an athletic hold the a 75 kg shot up in his hand with his lower arm horizontal. his lower arm has mass of 2.8 kg and it center of gravity and is 12 cm from the elbow-joint pivot. how much force must be exerted

1. 100

2. 500
3. 1000
4.1500

To determine the force required to hold the shot, we need to consider the balance of torques acting on the athlete's arm.

First, we need to calculate the torque exerted by the weight of the shot:

Torque = Force × Distance

The distance in this case is the distance from the elbow-joint pivot to the center of gravity of the shot, which is given as 12 cm or 0.12 meters. The force is the weight of the shot, which we can calculate using the formula:

Weight = mass × acceleration due to gravity

Weight = 75 kg × 9.8 m/s² (approximating the acceleration due to gravity as 9.8 m/s²)

Weight = 735 N

Now, let's calculate the torque exerted by the weight of the lower arm:

Torque = Force × Distance

The distance is again given as 12 cm or 0.12 meters, and the force is the weight of the lower arm. We can calculate this using the formula:

Weight = mass × acceleration due to gravity

Weight = 2.8 kg × 9.8 m/s²

Weight = 27.44 N

Now that we have the torques exerted by both the shot and the lower arm, we can set up an equation to find the force required to maintain balance:

Torque due to the shot = Torque due to the lower arm

(Force required to hold the shot) × (Distance from elbow-joint pivot to center of gravity of the shot) = (Weight of the lower arm) × (Distance from elbow-joint pivot to center of gravity of the lower arm)

Force required = (Weight of the lower arm) × (Distance from elbow-joint pivot to center of gravity of the lower arm) / (Distance from elbow-joint pivot to center of gravity of the shot)

Force required = (27.44 N) × (0.12 m) / (0.12 m)

Force required ≈ 27.44 N

Therefore, approximately 27.44 Newtons of force must be exerted to hold the shot in the athlete's hand.