Physics! An athlete at the gym holds a 2.5 steel ball in his hand. His arm is 70 long and has a mass of 4.0 .?

Question

An athlete at the gym holds a 2.5 kg steel ball in his hand. His arm is 70 cm long and has a mass of 4.0 kg. 
a) What is the magnitude of the torque about his shoulder if he holds his arm straight out to his side, parallel to the floor? 
b) What is the magnitude of the torque about his shoulder if he holds his arm straight, but 60 degrees below horizontal? 
Thank you.

Anonymous · Accepted Answer

Torque is massghorizontal distance to weight.  For the arm, its weight is considered concentrated at the center of mass, or half the length of the arm.
Arm straight out, the torque is 2.5g0.070 + 4.0g0.035 = 3.09 N-m
Arm at 60º below horizontal, the distances are multiplied by cos60º, or 0.5.  So the torque is half of 3.09 N-m

Haylee Thompson · Answer

I'm assuming acceleration due to gravity is 10 m/s^2

a) torque = r * F sin (theta)

For his arm, we assume the mass is evenly distributed, so we calculate the arms torque component based on the center of mass (which is the center of his arm .35 m from his shoulder)

torque = ((.35 (4.010)) + (2.5 (.7010)))(sin (90))

torque = ((.3540) + (2.57)) (1)

torque = 14 + 17.5

torque = 31.5 N*m

b) torque = r * f sin(theta)

The only thing changing here is theta, which is 30 degrees when his arm is 60 below horizontal

torque = 31.5 N*m (sin (30 degrees))

torque = 15.75 N*m

Anonymous · Answer

Center of mass = [4 x 0.35 + 2.5 x 0.7]/(4+2.5) = 0.48 OR 48 cm 
By τ = F X r
=>τ = F x r x sinθ
=>(a):- τ = F x r x sinθ
=>τ = (4+2.5)g x 0.48 x sin90*
=>τ = 6.5 x 9.8 x 0.48
=>τ = 30.97 N-m
=>(b) :-τ = 6.5 x 9.8 x 0.48 x sin30*
=>τ = 15.44 N-m

Prof. Mitchell Fay

Physics! An athlete at the gym holds a 2.5 steel ball in his hand. His arm is 70 long and has a mass of 4.0 .?

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