In Fig. 12-44, a 15 kg block is held in place via a pulley system. The person’s upper arm is vertical; the forearm is at angle$\mathit{\theta }\mathbf{=}{\mathbf{30}}^{\mathbf{o}}$ with the horizontal. Forearm and hand together have a mass of 2.0 kg, with a center of mass at distance ${\mathit{d}}_{\mathbf{1}}\mathbf{=}\mathbf{15}\mathbf{\text{\hspace{0.17em}cm}}$from the contact point of the forearm bone and the upper-arm bone(humerus). The triceps muscle pulls vertically upward on the forearm at distance${\mathit{d}}_{\mathbf{2}}\mathbf{=}\mathbf{2}\mathbf{.5}\mathbf{\text{\hspace{0.17em}cm}}$ behind that contact point. Distance${\mathit{d}}_{\mathbf{3}}$is 35 cm. What are the (a)magnitude and (b) direction (up or down) of the force on the forearm from the triceps muscle and the (c) magnitude and (d) direction (up or down) of the force on the forearm from the humerus?

i) Mass of block, M= 15 kg.

ii) The forearm makes an angle with the horizontal, $\theta ={30}^{\circ }$

iii) Mass of forearm and hand together, m = 2kg.

iv) Distance of center of mass of forearm and hand together from the contact point of forearm bone and upper bone,$d_1=15\text{\hspace{0.17em}cm}$ .

v) Distance of force by triceps muscles from the contact point of forearm bone and upper bone, $d_2=2.5\text{\hspace{0.17em}cm}$.

vi) The distance between fingers holds the rope and the contact point, $d_3=35\text{\hspace{0.17em}cm}$.

You can use the concept of static equilibrium and torque to find the force from the triceps on the forearm. Also, you can use the net force along the x and y directions for equilibrium to find the forces of the triceps and humerus.From the resulting answer, you can determine the direction of the forces.

The given forces are all vertical, and the distances are measured along the x-axis at theangle$\theta ={30}^{\circ }$.

Here, if we use trigonometry, the trigonometric factors cancel out. So, we can write the torque about the contact point and solve it for the force.

The net torque is calculated as,

$-F_{triceps}{d}_2-2\left(9.8\right)d_1+15\left(9.8\right)d_2=0$

Rearranging for$F_{triceps}$we get,

$F_{triceps}=\frac{15\left(9.8\right)d_2-2\left(9.8\right)d_1}{d_2}$

$\begin{array}{c}{F}_{triceps}=\frac{15\text{\hspace{0.17em}m}\times \left(9.8{\text{\hspace{0.17em}m/s}}^{\text{2}}\right)\left(0.35\text{\hspace{0.17em}kg}\right)-2\text{\hspace{0.17em}m}\times \left(9.8{\text{\hspace{0.17em}m/s}}^{\text{2}}\right)\left(0.15\text{\hspace{0.17em}kg}\right)}{0.025\text{\hspace{0.17em}m}}\\ =1.94\times {10}^3\cdot (1\text{\hspace{0.17em}kg}\cdot {\text{m/s}}^{\text{2}}\times \frac{1\text{\hspace{0.17em}N}}{1\text{\hspace{0.17em}kg}\cdot {\text{m/s}}^{\text{2}}})\\ =1.94\times {10}^3\text{\hspace{0.17em}N}\end{array}$

From part (a), we got a positive value. So, the direction is upwards.

Thus, the force on the triceps muscle will be in the upward direction.

For equilibrium, the vertical forces are balanced. We can thus write the equation as:

$\begin{array}{c}{F}_{triceps}+F_{humerus}+15\left(9.8\right)-2\left(9.8\right)=0\\ 1.94\times {10}^3+F_{humerus}=19.6-147\\ F_{humerus}=-127.4-1.94\times {10}^3\\ =-2067.4\text{\hspace{0.17em}N}\\ =-2.1\times {10}^3\text{\hspace{0.17em}N}\end{array}$

Thus, the magnitude is,$F_{humerus}=2.1\times {10}^3\text{\hspace{0.17em}N}$.

The sign of force of the humerus is negative, so the force points downward.

Thus, the force on the humerus will be in a downward direction.