Question

Torque problem:
A 1.61kg bowling trophy is held at arm's lenght, a distance of .605m from the shoulder join. What torque does the trophy exert about the shoulder if the arm is a) Horizontal, or b) at an angle of 22.5(degrees) below the horizontal?

Answer 1

\[\tau=F.d\] where \[\tau\] is dot multiplication of two vectors Force and distance from the point of rotation to the point of applied force, and dot multiplication yeilds a scalar quantity.

when the arm is horizontal, it means 0 degree with horizontal, and the force is acting downwards, therefore:
\[\tau = F.d = mg.d.cos\theta = 1.61(9.8) * 0.605 cos(0) = 9.54569 \space N.m\]

now in the second case, it is 22.5 below horizontal (his arm got tired :P), using same formula:

\[\tau=F.d=mg.d.cos\theta=1.61(9.8) * 0.605cos(22.5) = 8.81906 \space N.m\]

notice, 22.5 below horizontal means below x-axis which is supposed to be -22.5 degrees, but no matter you give 22.5 or -22.5 to cosine, answer will be positive because cosine is positive in first and fourth quadrant