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?

I mostly was confused as to why you would use COS(22.5) to solve it, not SIN(22.5). If you can clarify that it's all I don't understand.

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?

I mostly was confused as to why you would use COS(22.5) to solve it, not SIN(22.5).  If you can clarify that it's all I don't understand.

vincent-lyon.fr · Answer

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Torque = weight x distance from joint = W x d

cos 22.5 = d/l , hence d = l x cos 22.5