Answer:
[tex]\tau =37.34\ N m[/tex]
Explanation:
given,
mass of the weight = 8 Kg
distance = 0.55 m
angle below horizontal = 30°
torque about shoulder
[tex]\tau = \vec{r} \times \vec{F}[/tex]
[tex]\tau = r \times F cos \theta [/tex]
[tex]\tau = 0.55 \times 8 \times 9.8 \times cos 30^0 [/tex]
[tex]\tau =37.34\ N m[/tex]
torque about his shoulder join is equal to [tex]\tau =37.34\ N m[/tex]
