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Torque, Cross product

  1. Sep 9, 2010 #1
    1. The problem statement, all variables and given/known data

    the question is sample problem 11-3 of 'fundamentals of physics' 7th edition.

    Three forces, each of magnitude 2.0N, act on a particle. the particle is in the xz plane at point A given bz position vector r, where r0 3.0M and theta=30°. Force f1 is parallel to the x axis, force f2 is parallel to the z axis, and force f3 is parallel to the y axis. What is the torque, about the origin o, due to each force.


    2. Relevant equations
    I'm trying to use t= RxF
    the book is using tx= rf1sin(theta)


    3. The attempt at a solution

    I broke the position vector into components = (2.6, 0 , 1.5) and applied cross product.

    (2.6, 0 , 1.5) X (-2,-2,-2)= (3, 2.2, -5.2)


    The book gets an answer of (3, 5.2, 6)


    I cant for the life of me figure out where I went wrong. I know that the book is right. Because if there is no position in the y direction, so the moment arm is 3 meters. therefore a 2N force should give a torque of 6N. but I have double checked my calculations over and over and cant find my mistake.

    Help!
     
  2. jcsd
  3. Sep 9, 2010 #2
    Each force applies a torque on its own. The total torque on the mass is the sum of the individual torques, not the cross product of the mass' position relative to the origin (It is only relevant in this question because it is the same as the point of application of the forces) with the net force.
    The net force is not an interesting quantity when you're dealing with rotational motion.

    [tex]\vec \tau _{total} = \Sigma \vec r_i \times \vec F_i[/tex]
    The position vectors in question go from the point you're measuring the torque about, to the point where the force is applied. In the question you provided, all the forces are applied to the same point, which happens to be the position of the object relative to the origin.

    Note that you were asked for the torque due to each force, not the net force. You found the net torque, the answer provided in the book is not broken down according to axes like yours is, but in terms of the force which is the source of each torque.

    Torque is measured in Nm, not N, just as an aside. :)
     
  4. Sep 9, 2010 #3
    Thanks for your reply, but I still don't quite understand my error.

    I don't see how

    [tex] \vec \tau _{total} = \Sigma \vec r_i \times \vec F_i [/tex]

    Is different from

    [tex] \vec \tau _{total} = \vec r \times \vec F_{net} [/tex]

    What is the difference mathematically, and what is the difference in intuition? I'm having a hard time picturing the two different situations.

    Thank you!
     
  5. Sep 9, 2010 #4
    Imagine you have a square sheet lying flat initially in the xy plane. You could apply a force to it at any of the points of its circumference. So let's say I have one force acting on the left side of the square, and always pointing up, perpendicular to the plane of the square sheet, and another force, equal in magnitude, but opposite in direction, acting on the right side of the square. The square will rotate about its center of mass.
    Imagine if you were turning a screw with your forefinger and thumb. You are applying a pair of forces, opposite in direction and equal in magnitude to the screw, and each of them acts on a different point on the screw.

    What is [tex]\vec r[/tex] ? Is it the one going to the point at which the left force acts, or the right one? And for that matter, [tex]\vec F_{net}[/tex] is 0! Your expression makes no sense since you cannot meaningfully define what [tex]\vec r[/tex] is, and since it does not require a net force to provide a net torque.

    Each force provides a torque dependent on its particular lever arm! In a general scenario, each force is applied to a different point on the object, and therefore has a distinct lever arm relative to the point you are calculating torques about. Your expression coincides with the real one only for a system consisting of a single point mass.
     
  6. Sep 14, 2010 #5
    But isn't that the system that I have??? It's just one particle, with a single distance r from the origin, and three forces applied at one point-which can be modeled as a net force.

    I don't see where the problem is, but it gives the wrong answer.

    I understand your reasoning though, that net force is generally unimportant because a 0 net force can still give a torque. But here the net all three forces are applied at a point with distance R from the origin
     
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