Question

What is the magnitude of the torque about his shoulder if he holds his arm straight out to his side, parallel to the floor

Answer 1

Complete Question

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 4.0 kg. Assume, a bit unrealistically, that the athlete's arm is uniform.

What is the magnitude of the torque about his shoulder if he holds his arm straight out to his side, parallel to the floor? Include the torque due to the steel ball, as well as the torque due to the arm's weight.

Answer:

The torque is  $\tau = 34.3 \ N\cdot m$

Explanation:

From the question we are told that

   The mass of the steel ball is  $m = 3.0 \ kg$

    The length of arm is  $l = 70 \ cm = 0.7 \ m$

    The mass of the arm is $m_a = 4.0 \ kg$

Given that the arm of the athlete is uniform them the distance from the shoulder to the center of gravity of the arm is mathematically represented as

       $r = \frac{l}{2}$

=>    $r = \frac{ 0.7}{2}$  

=>    $r = 0.35 \ m$  

Generally the magnitude of torque about the athlete shoulder is mathematically represented as

      $\tau = m_a * g * r + m * g * L$

=>    $\tau = 4 * 9.8 * 0.35 + 3 * 9.8 * 0.70$

=>    $\tau = 34.3 \ N\cdot m$