# Work and free body diagrams

I am a bit confused with how to find work when there is a free body diagram. I am trying to work out this problem, and in it a box is being pulled at a constant speed by a rope at a constant angle above the horizontal. I am given mass, coefficient of kinetic friction, and the angle.I have drawn my free body diagram, and I think it's pretty accurate. I get the following equations, where $P$ is equal to force being pulled:

\begin{align} \Sigma F_x&=T\cos\theta-f=ma=0 \\ \Sigma F_y &= T \sin \theta + N - mg = ma = 0 \end{align}

So my unknowns are $P$, $T$, and $N$. I know $f=(\text{coefficient of kinetic friction})N$. I also know the distance the box is pulled.

The work I am trying to solve for is the work being done by the man. I solved for $T$ by isolating $N$ in the second equation and plugging it into the first. Then I'm thinking that $T\cos\theta$ would be the force that I would have to multiply by the distance in order to get work, but it isn't giving me the right answer. Is it because the rope is being pulled up at an angle? Do I need to account for that somehow to find the work done by the man?

• If the box is sliding.... it can't be static friction. – G. Paily Nov 5 '14 at 22:28
• oh sorry, right. It's kinetic friction my bad – K P Nov 5 '14 at 22:29
• Hmm, the $\cos\theta$ in $T\cos\theta$ should take care of that angle... can you include a picture if you have one? – G. Paily Nov 5 '14 at 22:31
• Problem didn't come with a picture, but I could include the free body diagram that I drew if that'd be helpful – K P Nov 5 '14 at 22:33
• Is there a different equation for kinetic friction compared to static because I was using the one for static friction, which could be the problem – K P Nov 5 '14 at 22:34

There is no acceleration in Y direction. If you consider the X and Y axis like this. Maybe this will help. Work done is P*d. Remember the net force will always be zero because there is no acceleration.

It depends on what work you are looking for. That is, it depends on the exact wording of the question. One could ask for the work done by individual forces, each of which would be non-zero in this case. But if you simply ask for the work done on the box, or more precisely, the net work done on the box, then the answer in this case is zero, as you have concluded.

Pay very close attention to the exact wording of the question. (Good general advice. Many times students give correct answers to questions that have not been asked!)

• Okay, that makes sense. Now that I look at it closer, it asks for the work down by the man pulling the box. Which makes much more sense than overall work, thank you! – K P Nov 5 '14 at 22:03

If your net force on the object is zero, then the total work on your object will be zero. Another way to see this is that by the work-energy theorem: $$W_{tot} = \Delta K = K_{f}-K_{i}$$ So if it moves with a constant speed, the total work done on it is zero. However, all that means is that your tension does positive work, and friction does negative work, and so the total is zero. On a different note, does the box see the force of the pull directly? Or does the pull act on the rope, and then the rope act on the box? If the latter, then only $T, f, N$ and $mg$ should appear in the Free Body Diagram of the box. The force $P$ would appear in the FBD (Free Body Diagram) of the rope.

• Okay, thank you! I was confused as to if they were the same force or different forces, thank you for clarifying because the box is pulled through the rope. – K P Nov 5 '14 at 22:02

Work is calculated by force multiply by distance. So the work that you want to calculate is:$$W=mgd\left(f\cos\theta+\sin\theta\right)$$ That d is the distance the box is pulled.

## protected by Qmechanic♦Jul 17 '16 at 13:29

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).