Question: What is the torque about the shoulder if the arm is held in an abducted position at 60 degrees from the body in the frontal plane while holding a 10 kg dumbbell? Assume that the mass of the arm is 6 kg, its center of mass is located 38 cm from the shoulder joint center, and the arm's total length is 80 cm.
The only thing I can think about starting with is the formula for torque, which I believe is $\tau = r \times F \times \sin(\theta).$
Where could I go from here?
Draw a picture:
$T = rF\sin\theta$ is the apropriate formula. but, we have two masses and two distances referenced
$T = r_1F_1\sin\theta+ r_2F_2\sin\theta$