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A soccer player extends her lower leg in a kicking motion by exerting a force with the muscle above the knee in the front of her leg. she produces an angular acceleration of 37.5 rad/s2 and her lower leg has a moment of inertia of 0.750 kg · m2. what is the force exerted by the muscle, in newtons, if its effective perpendicular lever arm is 3.30 cm?

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Answer:

The force exerted by the muscle is 852.27N

Explanation:

Force F exerted by the muscle is expressed as a function of the torque τ and the effective perpendicular arm r.

F = τ/r ... (1)

Where τ = moment of inertia I × angular acceleration α

τ = Iα ... (2)

Substituting equation 2 into 1 to get F will give;

F = Iα/r

Given the following parameters

I = 0.75kgm²

α = 37.5rad/s²

r = 3.30cm = 0.033m

F = 0.75(37.5)/0.033

F = 28.125/0.033

F = 852.27N

The force exerted by the muscle is 852.27N

Answer:

Force exerted by muscle = 852.27N

Explanation:

We are given that;

lever arm; d = 3.3cm = 0.033m

Angular acceleration; α = 37.5 rad/s²

Moment of inertia of Leg; I = 0.75 kg.m²

Now, we know that torque is given by the relationship ;

τ = Iα

Where, I is moment of inertia and α is angular acceleration.

Thus,

τ = 0.75 x 37.5 = 28.125 N.m

Now, the relationship between torque and force is given by;

Torque = Force x perpendicular distance

We have torque = 28.125 N.m

We also have perpendicular distance as the lever arm = 0.033m

Thus,

28.125 N.m = Force x 0.033m

Thus,

Force = 28.125 N.m/0.033m

Force = 852.27N

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