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Torque-Related Problem

  1. Oct 30, 2007 #1
    First off, I'd like to say I admire you all very much. At first, I thought I was pretty advanced in physics, and I am at my school. But looking through these forums... damn. Now my question seems stupid :rolleyes:

    I had a problem in my AP Physics class which I just couldn't get. It was related with Torque. (Chapter 9, Page 270, #63 for those who have the 5th Edition Cutnell book). Basically, the problem has a boulder of weight 695.8N on a plank of length 2 meters. The boulder is 1.4 meters from one end. Person A and B support it at each end so that it is horizontal. Person A is nearest the boulder. If weight of the board is negligible, what force does person A apply to the board, and what is the force person B applies to it?

    I got part of it correct. I found the ratios of the forces by using Tnet=Tcwise+Tccwise. That ended up giving me 0=-F(A)*1.4+F(B)*0.6, which gives me the ratio of FpersonA=2.33FpersonB. Now I don't know how to find the actual forces. A friend of mine in the class just set one of the endpoints as a fulcrum or something, and solved for one of the forces. I didn't understand this at all; I may not even have his basic concept down.

    Some help would be appreciated. Thanks in advance.
  2. jcsd
  3. Oct 30, 2007 #2


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    Think about the requirements of equilibrium. What three things have to be true for something to be in equilibrium?

    1. Vector sum of all external forces that act on the body must be zero.
    2. Vector sum of all external torques that act on the body, measured about any possible point, must be zero.
    3. The linear momentum of the body must be zero.

    Since the boulder, plank and people aren't moving, the system is in static equilibrium.

    Draw a FBD of the problem. You should see that the net force acting on the system is:

    F1 + F2 - mg = 0

    Where F1 is person A on the left and F2 is person B on the right. mg is the weight of the boulder in newtons.

    Since you have two unknowns, you must use fact number 2 above. That is, the net torque acting on the body must be zero as well. Hence...

    0*F1 - 1.4*695.8 + 2*F2 = 0

    Solve that equation for F2 and then plug F2 back into the first equation (F1 + F2 - mg = 0) to find F1.

    Make sense?
  4. Nov 1, 2007 #3
    Ah, yes sir, I forgot to put in mg of the boulder. Thank you. I've got my chapter 1-10 exam tomorrow. Cram, cram, cram. I'll post some more questions soon.
  5. Nov 1, 2007 #4

    Back, with a very similar question. This time it's an airplane. I need to find the normal force of each of the two back wheels and the one front wheel.

    The two back wheels are 15m behind the front wheel. The center of gravity is 12.6m behind the front wheel. It is a plane of weight 1.6x10^6 N.

    So, I did as Stewart said. I've set F1 (back wheel) + F2 - mg = 0. Then I did the t1+t2-tmg=0, and set the "measured point" as F2. That makes it:

    0*F1+15*F2-12.6*1.0x10^6=0, and therefore F1=8.4x10^5N. This is wrong. It is double of what it should be, according to the answer book. One possible explanation is that I somehow inverted it, where F2 should have been the "measured point". That means that I would have to divide it by two, for I need the force of one wheel.

    Would you also be so kind as to explain how exactly you get the other information once you set your measured point? Using the last example, I'm assuming that it's just 1.4*695.8 because it is 1.4m away from the non-measured point. If it were switched, would this be 0.6*695.8?

    Thanks in advance.
  6. Nov 2, 2007 #5


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    Look back at condition number 2. It says that the vector sum of all external torques that act on the body, measured about any possible point, must be zero.

    So, in that example, I picked the left end of the beam (you could have picked the right if you wanted to instead). The distance from the applied force (F1) is 0 meters, hence, 0*F1 = T0. Now for the boulder, it is 1.4 meters from the reference point I picked. Hence, 1.4*mg = T1 about the reference point due to the boulder. However, it is applied in the downward direction, so using the standard sign convention for torques, it will be negative (i.e. clockwise = negative, and counter-clockwise = positive). The last point was the length of the beam (the end point on the right where F2 was applied). Hence, 2*F2 = T2. Now, remembering condition 2 again, the vector sum of those must be zero, so set the vector sum of those to 0 and solve for the unknown variable.

    This particular question indicated that the forces were perpendicular to the beam, so it works out nice and easy. If they were not perpendicular, the you would have to multiply by the cos(theta). Where theta is the angle the force is applied at.

    BTW, please post your new question's under a new topic if it is not related to the old one (i.e. it's a new question). You'll get more responses. Just don't double post in other forums, you'll get yelled at.
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