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Cylinder Equilibrium (2012 F = ma #12)

  1. Jan 23, 2015 #1
    1. The problem statement, all variables and given/known data
    The problem is #12 found here: https://www.aapt.org/physicsteam/2013/upload/exam1-2012-unlocked-solutions.pdf.

    A uniform cylinder of radius a originally has a weight of 80 N. After an off-axis cylinder hole at 2a/5 was drilled
    through it, it weighs 65 N. The axes of the two cylinders are parallel and their centers are at the same height.
    a 2a/5
    A force T is applied to the top of the cylinder horizontally. In order to keep the cylinder is at rest, the magnitude
    of the force is closest to:
    (A) 6 N← CORRECT
    (B) 10 N
    (C) 15 N
    (D) 30 N
    (E) 38 N

    2. Relevant equations
    Tau = r x F
    M_total x_cm = sum m_i x_i

    3. The attempt at a solution
    This problem has been bugging me because I keep getting the wrong answer.
    So i found the center of mass of the new cylinder to be at -6a/65 with the origin being at the center. (0 = 65x + 15*2a/5 -> 65x = -6a -> x = -6a/65.)
    The torque due to gravity must be -6/65a*65 = -6a relative to the point of contact. Therefore, the torque due to the horizontal force T must be 6a. 6a = 2aT -> T = 3N which does not agree with the answer of 6N. What have I done incorrectly?
     
  2. jcsd
  3. Jan 23, 2015 #2

    TSny

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    Your work looks correct to me. You have wisely taken the torques about the point of contact. I agree with your answer. Hopefully, others will confirm this or else point out where we are overlooking something.

    (I think the answer of 6 N would be correct if the cylinder were mounted on a horizontal, frictionless axle running along the central axis of the cylinder.)
     
    Last edited: Jan 23, 2015
  4. Jan 23, 2015 #3
    I wasn't able to follow what you did, but here is what I did. Drilling out the hole centered at x=2a/5 is equivalent to putting an upward force of 15 N at x = 2a/5 in the original undrilled cylinder. The moment of this force about the center of the cylinder is 6a N-m. So the force to balance this moment at r = a has to be 6 N.

    Chet
     
  5. Jan 23, 2015 #4

    haruspex

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    I agree with 3N, and there's an easier way to get it.
    Think of the drilling the hole as, instead, adding a negative weight (80N-65N=15N) at the hole's mass centre.
    This has torque 15N*(2a/5) = 6aN about the point of contact with the ground. The countering force is 2a perpendicularly from that point.
     
  6. Jan 23, 2015 #5

    haruspex

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    Did you check the diagram? No axle is indicated for the cylinder. Rather, it is shown lying sideways on a surface.
     
  7. Jan 23, 2015 #6
    You're right. The key words in the problem statement are "at rest." The force at the top has to be 3N, and a static friction force of 3N has to develop at the table top to balance the horizontal force (to prevent linear acceleration) and to provide the required moment (couple). In this way, the moment about the center of mass and the moment about the contact point at the table top are the same. Very cute.

    Now, back to the problem statement. They asked which answer the force is closest to. 3N is closest to the 6N choice. I'm guessing they were allowing people who made my misinterpretation to get credit for a "correct" answer.

    Chet
     
  8. Jan 23, 2015 #7

    TSny

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    Could be. But they could have achieved the same objective by using the correct answer of 3 N instead of 6 N.
     
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