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# An athlete at the gym holds a 3.0 kg steel ball in his hand. His arm is 70 cm long and has a mass of 3.8 kg, with the center of mass at 40% of the arm length from the shoulder.Part A: What is the magnitude of the torque about his shoulder due to the weight of the ball and his arm if he holds his arm straight out to his side, parallel to the floor?Part B: What is the magnitude of the torque about his shoulder due to the weight of the ball and his arm if he holds his arm straight, but 45° below horizontal?

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Problem A Solution

The problem applies the concept of torque, where torque is the cross product between the displacement from the line of action or lever arm and the force acting on it. In this case, the cross product takes the product between the force and the perpendicular component of the length.

In the first part of the problem, the length vector from the point of action is perpendicular with the weight of the bodies, so we take the sum of the products for each body that contributes torque.

\begin{align*} \Sigma \tau &= \Sigma r \times F \\ &= (0.4r_a)(m_ag) + (r_a)(m_bg) \\ &= 0.4(0.7 \ \text{m}) (3.8 \ \text{kg} )g +(0.7 \ \text{m}) (3 \ \text{kg} )g \\ &= 10.43 \ \text{N} \cdot \text{m} + 20.58 \ \text{N} \cdot \text{m} \\ &=\boxed{ 31.01 \ \text{N} \cdot \text{m} } \end{align*}

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