Question

An athlete at the gym holds a 1.5 kg steel ball in his hand. His arm is 70 cm long and has a mass of 4.0 kg .

Answer 1

Answer:

torque τ   = 24 N-m

torque  τ  = 11.98 N-m  

Explanation:

given data

mass of the ball = 1.5 kg

arm length l = 70 cm = 0.7 m

mass of arm = 4 kg

solution

we get here  center of mass that is

center of mass = $\frac{m1*x1+m2x2}{m1+m2}$  ..........1

center of mass = $\frac{4*.35+1.5*0.7}{4+1.5}$  

center of mass = 0.44545 m  

and

torque will be

torque τ   =  F × r ×sinθ  ...........2

torque τ   =  (4+1.5)9.8 × 0.4454 × sin90

torque τ   = 24 N-m

and now we get magnitude of the torque  

magnitude of the torque  τ  = F × r ×sinθ

magnitude of the torque  τ  = (4+1.5)9.8 × 0.4454 × sin60

magnitude of the torque  τ  = 11.98 N-m