Q71P - University Physics with Modern Physics

Question

A physical therapy patient has a forearm that weighs 20.5 N and lifts a 112.0-N weight. These two forces are directed vertically downward. The only other significant forces on this forearm come from the biceps muscle (which acts perpendicular to the forearm) and the force at the elbow. If the biceps produce a pull of 232 N when the forearm is raised 43^° 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.)

Step-by-Step Solution

Answer

The magnitude of force that elbow exerts on the foreman is, E = 160 N.

The direction of force that elbow exerts on the foreman, $\theta ={13}^{°}$.

Step 1: Identification of given data

• The weight of the foreman is,${\mathrm{W}}_{\mathrm{F}}=20.5\mathrm{N}.$.
• The weight of the lift weight is, ${\mathrm{W}}_{\mathrm{w}}=112.0\mathrm{N}.$
• The force exerted by biceps that the elbow exerts on the foreman is, B = 232 N.

Step 2: Concept of force

When applied externally, force has the ability to alter a body's rest or motion state. It has a size and a direction.

Force is a term that refers to the exertion or application of strength or energy.

Step 3: Determine magnitude and direction of force exerted on elbow by foreman

The magnitude of force can be evaluated by,

$$\begin{gathered} \overline{E}+\overline{B}=\overline{R} \\ \overline{E}=\overline{R}-\overline{B} \\ \mathrm{The} \mathrm{component} \mathrm{of} \mathrm{force} \mathrm{that} \mathrm{the} \mathrm{elbow} \mathrm{exerts} \mathrm{is}, \\ {\mathrm{E}}_{\mathrm{x}}={\mathrm{R}}_{\mathrm{x}}-{\mathrm{B}}_{\mathrm{x}} \\ {\mathrm{E}}_{\mathrm{Y}}={\mathrm{R}}_{\mathrm{Y}}-{\mathrm{B}}_{\mathrm{Y}} \\ \mathrm{The} \mathrm{force} \mathrm{exerted} \mathrm{by} \mathrm{biceps} \mathrm{that} \mathrm{the} \mathrm{elbow} \mathrm{is}, \\ {\mathrm{B}}_{\mathrm{x}}=-\mathrm{Bsin}{43}^{°} \\ {\mathrm{B}}_{\mathrm{x}}=\left(232\mathrm{N}\right)\sin {43}^{°} \\ {\mathrm{B}}_{\mathrm{x}}=-158.2\mathrm{N} \\ {\mathrm{B}}_{\mathrm{y}}=\mathrm{Bcos}{43}^{°} \\ {\mathrm{B}}_{\mathrm{y}}=\left(232\mathrm{N}\right)\cos {43}^{°} \\ {\mathrm{B}}_{\mathrm{y}}=169.7\mathrm{N} \\ \mathrm{The} \mathrm{component} \mathrm{of} \mathrm{the} \mathrm{resultant} \mathrm{force} \mathrm{is}, \end{gathered}$$

$$\begin{gathered} {\mathrm{R}}_{\mathrm{x}}=0 \\ {\mathrm{R}}_{\mathrm{y}}=132.5\mathrm{N} \\ \mathrm{Substitute} \mathrm{the} \mathrm{above} \mathrm{value} \mathrm{in} \mathrm{force} \mathrm{exerted} \mathrm{on} \mathrm{elbow}, \\ {\mathrm{E}}_{\mathrm{x}}=0-(-158.2\mathrm{N}) \\ {\mathrm{E}}_{\mathrm{X}}=158.2\mathrm{N} \\ {\mathrm{E}}_{\mathrm{y}}=\left(132.5\mathrm{N}\right)-\left(169.7\mathrm{N}\right) \\ {\mathrm{E}}_{\mathrm{y}}=-37.2\mathrm{N} \\ \mathrm{The} \mathrm{resultant} \mathrm{magnitude} \mathrm{of} \mathrm{force} \mathrm{exerted} \mathrm{on} \mathrm{elbow} \mathrm{is}, \\ \mathrm{E}=\sqrt{{\mathrm{E}}_{\mathrm{x}}^2++{\mathrm{E}}_{\mathrm{y}}^2} \\ \mathrm{E}=\sqrt{{\left(158.2\mathrm{N}\right)}^2+{(-37.2\mathrm{N})}^2} \\ \mathrm{E}=160\mathrm{N} \end{gathered}$$

Thus, the magnitude of force that elbow exerts on the foreman is,160N.

The direction of force that elbow exerts on the foreman is,

$$\begin{gathered} \mathrm{tan\theta }=\left|\frac{{\mathrm{E}}_{\mathrm{y}}}{{\mathrm{E}}_{\mathrm{x}}}\right| \\ \mathrm{tan\theta }=\left|\frac{37.2\mathrm{N}}{158.2\mathrm{N}}\right| \\ \mathrm{tan\theta }=0.2351 \\ \mathrm{\theta }={\tan }^{-1}(0.2351) \\ \mathrm{\theta }={13}^{°} \end{gathered}$$

Thus, the direction of force that elbow exerts on the foreman is ${13}^{°}$,

The resultant upward force is less than the upward component, then the resultant force must be downwards.