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Torque when to use cos and sin

  1. Dec 5, 2012 #1
    I have a issue with determining when to use cos and sin when calculating torque;

    For example. Referring to the attached picture. The SEESAW asks evaluate the torque of m2 (m2>>m1).

    For second part evaluate the torque...

    MY question:

    WHY in the first part (SEESAW) do they use cos and why in the second part do they use sin? How do you determine; when to use cos or sin???

    20121205_125348-1.jpg
     
  2. jcsd
  3. Dec 5, 2012 #2
    The selection of theta is what is causing the confusion here. Theta lies in between the vectors r and F.
    In the case of seesaw the radial vector is pointing from the center towards any of the mass and the corresponding force is pointing downwards or upwards. so in that position the angle should've been an acute angle. when the seesaw becomes parallel with the ground, the angle is 90degree causing a maximum torque.
    this is the actual and only correct angle
    But instead of selecting that angle, we can choose others but a trigonometric convertion is to be made. like sin(90-x)=cosx

    let θ+∅=90degrees:

    you chose the angle between perpendicular line from ground and seesaw's perpendicular axis. which is zero when the seesaw is in parallel state and swipes an angle each time it oscillates to both sides. this is infact the 90-∅ of the angle we need (∅)
    the cross product being sine ∅ now becomes sine ∅= cos 90-∅ = cos θ
     
  4. Dec 5, 2012 #3

    PhanthomJay

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    There are several ways to calculate the torque of a force about a point. You can use T =r X F = rF sintheta, where r is the magnitude of the position vector and theta is the included angle betwen the force and position vector.

    Or you can use T = force times perpencicular distance from line of action of the force to the pivot point.

    Or you can resolve the force into its components parallel and perpendicular to the position vector and use T = (F_perp)(r).

    Watch plus/minus signs.

    All roads lead to Rome.
     
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