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Homework Help: Calculus Questions with Vectors (cross product)

  1. Aug 29, 2013 #1
    1. The problem statement, all variables and given/known data

    Calculate the net torque about O at P,assuming that a 30-kg mass is attached at P [Figure 21(B)]. The force Fg due to gravity on a mass m has magnitude 9.8m m/s2 in the downward direction.

    2. Relevant equations

    The torque about the origin O due to a force F act- ing on an object with position vector r is the vector quantity τ = r × F. If several forces Fj act at positions rj , then the net torque (units: N-m or lb-ft) is the summation of r cross product F.

    3. The attempt at a solution

    Have solution for no mass hanging that is: r=OP =10(cosθi+sinθj) The angle between the force vector F and the x-axis is (θ + 125◦), hence, F = 25 (cos (θ + 125◦) i + sin (θ + 125) j)
    The torque τ about O acting at the point P is the cross product τ = r × F. We compute it using the cross products of the
    unit vectors i and j:
    τ =r×F=10(cosθi+sinθj)×25((cosθ+125◦)i+sin(θ+125◦))j = 250 (cos θ i + sin θ j) × ((cos θ + 125◦) i + (sin θ + 125◦) j)= 250 (cos θ sin (θ + 125◦) k + sin θ cos (θ + 125◦) (−k))
    = 250(sin(θ +125◦)cosθ −sinθ cos(θ +125◦))k
    We now use the identity sin α cos β − sin β cos α = sin(α − β) to obtain
    τ = 250sin(θ +125◦ −θ)k = 250sin125◦k ≈ 204.79k

    Attached Files:

    Last edited: Aug 29, 2013
  2. jcsd
  3. Aug 29, 2013 #2


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    It's really hard to follow your calculations because you have a LOT of these question mark thingies in the OP.
  4. Aug 31, 2013 #3
    The method looks good but wouldn't you want the force to be:

    25 (cos (θ + 125◦) i + sin (θ + 125) j)- (30*g) j

    The added term would account for the mass trying to rotate the arm itself.

    Or is the 25 N force the net force on the mass (I thought it was an external force in addition to that of gravity)?
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