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Homework Help: Help with this one force/torque problem

  1. Nov 14, 2007 #1
    here is the Q i just cannot solve, i have tried sooo much:
    I tried my best to make the figure from paint, so plz help.
    http://img207.imageshack.us/img207/6315/phyhv3.jpg [Broken]

    Find the forces T,V, and H for the system in equilibrium shown above. Note the beam is homogeneous, uniform, 4 meters long and has a mass of 2kg.

    i think it has something to do with torque: because >> Torque=Fd, where F is the force and d is the distance, so i think im supposed to find bunch of torques to find them forces or something, and i think torque in equilibrium system is 0 ... also the beam is the darkest line i think, like a rod and that round thing is the axis, its not very very complicated or complex since its only the 1st year college physics.

    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Nov 14, 2007 #2
    Right, so we can use the equation for the net torque, and Newton's Second Law to find the two hinge forces as well as the tension. Can you show me how you would set up the three equations?
  4. Nov 14, 2007 #3
    well i tried many things but right now i just did:
    Torgue 1 + torque 2 = torque 3
    4(4 x 9.8) [from the mass of 4 kg]+ 2 (2 x 9.8) [the beam] = 2 T [tension], so i got T = 98 N
    i am really confused because net torque = I angular acceleration and isnt net torque suposed to b zero since the system is equlibrium. i am not sure how i can do this :(
  5. Nov 14, 2007 #4
    Ok, well lets be careful about what we're saying. The equilibrium will cause the net forces in the x and y directions to be zero. Secondly, what coordinate system are you choosing for the torques, in other words, would you like to define the clockwise or the counter clockwise direction as positive?
  6. Nov 14, 2007 #5
    clock wise as positve, meaning mg of the 4kg block would b positive
  7. Nov 14, 2007 #6
    Ok, so you're going to have 3 equilibrium equations, and 3 forces to solve for. Let's start with the torque equation, defining the point where the tension acts on the rod to be the axis of rotation.

    [tex]\Sigma \tau = \tau_{Mass} - \tau_{H} + \tau_{V} = 0 [/tex]

    where we define the torque as [tex] | \tau | = FRsin \theta [/tex]

    Can you show me how you would set the the two force equations in terms of the variables?
  8. Nov 14, 2007 #7
    torque = T distance sin 30

    is the distance 2? and is it sin 30?

    i think i get the picture, its like newtons 2nd law, so you find the torque of the mass and then subtract the torgue of H and v
  9. Nov 14, 2007 #8
    sprry but the reason i am a little fristaretd and in a hurry is because i have this class in exactly one hour and this is my last Q of the homework, and the instructor is very strict about homeworks
  10. Nov 14, 2007 #9
    the torque of the mass is mg(d)
    t = 4(9.8)(d) << so what's d .. am i on right track, somewhat?
  11. Nov 14, 2007 #10
    One hour is plenty of time, but there's no point in rushing and getting it wrong =)

    If you don't write out the equations for the net forces, and plug in the variables to the torque equation, I'm afriad you may get the wrong answer. I'm trying to work it out with you so that you understand it, but if you're that pressed for time, give me your final answer and I'll tell you if it checks with mine.

    Edit: I beg of you to deal with variables first! lol. It can become very confusing if we begin pluggin in values right away. If you have the time, I would suggest typing out the equations. I don't know if you're on the right track because you're not showing your work...
    Last edited: Nov 14, 2007
  12. Nov 14, 2007 #11
    Torque H = RFcos theta
    Torque V = RFsin theta
    Torque mass = RF

    ^^ do these equations make sense?
  13. Nov 14, 2007 #12
    i wish i could get to the answers but i have no idea how to do this problem, thats y i posted the Q here, so i guess ill have to go step by step with u
  14. Nov 14, 2007 #13
    Those equations make sense to a degree, but I'll show you how I set it up, and if you have any questions, just ask.

    Just to be clear, I take the axis of rotation to be the point where the tension acts, I take clock wise to be positive, and I assume that the tension acts at the point of the center of mass, becasue I didn't see specification of the distance that the tension occurs. We have:

    [tex]\Sigma \tau = (Mass of block)(2)(sin60) + V(2)sin60 - H(2)sin30 = 0[/tex]

    [tex] F_x = H - Tcos60 = 0 [/tex]

    [tex] F_y = Tsin60 + V - (mass of beam)g - (mass of block)g = 0[/tex]

    Now you can solve for the three unknowns.
  15. Nov 14, 2007 #14
    ok .... its making a little sense, so where do you go from here ...
  16. Nov 14, 2007 #15
    Solve the third equation for T in terms of V, and plug that into the second equation. Then solve the second equation for H in terms of V, and you can plug that into the first equation and solve for V. Then you should be able to solve for the other two varibles knowing V.
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