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Question about torque by gravity

  1. Feb 10, 2013 #1
    1. The problem statement, all variables and given/known data

    Torque = Fr sin theta, but is it always sin?
    If I am looking for torque by gravity , will it by mg cos theta instead?
     
  2. jcsd
  3. Feb 10, 2013 #2

    mfb

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    That just depends on your definition of the angle theta.
    Gravity has no special role here in any way.
     
  4. Feb 10, 2013 #3
    i am doing one question, but the teacher said torque by gravity is mg cos theta..

    The question is
    A trap door, of length and width 1.65 m, is held open at an angle of
    65.0o with respect to the floor. A rope is attached to the raised edge of
    the door and fastened to the wall behind the door in such a position that the rope pulls perpendicularly to the trap door. If the mass of the trap the trap door is 16.8 kg, what is the torque exerted on the trap door by the rope?


    There is a torque by rope and also a torque by gravity
     
  5. Feb 10, 2013 #4
    Since the trapdoor is the lever arm, the rope is exerting force upward, and the gravity is downward. what is the angle with respect to?
     
  6. Feb 10, 2013 #5

    rcgldr

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    It's always sin(theta) when theta is defined as the angle from the radial direction of the applied force, so that theta = ±90° when the force is tangental (perpendicular to the radial direction).

    The problem states that the tension in the rope is perpendicular to the door, which would mean that theta as used in the torque equation is ±90°.

    The angle of the trap door with respect to the floor (the angle from horizontal) is not the same as the angle used in your torque equation, torque = F r sin(theta). The angle of the trap door with respect to the floor is a different "theta".
     
  7. Feb 10, 2013 #6

    "with respect to the floor" do u mean the angle between the door and the floor (which is the force of gravity"? I do understand the angles are different, but i dont know how to get those different angles.. can u show me how do i decide on which angles to use? thanks
    so since it is always sin, i guess my teacher means u can use cos if the angle is opposite..so u dont need to convert the angle 180-theta
     
  8. Feb 10, 2013 #7

    rcgldr

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    It would probably be easier to state the angles relative to the trap door and the radius relative to the hinge of the trap door, so that you always use torque = force x radius x sin(angle of force). What is the direction of gravity relative to the trap door? What is the direction of the rope relative to the trap door?

    That should be 90 - theta.
     
  9. Feb 10, 2013 #8

    oh yea srry , i meant 90-theta

    Gravity is acting down, direction of the rope is acting upwards

    0 = Tr + Tg

    Tr= Fr sin 90
    (i dont know what the force is here, am i suppose to find that
     
  10. Feb 10, 2013 #9
    I found the answer online

    It uses cos65 x Fg x r/2
     
  11. Feb 10, 2013 #10

    rcgldr

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    You should be able to determine what the force from gravity is. Your class or text book should have explained that you can assume that gravity acts as a downwards force applied to the middle of the trap door.
     
  12. Feb 10, 2013 #11
    yes.. i know all those already to whatever u have said so far
     
  13. Feb 10, 2013 #12

    rcgldr

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    That should be T = mg x r/2 x cos(65°). 65° is the angle from the floor. The angle of gravity force from the door is 25°. If you used this, then the equation is T = mg x r/2 x sin(25°).
     
  14. Feb 10, 2013 #13
    but the question is asking for the torque on the rope exerted on the door...is it because torque of rope = torque of gravity since it is an equilibrium torque? if so, it is also possible to find the answer using torque of rope give sin90 = 1, but the only thing missing is the force applied that's why it is not possible to solve the torque of rope by itself?
     
  15. Feb 10, 2013 #14

    rcgldr

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    Yes, the torques are equal in magnitude but opposing direction (equilibirum). The question is asking for the force in the rope. What is the equation for torque of the rope in terms of force from the rope?
     
    Last edited: Feb 10, 2013
  16. Feb 10, 2013 #15
    T = rF sin theta

    Force i'm not sure
    Fn + Ty - Fg = 0..?
     
  17. Feb 10, 2013 #16

    rcgldr

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    You've already stated that theta for the rope is 90°. So for the rope, T = F r sin(90°), and for gravity T = mg x r/2 x sin(25°). You're also told that gravity = 9.8 m / s^2, and the mass of the trap door is 16.8 kg, so how many Newtons of force does gravity exert on the door? With this information, you should be able to solve for F, the force of the rope.
     
  18. Feb 10, 2013 #17
    huh..? setting them equal to each other?
     
  19. Feb 10, 2013 #18

    rcgldr

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    Yes, you stated that the torques were in equilibrium. Technically it's the sum of the torques that is zero, so

    Tgravity + Trope = 0

    The force from gravity is negative, and the force from the rope is positive.
     
  20. Feb 10, 2013 #19

    umm yea, i think i know wat to do..i will try
    thanks
     
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