José is pinned against the walls of the Rotor, a ride with a radius of 3.00 meters that spins so fast that the floor can be remo
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Answer:
The answers to the questions are;
The force the biceps must exert to hold a 850 g ball at the end of the forearm at distance dball = 34.0 cm from the elbow, with the forearm parallel to the floor is 166.77 N Upwards.
The force the elbow must exert is 145.6785 N downwards.
Explanation:
We are required to analyze the system using the principle of moments as follows.
Mass of forearm = 1.30 kg
Weight of forearm = mass × acceleration due to gravity = 1.30 kg × 9.81 m/s²= 12.753 N
Location of point of action of weight of forearm = midpoint 17 cm
Forearm to biceps distance = 3.00 cm from the elbow.
Mass of the ball = 850 g
Weight of ball = mass × acceleration due to gravity = 0.850 kg × 9.81 = 8.3385 N
Position of the ball = end of the forearm
Therefore, taking moment about the elbow, we have
∑Mₓ = 0, we take clockwise moment to be positive, therefore
8.3385 N× 34.0 cm + 12.753 N× 17 cm - ×3.00 cm = 0
Therefore ×3.00 cm = 500.31 N cm
Therefore = (500.31 N·cm)/(3.00 cm) = 166.77 N Upwards
The force exerted by the elbow is given by,
Taking moment about the point of contact between the biceps and the forearm, we get
× 3.00 cm = 8.3385 N× 31.0 cm + 12.753 N× 14 cm
= 145.6785 N downwards.