I am having issues with this problem. I think it has to deal with the FBD, which includes the mg acting at a 22.62 degree angle with the normal force and the 110 lb force in the wire. I have a feeling that is where I messed up because there is a double wire. Any suggestions?
5 comments:
has anybody noticed that the problems are assigned in a section that doesn't even talk about Potential Energy until the next section...but in the previous problem we had to use it to get the correct answer? I'm just saying that were technically using something that "hasn't been taught" yet. Any comments?
Prof. Krousgrill went over Potential Energy as well as Kenetic Energy in class. Look in your notes on page 16-17. For this problem, repeat the same process as in problem 3/107. For the conservative and non-conservative forces you can find the potential due to each force by dotting them with the et vector and taking partial integrals over the path. Hope this helps.
would you mind explaining the integral part of the problem? I guess I just don't get why U(1 to 2) is not just 110lbs * 10ft?
I definitely didn't do any partial integrals to solve this one, or 107 for that matter, but if you say that work equals delta KE, you have to include the 220 total pounds from the cables, and the 300lb force in the downward direction. Personally I just summed up force times distance in x: 220cos(theta)*10cos(theta) and then in y: (220sin(theta)-300)*10sin(theta) and said this was equal to the change in kinetic energy. This worked out nicely, and it was quick.
There are a number of perspectives (and methods) to use on this problem.
Consider System I shown in figure that is made up of block B. The nonconservative forces acting on B are the two tension forces, each of magnitude F = 110 lb. The work done on this system by nonconservative forces is 2*F*sB, where sB is the distance traveled by B.
Or, consider System II also shown in figure above made up of block B and the cable. Here the work done by nonconservative forces is that due to F acting at A: F*sA, where sA is the distance traveled by A. Note that from kinematics, sA = 2*sB. Therefore, this gives the same work as on System I.
Or, consider the other suggestions given in the above comments.
Choose the one that makes the most sense to you.
Let us know if you have questions. These are all very good points.
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