How do I find the necessary force to start a crate moving on a rough floor?

[JennV]

Homework Statement

To move a large crate across a rough floor, you push on it with a force F at an angle of theta=21.0 degrees below the horizontal. Find the force necessary to start the crate moving, given that the mass of the crate is 32.0 kg and the coefficient of static friction between the crate and the floor is 0.600.

Homework Equations

Fcos(theta)=Coefficient of static friction*mg

The Attempt at a Solution

F= [(0.600)(32)(9.81)] / cos(21)
F= 202N

I entered this answer, but the system says that it is wrong and it told me that "the normal force is not equal to the weight here." So I am not sure what I'm doing wrong, but I'm close to the right answer?
Thank you very much.

 

[vela]

Start by drawing a free-body diagram for the crate and then tell us the equation of motion you get for the crate for the vertical direction.

 

[JennV]

Start by drawing a free-body diagram for the crate and then tell us the equation of motion you get for the crate for the vertical direction.

Vertical direction: N=mg-Fsin(theta) ?

 

[vela]

Not quite. You know N points up, mg points down, and the vertical component of F, F sin θ, points down. So sum these forces and set it equal to 0. What equation do you get?

 

[JennV]

Not quite. You know N points up, mg points down, and the vertical component of F, F sin θ, points down. So sum these forces and set it equal to 0. What equation do you get?

0=N-mg-Fsin(theta) ?

 

[vela]

Right! So if you solve for N, you get N=mg+F sin θ. Hopefully that makes sense to you. Since you're pushing down on the crate, the ground pushes back up with more force.

 

[JennV]

Right! So if you solve for N, you get N=mg+F sin θ. Hopefully that makes sense to you. Since you're pushing down on the crate, the ground pushes back up with more force.

So N is the necessary force to move the crate?

 

[vela]

No, N is the normal force. It's the force the ground exerts on the crate. To move the crate, you have to overcome the force of static friction. The maximum amount of static friction, in turn, depends on the normal force, so you had to find the normal force first.

 

[JennV]

No, N is the normal force. It's the force the ground exerts on the crate. To move the crate, you have to overcome the force of static friction. The maximum amount of static friction, in turn, depends on the normal force, so you had to find the normal force first.

But the Force for Fsinθ is unknown...

 

[vela]

Yes, so you're going to have to do a bit more math to get the answer. So far you have the equation for the vertical direction. What equation do you get for the horizontal direction?

 

[JennV]

Yes, so you're going to have to do a bit more math to get the answer. So far you have the equation for the vertical direction. What equation do you get for the horizontal direction?

In the horizontal direction the forces that act are; applied force, static friction.
So 0=Fcosθ-(uk*Fn) ?

 

[vela]

Use either N or Fn to denote the normal force, not both. Also, it's static friction, right? So you want μ_s, not μ_k.

You have these equations:

0 = N - mg - F sin θ
0 = F cos θ - μ_sN

Solve for N in the first equation and plug the result into the second equation. Then you'll be able to solve for F.

 

[JennV]

Use either N or Fn to denote the normal force, not both. Also, it's static friction, right? So you want μ_s, not μ_k.

You have these equations:

0 = N - mg - F sin θ
0 = F cos θ - μ_sN

Solve for N in the first equation and plug the result into the second equation. Then you'll be able to solve for F.

Thank you so much for helping me. =) I finally got the answer correct.