Aperson is sitting with one leg outstretched and stationary, so that it makes an angle of θ = 27.5° with the horizontal, as the drawing indicates. the weight of the leg below the knee is 40.1 n, with the center of gravity located below the knee joint. the leg is being held in this position because of the force m applied by the quadriceps muscle, which is attached ℓ1 = 0.105 m below the knee joint (see the drawing). obtain the magnitude of m. (assume the angle α = 30.0° and the distance ℓ2 = 0.150 m.

The rotational balance condition allows finding the result for the force applied by the muscle to keep the leg in balance is:

      F = 30.5 N

Given parameters

To find

Newton's second law for rotational motion gives a relationship between torque, moment of inertia, and angular acceleration. In the special case that the acceleration is zero, it is called the rotational equilibrium condition.

          Σ τ = 0

          τ = F x r

Where the bold letters indicate vectors, τ is the torque, F the force and r the displacement vector.

The module of this expression is:

         τ = F r sin θ

In the attachment we have a free body diagram where the force and its distance from the turning point are indicated, we fear the counterclockwise as positive.

We write the  rotational balance condition, with respect to the knee as the turning point.

Let's look for every torques.

we substitute

       F l₁ cos 30 = W l₂ sin 27.5

       F =

Let's calculate

        F =    

        F = 30.5 N

In conclusion, using the rotational equilibrium condition we can find the result for the force applied by the muscle to keep the leg in balance is:

      F = 30.5 N

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here we will use the torque balance about the knee joint

here we can say that

here torque due to weight is given as

now torque due to applied force of muscle

now by torque balance we will have

so here the magnitude of m will be 173 N

answer: a) y=-8x+5

a) m= -8

b) m= -2

c) m= 7

d) m= 6