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Free-body diagrams and torque

  1. Dec 23, 2007 #1
    1. The problem statement, all variables and given/known data
    Two friends are carrying a crate of mass 200 kg up a flight of stairs. The crate has length 1.25 m and height 0.50 m, and its center of gravity is at its center. The stairs make a 45.0 angle with respect to the floor. The crate also is carried at an 45.0 angle , so that its bottom side is parallel to the slope of the stairs . The force each person applies is vertical.

    Find the force that each person contributes with.

    3. The attempt at a solution
    I want to find the torque around both points. But I do not know, how the free-body diagram looks like.

    There's a gravity at CM, and we have the forces T_b and T_u (bottom and up). Where are T_b and T_u directed?
    Last edited: Dec 23, 2007
  2. jcsd
  3. Dec 23, 2007 #2

    Shooting Star

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    > The force each person applies is vertical.

    That is given in the problem.

    Take torque of the two forces about the CM. The calculation is simpler that way.
  4. Dec 23, 2007 #3
    Ok, since I have your attention, Shooting Star, I must ask you: When we have following setup, what is the direction of the normalforce?

    (the dot in the middle is the CM and CG in this case).


    - and btw, if the forces are vertical, arent they equal in magnitude?

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  5. Dec 23, 2007 #4

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    I can't see the picture yet, but in the problem it's given that the crate has a "height". If by height, thickness is meant, then at which points does each person hold the crate? At the bottom corners? But I'll wait for the pic.

    Normal force is always normal to a surface. Why do you ask?

    And remember from last time -- clearly write the problem.
  6. Dec 23, 2007 #5
    I solved the first problem (with the two guys lifting the crate) - so that one is out of the picture. I hadn't seen the forces were vertical (even though I typed in in.. do'h!)

    I have a new question. We have a block - a rectangular stone - which is tipped at an angle theta (the picture is showing this setup). I just want to know, what the normal force is in this case?
  7. Dec 23, 2007 #6

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    So it is just resting on a flat horz surface on one of its corners as shown? In equilibrium or not?
  8. Dec 23, 2007 #7

    No, but a handyman actually has attached a string to the upper end and he is holding the string so yes - it is in equilibrium, but only because of the string and handyman - not because of gravity.
  9. Dec 23, 2007 #8

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    Why do I have to interrogate you like an infiltrating spy? :) You could have given all the info right at the beginning.

    So, it IS in equilibrium, and because of the handyman and gravity and normal reaction, all three. The normal reaction of the ground on the block is the force that's acting upward on the stone at the bottom corner.

    Since the vertical components add up to zero, equate the sum of the three vert forces to zero. If you don't know the pull by the handyman, then take the moments of the weight and the normal reaction about the top corner and equate to zero.

    You have to know the angle at which it's tipped and the dimensions of the block, or at least the ratio of length to breadth.
    Last edited: Dec 23, 2007
  10. Dec 23, 2007 #9

    I'm sorry, from now on I will write the whole text and all my own reasonings. I know the angle it is tipped (20 degrees), the angle that the string makes with the horz. (58 degrees) and the dimensions of the block.

    Ok, so the normal-force acts upward (the angle with horizon is 90 degrees) - does the magnitude of the normal-force equal the magnitude of the gravity + the vert. component of the string tension in this case as well? If yes, then the reason for the stone to tip if the handyman wasn't in the picture is that center of gravity acts on a point that is outside the area of support?

    Also, there would be an acceleration in the x-direction?

    - By the way, I just thought of something: I can measure the torque around any reference point, right? This goes for static as well as non-static problems? But in static problems, torque has to be zero regardless of which point we measure it about, but it is not the same when we measure the torque about different points in non-static problems?
    Last edited: Dec 23, 2007
  11. Dec 24, 2007 #10

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    Everything you have said is correct, except the following which I didn't understand:

    >Also, there would be an acceleration in the x-direction?

    The whole block is static, right? Then why should it accelerate?
  12. Dec 24, 2007 #11
    Since the only x-component in the system is the one coming from the string-force?
  13. Dec 24, 2007 #12

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    Yes, it occured to me just after posting. So, it seems that there has be friction with the floor. What exactly was the question, to find the direction of normal reaction or reaction?
  14. Dec 24, 2007 #13
    I have to find the tension in the string. The whole text is:

    A rectangular stone is tupped at an angle theta above the horizontal using a rope. If theta = 20 degrees, what force does the handyman have to exert on the rope?

    The heigh of the stone is 3,75 m and legth is 1,75. They say nothing about friction.

    The angle between the rope and the horz. is 58 degrees.
  15. Dec 24, 2007 #14

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    They say nothing about friction, and you too in the final answer have to say nothing about friction. There will be T acting, presumably (because you didn't tell me) on the upper most corner, W at the CM vert downward, F and N at the point of contact with the floor.

    Just take moments of T and W about the point of contact, so that F and N doesn't even come into the picture. This moment must be zero. Find T in terms of W. (Notice how you've managed to withhold the weight of the block from me up to now. Just remember this the next time you post!)
    Last edited: Dec 24, 2007
  16. Dec 24, 2007 #15
    T is acting at the "lower" corner.. sorry, I forgot you don't have a picture.

    Actually, I am not told the weight of the object. But as you said, I can just find the tension in terms of W.

    One last thing - how do I find the perpendicular distance between the gravity-vector and the point of contact? Will I just use alot of trigonometry for this?

    - and have a merry christmas, and thank you for being so patient :-)!
  17. Dec 24, 2007 #16

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    Not a lot, but some.

    After many many posts, I discover that T acts at the lower corner!

    Well, it is the time for forgiving... And a merry Christmas to you too.
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