Question

A man in a gym is holding an 8.0-kg weight at arm's length, a distance of 0.55 m from his shoulder joint. What is the torque about his shoulder joint due to the weight if his arm is held at 30° below the horizontal?

Answer 1

Answer:

  $\tau =37.34\ N m$

Explanation:

given,

mass of the weight = 8 Kg

distance = 0.55 m

angle below horizontal = 30°

torque about shoulder

  $\tau = \vec{r} \times \vec{F}$

  $\tau = r \times F cos \theta$

  $\tau = 0.55 \times 8 \times 9.8 \times cos 30^0$

  $\tau =37.34\ N m$

torque about his shoulder join is equal to   $\tau =37.34\ N m$