 0.3.44. If a person pushes the crate by pushing down at an angle of 24°, what is the minimum force it would take to get the crate moving and then keep it moving? Suppose instead a rope is tied to the box and it is pulled at an angle of 24° above horizontal, what minimum force is now required to start and to keep the crate moving?

2. Relevant equations
ƩFy=ma_y
Ʃx=ma_x

3. The attempt at a solution
Diagram

View attachment 56753

FBD

so

Ʃy = ma_y

the forces acting in the y direction are N....mg...and mgcosθ

So wouldnt that be

N - mgcosθ - mg = 0? because there is no acceleration in the y direction?

and

ƩFx = ma_y

the forces acting in the x direction are mgsinθ and friction

so would that be

mgsinθ - static friction = 0? Because we just want to get it moving not accelerate?

i know static friction is μ_s N

just a little confused on how to start here...any hints would be greatly appreciated

2. Mar 15, 2013

### CollegeStudent

typo in the equations

ƩFx = ma_x

and

diagram didn't post

3. Mar 15, 2013

### CollegeStudent

no hints with this one?

4. Mar 15, 2013

### ap123

Why have you got mg twice here? The second term should involve the applied force.
(same problem with the x-components)

Also, I'm not sure that the angle is the right way round - should it not be the other angle in your triangle?

Think of what is happening if you increase F until the crate is just about to move.

Last edited: Mar 15, 2013
5. Mar 15, 2013

### CollegeStudent

i thought that since the force of gravity is acting straight down m*g ...that it would be in the y-component equation?

as far as the angle...i'm not too sure...we just started adding the angle component into the questions...before it was just horizontal and vertical...

then what should the angle be? did i put the (dashed lines) components in the wrong direction?

i'm having a hard time with the angle portion of it.....i tried to look for examples of it...but it didn't help TOO much

6. Mar 15, 2013

### ap123

Yes, that's the -mg part.
The -mgcosθ part should be -Fcosθ, where F is the force applied by the person.
You also need to make a similar change to the x-component equation.

For the angle, just keep it the way it is in your free-body diagram for the moment.

Then you can combine both equations to solve for the applied force F.

7. Mar 15, 2013

### PhanthomJay

For clarification, yes, mg, the weight of the crate, acts down in the y direction, and the Normal force, N, acts up in the y direction. But as ap123 pointed out, the applied force at the corner..the person's pushing force.. that should be designated as F, with components F sin theta and F cos theta, not mg sin theta and mg cos theta which you incorrectly indicated on the sketch
Again as ap123 has noted, the angle you have shown as 24 degrees is 66 degrees , since the 24 degree angle is presumably with the horizontal, not the vertical, although perhaps not so clearly stated in the problem.

Solve the 2 equations for the 2 unknowns, F and N.