Complete Question
An athlete at the gym holds a 3.0 kg steel ball in his hand. His arm is 70 cm long and has a mass of 4.0 kg. Assume, a bit unrealistically, that the athlete's arm is uniform.
What is the magnitude of the torque about his shoulder if he holds his arm straight out to his side, parallel to the floor? Include the torque due to the steel ball, as well as the torque due to the arm's weight.
Answer:
The torque is 
Explanation:
From the question we are told that
The mass of the steel ball is 
The length of arm is 
The mass of the arm is 
Given that the arm of the athlete is uniform them the distance from the shoulder to the center of gravity of the arm is mathematically represented as

=>
=>
Generally the magnitude of torque about the athlete shoulder is mathematically represented as

=> 
=> 
The maximum force of static friction is the product of normal force (P) and the coefficient of static friction (c). In a flat surface, normal force is equal to the weight (W) of the body.
P = W = mass x acceleration due to gravity
P = (0.3 kg) x (9.8 m/s²) = 2.94 kg m/s² = 2.94 N
Solving for the static friction force (F),
F = P x c
F = (2.94 N) x 0.6 = 1.794 N
Therefore, the maximum force of static friction is 1.794 N.
Answer:
30ms
Explanation:
you need to multiple the 10ms by 3s which gives you 30ms
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