17 thoughts on “Homework #8”

  1. slashley

    For the online hw.. I’m aware that we did a problem kind of like 8.38.. but how does the forearm mass &15 degree angle come into play? I can’t figure out how to use any of the information in the problem in the equation that is given in the book.
    Also, are all the masses to be converted into N?

  2. physics131-willocq Post author

    Problem 8.38 is just like the example with the barbell I did in class. The only difference is that the forces due to the weight of the ball and the tendon make an angle of 15 degrees relative to the arm. What is relevant to calculate the torque is the force component perpendicular to the arm. So here you need to take the 15 degree angle into account to calculate the total torque exerted on the arm.

    Another difference with the example from class is that in this case we cannot neglect the weight of the arm since it is not small compared with that of the ball. So consider that the weight of the arm acts at the arm’s center of gravity. This weight creates a torque that needs to be included in the calculation of the net torque acting on the forearm.

  3. slashley

    Where in the equation am I supposed to take the 15 degrees into account? I’m not really seeing how similar the problems are because I don’t know where to plug numbers in or even how to do the problem, period.
    Are all the masses supposed to be converted in N to use it in whatever equation?

  4. physics131-willocq Post author

    The torque corresponding to each force F is computed with the equation: tau = r F_perp, where r is the distance between the pivot point and the point on the arm at which the force is applied, and F_perp is the component of the force that is perpendicular to the arm.

    Masses do need to be turned into weights since forces enter the calculation of the torque.

    The 15 degrees need to be taken into account when calculating F_perp.

  5. rsampaio

    For the written hw, how do we calculate the force from the right hand acting on the spatula (part c)?

  6. physics131-willocq Post author

    Each force acting on the spatula induces a torque (provided the force doesn’t act at the pivot point in which case it has torque=0). For the spatula to remain in equilibrium, you need to require that the net torque on the spatula is zero. Since torque involves force, you can then find the value of the force.

  7. mstaples

    I am not sure how to go about solving problem 8.1 in the online homework. I know that I must break up the problem in to two parts with two different pivot points but I am confused as what equation to use. Could you please offer some assistance?

  8. physics131-willocq Post author

    You need to apply the requirement that the net torque on the board be equal to zero. This requirement must hold for any pivot point. So it is a matter of pivot choice to simplify the equation such that it contains only one unknown force.

  9. rdetroy

    I seem to be stuck on problem 8.6. I am not sure how we can obtain the distance from the pivot point. Since the weight is actually on top of the pivot point, how do we measure the lever arm from that pivot point?

  10. awilcins

    For the written homework, point A, if the r for that position is zero, and the net torque is zero, then it seems you cannot find the force acting on it (N). Also, for point B, if the opposing force (keeping the spatula down) is of negligible mass then how do we determine the force keeping the spatula up?

  11. mspier

    I am still having problems with the online homework #8.38 and none of the examples in the book seem to explain it adequately. I have written down from class that we need to use the cosine of of the angle in the problem but I am not sure how.

  12. physics131-willocq Post author

    The cosine of 15 degrees is needed to calculate the component of each force perpendicular to the forearm (as required for the calculation of the torque due to each force). See also comment #2.

    You may also want to look at example 7.3 in the textbook to see how to calculate the torque for a force that is not perpendicular to the object it is acting on.

  13. physics131-willocq Post author

    For problem 8.6, you may want to figure out the distance d such that the center of gravity is exactly above the pivot point. Another approach is to calculate the net torque produced by the two objects and require it to be zero. The weight of each object acts at the object’s center of gravity.

  14. lross

    I am completely stuck on 8.39. I can’t figure out where to put the pivot point, and also I am having trouble determining the force exerted by the hinge.

  15. physics131-willocq Post author

    Since the hinge is the pivot point, the force exerted by the hinge has torque=0. This means you do not need to know the force exerted by the hinge to calculate the net torque exerted on the pole.

    Here you need to calculate the perpendicular component of each of the forces acting on the pole and get the distance r between the pivot point and the point at which each of the forces act on the pole. Then torque for a given force = r F_perp.

    Also, remember that the torque has different sign for forces that tend to make the pole rotate clockwise or counter-clockwise.

  16. jkern

    I was able to solve 8.1 by observing a similar problem in the book. However, on 8.2 I am not sure how to incorporate the weight of the board into the same situation. How exactly does it fit into the equation for the net force and the net torque?

  17. physics131-willocq Post author

    You should consider that the weight of the board acts at the center of gravity of the board (i.e. at the half-way point along the length of the board). This weight force then contributes to the net torque exerted on the board.

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