A student in the Biomechanics class has decided that she would like to make her arms
stronger. She has a mass of 63 kg, She chooses to complete some elbow flexion exercises
using a kettlebell. For this problem, consider the hand and forearm to be a single segment.
The distance from her elbow to her wrist is 22.86 cm.
The force from the kettlebell is applied to her hand, which is 30.48 cm from her elbow joint.
She knows that the moment arm of the elbow extensor muscles about the elbow axis is

The moment of force about a point is the product of applied force and the perpendicular distance of the force action. The moment of force about a point is equal to the torque exerted by the force.

τ = Fr

where;

F is the resistance force of the kettle
r is perpendicular distance

F₁r₁ = F₂r₂

925(22.86) = F₂(30.38)

696 N = F₂

in kg, m = 696/9.8 = 71 kg

Thus, the resistive force from the applied force and distance is 71 kg.

The complete question is below:

If she wants the elbow flexors to exert 925 N of force during an isometric contraction, what should the resistance force be at 90 deg of flexion (the upper arm is vertical and the forearm is parallel to the ground) in kg

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