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

`\tau =37.34\ N m`

Step-by-step explanation:

given,

mass of the weight = 8 Kg

distance = 0.55 m

angle below horizontal = 30°

torque about shoulder

`\tau = \vec{r} * \vec{F}`

`\tau = r * F cos \theta`

`\tau = 0.55 * 8 * 9.8 * cos 30^0`

`\tau =37.34\ N m`

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