# Finding the value of support and friction force?

Here is the problem I was given:
The worker pushes downward and to the left on the box at an angle of 35°. The box has a mass of 135 kg. The box moves horizontally across the floor to the left at a constant velocity.

If the force applied by the man is 2030 N∠215°, what is the value of the support force?

The equation given for this problem was (m · g) + Fsupport.

Here is my work for the problem:
(135 kg)(-9.81 N/kg) + Fsupport
(135 kg)(0 m/s^2) - 1324.35 N + Fsupport
Fsupport = 1324.35 N

Would this be correct? I am uncertain because I was given information that seemed like it should be plugged into an equation to solve the problem, the applied force of 2030 N∠215°, but my options for answers include the answer I had gotten for Fsupport. My options on my study guide are the following:

a. 1164 N
b. 1324 N
c. 1663 N
d. 2489 N

Did I make a mistake in trying to find this support force, or is it correct?

With the same information, I am to find the friction force. I am given a few equations, but I feel as though I do not have enough information to plug into the any of the equations to solve for the friction force. Here are the equations to solve for the value of the friction force:

f = μ · Fsupport

I'm not sure what information I would plug in, because to the best of my knowledge I only have the information for the Fsupport. Plus I need both the Fsupport and force friction to solve for the acceleration. If it isn't obvious.. I'm learning on my own and am at a bit of a loss.

Any help is greatly appreciated! Clarification will work wonders. :) Thank you.

## Answers and Replies

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What you need to do is sum your forces in the y direction. Since you have no acceleration in the y direction the sum of forces must equal zero. So you have:

Fsupport-Weight-Fsin35= m*a= 0

Where Fsin35 is the y component of the applied force. We know its negative because of the given direction. So now you just would need to solve for the only unknown Fsupport.
For the second part to calculate the friction force you would use the equation:

Ffriction=(friction coeff.)(Fsupport)

Also alittle advice for learning physics is to never just memorize formulas. About the only formula you really need to remember is F=ma along with a small handfull of others. Its more important to be able to properly draw free-body diagrams which show all the forces acting on a given body and then apply F=ma.

I think I have it (the value of support force)!

sin 35 x 2030.
.5736 x 2030 = 1164.4N.
135 x 9.81 = 1324.35N.
= 2488.75 N

Is this correct now?

That is correct. It makes sense that the support force will have to be greater than just the weight since there is also a force being applied down. Another way to think of it is that the support force feels the weight of (Fsin35+W).