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Statics: find torque at door hinges

  1. Apr 17, 2015 #1
    1. The problem statement, all variables and given/known data
    Door weighs 20kg, and is held open by force P at doorknob opposing 15kg weight attached at C and wrapped around pulley. Theta is 90 degrees, so door is on x axis. Find torques on hinges A & B



    2. Relevant equations
    P is 118.8N in -k direction, which agrees with answer in book.
    I am getting answers for torques at A & B that don't agree with the book



    3. The attempt at a solution
    Here are summations of torques at B:

    (1.75j + 1.2i)(1.2i – 1.2k)(147N) T along top (from C to E)

    (1.75j + 1.2k)(-147j) T from E to 15kg weight

    (.65j + 1.05i)(-118.8k) P; force on doorknob

    (.75j + .6i)(-196j) Weight of door

    (1.5j)(Axi + Ayj + Azk) Forces at hinge A


    This set of forces yields the following values:

    -308.7k -308.7i + 211.7j

    176.4i

    -77.2i – 127.7j

    -117.6k

    -1.5Axk + 1.5Azi


    Am I correct so far?
    Thank you....any help is appreciated.



     

    Attached Files:

  2. jcsd
  3. Apr 17, 2015 #2
    Small clarification: you are finding the reactions at A and B, not the torques.

    Also, what is the assumption made in the problem regarding A?
     
  4. Apr 17, 2015 #3
    I didn't check your numbers but:

    1) where do the numbers (1.2i-1.2k) come in from your first equation? also check your signs.
    2) it is confusing why you are including the 2nd equation since it does not apply when summing the torques about B.
    3) the signs on your 3rd equation might be off.
     
  5. Apr 18, 2015 #4
    Yes, my bad; we are looking at the reactions at A & B, not torques.
    The book says only that at A there is no axial force.

    1) The first equation is the force of the cable from C to E. So if we look at r cross F, then r (distance from B to C) would be the 1.75j + 1.2i component, and then F is the force of the cable. Since the 15kg wt pulls down, then on the other side of the pulley, the force must go in the opposite direction; it travels 1.2 to the right, and 1.2m into the paper, so its components would be 1.2i – 1.2k.

    2) In another problem involving a tower crane (I have included a diagram), the tension in EA exerted a torque on the pin B; so I assumed this was similar and included the influence of the 15kg weight. Was this incorrect? If so, why did it apply to the tower crane and not in this case?

    3) For the influence of the force at the doorknob, we go up .65 and to the right 1.05 to get from B to the doorknob; the force P is 118.8 into the paper, so it should be negative k: (.65j + 1.05i)(-118.8k)

    Thank you for helping...I look forward to your reply.
     

    Attached Files:

  6. Apr 18, 2015 #5
    1) That's not how to calculate components of a force. You know the magnitude T=147N, correct? Then the components would be from the unit vector 1/√2i-1/√2k.

    2) Your first equation already states the value of the force T=147N. So the 15kg mass is already accounted for here. In your tower crane problem the tension force EA is applied directly to the structure unlike the 15kg mass here. If you do a free body diagram of the door then the mass doesn't come into it.

    3) Your equation is right but the resultant signs are wrong.
     
  7. Apr 20, 2015 #6
    As Homer Simpson would say, "D'Oh!" I see now what a clumsy mistake I made--I forgot that the tension in the cable is the force times the unit vector.
    Once I corrected that, and knocked out the (unnecessary) forces in equation 2, then I ended up with the correct answers, as confirmed by the answer key in the back of the book.

    Thank you for pointing me in the right direction!:smile:
     
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