Question

an athlete’s forearm holding a steel ball. The forearm is supported by a
contraction of the biceps muscle, which causes a torque around the elbow. Assuming that the tension
in the biceps acts along the vertical direction given by gravity and the centre of gravity of the forearm
is half of L. What tension must the biceps muscle exert to hold the forearm at the position shown?
Sketch and label your free body diagram. (Let L = 33.5 cm, d = 3.8 cm, mass of steel ball = 280 g, mass
of the forearm = 2.5 kg)

Answer 1

We see from the free-body diagram that the x-component of the net force satisfies the equation

$$+F_x+T_x-w_x=0$$

and the y-component of the net force satisfies

$$+F_y+T_y-w_y=0$$

Next, we read from the free-body diagram that the net torque along the axis of rotation is

$$r_TT_y-r_wW_y=0$$

$$F+T-w=0\\r_TT-r_wW=0$$

$$T=\frac{r_w}{r_T}w=\frac{33.5}{3.8}(0.28)(9.8)=24.2\ N$$