Question

A person 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.

Answer 1

here we will use the torque balance about the knee joint

here we can say that

$\tau_g = \tau_m$

here torque due to weight is given as

$\tau_g = 40.1 cos\theta*(l_1 + l_2)$

$\tau_g = 40.1 cos27.5*(0.105 + 0.150)$

$\tau_g = 9.07 Nm$

now torque due to applied force of muscle

$\tau_m = M*sin\alpha * l_1$

$\tau_m = M*sin30* 0.105$

now by torque balance we will have

$9.07 = M*0.5*0.105$

$M = 173 N$

so here the magnitude of m will be 173 N