91Ó°ÊÓ

• 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.

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.

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,

$$\begin{gathered} \mathrm{³Ù²¹²Ôθ}=\left|\frac{{\mathrm{E}}_{\mathrm{y}}}{{\mathrm{E}}_{\mathrm{x}}}\right| \\ \mathrm{³Ù²¹²Ôθ}=\left|\frac{37.2\mathrm{N}}{158.2\mathrm{N}}\right| \\ \mathrm{³Ù²¹²Ôθ}=0.2351 \\ \mathrm{θ}={\tan }^{-1}(0.2351) \\ \mathrm{θ}={13}^{°} \end{gathered}$$

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.