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Elasticity Problem

  1. Oct 28, 2007 #1
    1. The problem statement, all variables and given/known data
    Figure is an overhead view of a rigid rod that turns about a vertical axle until the identical rubber stoppers A and B are forced against the rigid walls at distances rA = 7cm, rB = 4cm. Initially, the stoppers touch the walls without being compressed. Then force F of magnitude 220N is applied perpendicular to the rod at a distance R = 5cm. Find the magnitude of the force compressing stopper A and stopper B.


    2. Relevant equations
    I'm not sure how elasticity even plays in to this problem... Should I treat this as an equilibrium type problem?

    F = (+) FN1 = (-) FN2 = (+)
    R = (-) rA = (-) rB = (+)
    Tnet = 0, Fnet = 0
    T1 = RF
    T2 = rB(FN2)
    T3 = rA(FN1)
    T1 + T2 + T3 = 0

    RF + rB(FN2) + rA(FN1) = 0

    F + FN1 + FN2 = 0


    3. The attempt at a solution
    Not sure where to start really...

    Diagram:
    [​IMG]
     

    Attached Files:

    Last edited: Oct 28, 2007
  2. jcsd
  3. Oct 28, 2007 #2

    PhanthomJay

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    The system is in equilibrium, but you're missing some key points here. First, your rotational equilibrium torque equation is ignoring plus and minus signs for clockwise vs. counterclockwise moments. Second, your translational equlibrium equation for sum of forces not only ignores signage, but you're missing the force at the fulcrum. Third, you'll need another equation besides the equilibrium equations in order to solve the problem (Hint: since the rod is rigid, there is a ratio between the forces at the left and right walls).
     
  4. Oct 28, 2007 #3
    Shall I make it simpler?
    First thing, it is indeed a question of elasticty. You will need to apply Hooke's Law, I guess.
    Secondly.. consider hint given by "PhanthomJay" : "(Hint: since the rod is rigid, there is a ratio between the forces at the left and right walls)" -- I am adding one thing: ratio of the forces is related by the ratio of compressions through Hooke's Law.
     
  5. Oct 28, 2007 #4
    Great, thanks for the replies - I wrote the forces and torques as vector quantities, not magnitudes, so they have their signs in them. For this diagram, the coordinate system is based around the axis of rotation (axle), so up & right are positive, down and left are negative. And, like you said: counterclockwise is positive, and clockwise is negative.

    How should I express the force at the fulcrum - which direction would it be acting in? I thought any forces acting on the axis of rotation don't affect torque

    Hooke's Law is F = -kx, so
    FN1 = k(rA) = force of wall on A : rA is negative
    FN2 = k(rB) = force of wall on B : rB is positive

    Should I be looking at the force exerted by the wall on the rod, or by the rod on the wall - I know that they will have the same magnitude, but does it matter?

    are these the correct relationships?
     
    Last edited: Oct 28, 2007
  6. Oct 28, 2007 #5

    PhanthomJay

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    And you should choose a plus and minus direction for the torques, also.
     
  7. Oct 29, 2007 #6

    PhanthomJay

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    good, this establishes a sign convention

    you can assume a direction for now, but since you are summing torques about the fulcrum, you are correct that it won't enter into your torque equation.

    yes it matters. since you are looking at forces applied to the rod, you should be looking at the forces of the wall on the rod, and not vice versa, or else your signs are going to get messed. Now you're still missing one more geometric relationship between the displacements at both ends of the rods....you can get that using similar triangles.
     
  8. Dec 30, 2011 #7
    If you reason "220 minus 5/11 of itself is 120", you get the value of 120 newtons for FN1. Then you can use the 4 cm to 7 cm proportion to get 68N for FN2. The 11 in 5/11 comes from the sum of 4 cm and 7 cm.
    This works, but I don't know why.
    The equations that were derived above look valid, except for there not being any minus signs reflecting upward and downward forces, and clockwise vs. counterclockwise torque.
     
  9. Dec 31, 2011 #8
    The torque equation is fairly obvious: FN1*0.07 - F*0.05 + FN2*0.04 = 0.

    The obvious choice for the second equation would be to net out the vertical forces, but this involves three quantities. Better to find the two horizontal forces on the rod from the stoppers. These exist because the rod is slightly tilted. F, however is purely a vertical force and makes no showing here.

    The tilt creates similar right triangles having hypotenuses 0.07 meters and 0.04 meters respectively. The short legs are the distance y that each stopper gets compressed. Theta will be the angle between the horizontal and each short leg. Value "y" is the same for each stopper because of the rigidity of the rod.

    For the left hand stopper the horizontal force, FNx, is FN1*cos(theta). But cos(theta) is y/0.07, so we have FN1x = FN1*(y/0.07). Similarly, we get FN2x = FN2*(y/0.04). When we equate FN1x to FN2x (equilibrium condition), y cancels and we get FN1 = (7/4)*FN2. This value of FN1 can be substituted into the torque equation, and we only have one unknown.
     
  10. Dec 31, 2011 #9
    This is a slight rewrite of the above proof, which used "FNx" where "FN1x" should have been employed:

    The torque equation is fairly obvious: FN1*0.07 - 220*0.05 + FN2*0.04 = 0.

    The obvious choice for the second equation would be to net out the vertical forces, but this involves three quantities. Better to find the two horizontal forces on the rod from the stoppers. These exist because the rod is slightly tilted. F, however is purely a vertical force and makes no showing here.

    The tilt creates similar right triangles having hypotenuses 0.07 meters and 0.04 meters respectively. The short legs are the distance y that each stopper gets compressed. Theta will be the angle between the horizontal and each short leg. Value "y" is the same for each stopper because of the rigidity of the rod.

    For the left hand stopper the horizontal force, FN1x, is FN1*cos(theta). But cos(theta) is y/0.07, so we have FN1x = FN1*(y/0.07). Similarly, we get FN2x = FN2*(y/0.04). When we equate FN1x to FN2x (equilibrium condition), y cancels and we get FN1 = (7/4)*FN2. This value of FN1 can be substituted into the torque equation, and we only have one unknown.
     
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