perpendicularly to the forearm) and the force at the elbow. If the biceps produces a pull of 232 N when the forearm is raised $43^o$ above the horizontal, find the magnitude and direction of the force that the elbow exerts on the forearm. (The sum of the elbow force and the biceps force must balance the weight of the arm and the weight it is carrying, so their vector sum must be 132.5 N, upward.)
SOLUTION
Let the weight of the arm, $\vec{W_A}=20.5 \ N$, downward
weight of the object lifted, $\vec{W_8}=112.0 \ N$, downward
force-pull of biceps, $\vec{B}=232 \ N$, $43^o$ left of vertical
force-pull of elbow, $\vec{E}=$?
resultant $\vec{R}=\vec{E}+\vec{B}=\vec{W_A}+\vec{W_8}=132.5 \ N$. Figure 1 and 2 below shows a rough sketch of the situation. You can place the $\vec{E}$ arrow since you don't know where it will be pointing yet. |
| Figure 1 |
 |
| Figure 2 |
Since we don't know $\vec{E}$, then we will re-arrange our equation such that
$\vec{E}=\vec{R}-\vec{B}$.
The components of the given vectors where the angles used are measured from the +x-axis towards +y-axis are,
$R_x=(132.5 \ N) \ cos \ 90^o = 0$
$R_y=(132.5 \ N) \ sin \ 90^o = 132.5 \ N$
$B_x=(232 \ N) \ cos \ (90^o+43^o)=-158.22 \ N$
$B_y=(232 \ N) \ sin \ (90^o+43^o)=169.67 \ N$
then adding the x- and y-components separately,$E_x=R_x-B_x=0-(-158.22 \ N)=158.22 \ N$
$E_y=R_y-B_y=132.5 \ N - 169.67 \ N=-37.17 \ N$
We can now solve for the magnitude and direction,
$E=\sqrt{(E_x)^2+(E_y)^2}=160 \ N$ (final result is in 2-significant figures)
$tan \ \theta=\frac{E_y}{E_x}$
$\theta=-13^o$ measured clockwise from the +x-axis, or $13^o$ south of the horizontal.
