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I didn't even learn this yet but got hw on TORQUE!

  1. Mar 2, 2009 #1
    1. The problem statement, all variables and given/known data
    untitled.jpg

    1. (top picture) Mass A is 30 kg, Mass B 15 kg, Mass C 5.0 kg. Mass A is 0.50 m away from pivot and Mass C is 1.5 m away from pivot. seesaw at a 30 degree angle from horizontal. Calculate torques of A and C and where B should be placed for equilibrium

    2. (middle picture) assume uniform distribution of mass, calculate center of mass

    3. calculate the torque for the four positions shown in the picture. The dumbell is 3.5 kg

    2. Relevant equations
    1. torque = rFsin(theta)
    0=torqueA+torqueB+torqueC

    2. I know to divide the L into rectangles, and the center of mass are the intersection of the rectangles. But i do not know how to calculate this...

    3. torque = rFsin(theta)

    3. The attempt at a solution
    1. I got stuck right off. I do not know which way is positive and which way is negative for the angles. I know its not important for the equilibrium problem but for the single torque finding problems i need a bit of help:

    T_a=(30*-9.8)*-0.5m(sin(60))=+127 Nm
    T_c=(5.0*-9.8)*1.5m(sin(120))=-63.7 Nm
    0=T_a+T_c+T_b
    0=127-63.7+rFsin(120)
    -63.3=r(15)(-9.8)sin120
    -63.3/-127.305...=r
    r=.497=.50 m

    2. no idea, i can draw it out but i cant calculate it
    would this work? 1. putting the shape onto a coordinate system
    2. find the center of mass of the rectangles (L/2, W/2)
    3. find the slope and equation of the line with the points
    4. repeat 2 and 3 for another combination of rectangles
    5. sets equations equal to each other and solve for intersection point between the two lines
    6. answer

    3. again the same problem as number 1. But I am assuming here... thinking like a door with its hinge so that gravity is pointing to the right (positive direction)
    torque = .4(3.5)(9.8)sin0 = 0
    =.4(3.5)(9.8)sin(30)=6.86
    =.4(3.5)(9.8)sin(60)=11.9
    =.4(3.5)(9.8)sin(90)=13.7
    I dunno if these are supposed to be positive or negative. I thought positive since lifting up the dumbell = counter clockwise = positive. but i have no idea

    I have no idea how to do these because the teacher haven't gone over them yet. But he assigned the homework sooooo here I am
     
    Last edited: Mar 2, 2009
  2. jcsd
  3. Mar 2, 2009 #2

    rl.bhat

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    Homework Helper

    Torque = force X perpendicular distance between direction of force and the pivot.
    In this problem direction of force always downwards.
    Even if they do, which way is positive and which way is negative....
    Usually anticlockwise torque is taken as positive.
     
  4. Mar 2, 2009 #3
    which problem are you referring ot ?
     
  5. Mar 2, 2009 #4

    rl.bhat

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    Homework Helper

    Problem No. 1
     
  6. Mar 2, 2009 #5
    by my definition of the coordinate system for number 1, i would have a negative F (due to gravity) and a negative length for A. however, i am stuck on how to tell if the angle is positive or negative. i know to put the r vector and F vector tail to tail to get the angle, which is 60 and 120 respectively (A, c) both give the same number but the sign keeps me from progressing with this problem
     
  7. Mar 2, 2009 #6
    I added some thing to the first post, please help, i need to understand this by tomorrow in time for lab
     
  8. Mar 2, 2009 #7

    Doc Al

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    Staff: Mentor

    I didn't check your arithmetic, but your method looks fine. Don't get hung up on the sign of the torque. As rl.bhat said, usually counter-clockwise is taken as positive. (Don't give signs to r and g; use rFsin(theta) to find the magnitude of the torque and then give it a sign based on which way it acts to rotate.)


    If you had two point masses, how would you find their center of mass? You can divide that shape into two rectangles, replace each rectangle by a point mass at its center, then calculate the x and y coordinates of the center of mass.

    Looks good to me. Unless they specifically ask for the correct sign of the torque, it's reasonable to assume that they just want the magnitude of the torque.
     
  9. Mar 2, 2009 #8
    thank you so much, just a couple of last questions
    1. still confused as to how i can define the sign. Is it by the direction that the force is going at or is it be the "final motion". for example, both forces (gravity) are acting downwards on a seesaw OR since one is heavier, they will both go at a counterclockwise direction
    2. so basically i can say (Xcm, Ycm)=[m1(x,y)+m2(x,y)]/m1+m2. and defining 1cmx1cm as 1 mass unit? and those x,y as the center of the 2 rectangles?
    3. so since the dumbell is being lifted up in a counter clockwise position, i can assume that its positive? however, what if the picture is flipped around, wouldn't that be negative?
     
  10. Mar 2, 2009 #9

    Doc Al

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    Staff: Mentor

    Using the standard sign convention: The weight of box C exerts a clockwise torque (it tends to turn the seesaw in a clockwise direction) so its torque would be negative. Similarly, box A exerts a counter-clockwise torque, so would be positive.

    Sounds good.

    I suspect that all that they want is the magnitude of the torque. But the sign of the torque exerted by the leg, since the leg is pushing upward, would be positive (counter-clockwise). If you flip the picture so that the leg exerts a clockwise torque, it would be negative.
     
  11. Mar 2, 2009 #10
    ok i am kinda starting to get it, but how did you know the C exerts a clockwise torque? is it because its further out?

    I try to use the right hand rule, but i am no good at it
     
  12. Mar 2, 2009 #11

    Doc Al

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    Just ask yourself: If the only force acting on the seesaw was that due to the weight of C, which way would it turn?
     
  13. Mar 2, 2009 #12
    OHHH i get it, looking at it one torque at a time, thanks a bunch AGAIN :)
     
    Last edited: Mar 2, 2009
  14. Mar 2, 2009 #13
    yeek, last question:
    ok so when solving for B, i DO need to put the negative sign on the gravity constant right? since it is the difference between a + or - r (right or left of pivot)
     
  15. Mar 2, 2009 #14

    Doc Al

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    No. The reason that r is to the right of the pivot is that you need a clockwise torque to balance the torque from A and C.

    Ta + Tb + Tc = 0, so
    Tb = -(Ta + Tc)

    Since Ta + Tc is counter-clockwise, Tb must be clockwise.
     
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