Chapter 4, Problem 50QAP

To determine

(a)

The force $\stackrel{\rightharpoonup }{F_3}$ exerted by player $3$

Answer to Problem 50QAP

The force $\stackrel{\rightharpoonup }{F_3}$ exerted by player 3 in positive x direction is $\stackrel{\rightharpoonup }{F_{3_x}}=-209.73\text{ N}$ and The force $\stackrel{\rightharpoonup }{F_3}$ exerted by player 3 in in perpendicular to positive x direction $=$

  $\stackrel{\rightharpoonup }{F_{3_y}}=-33.76\text{N}$.

Explanation of Solution

Given:

  $\begin{array}{l}\stackrel{\rightharpoonup }{F_1}=100\text{ N}\\ \stackrel{\rightharpoonup }{F_2}=200\text{ N}\\ {\theta }_1=\text{ 60}{\text{.0}}^{\text{o}}\text{ with respect to +}x\\ {\theta }_2=\text{ 37}{\text{.0}}^{\text{o}}\text{ with respect to +}x\end{array}$

Formula used:

Newton's first law; In equilibrium ${\displaystyle \sum F_{ext}}=0$

Calculation:

Considering the equilibrium of forces in + x direction

Considering the equilibrium of forces in + y direction (perpendicular to x direction)

Conclusion:

Thus, component of force $\stackrel{\rightharpoonup }{F_3}$ in positive x direction $\stackrel{\rightharpoonup }{F_{3_x}}=-209.73\text{ N}$ and component of $\stackrel{\rightharpoonup }{F_3}$ force in perpendicular to positive x direction= $\stackrel{\rightharpoonup }{F_{3_y}}=-33.76\text{N}$

To determine

(b)

Redraw the diagram with force $\stackrel{\rightharpoonup }{F_3}$

Answer to Problem 50QAP

  ${\theta }_3={9.146}^{\text{o}}$

  $\stackrel{\rightharpoonup }{F_3}=212.43\text{ N}$

Explanation of Solution

Given:

  $\begin{array}{l}\stackrel{\rightharpoonup }{F_1}=100\text{ N}\\ \stackrel{\rightharpoonup }{F_2}=200\text{ N}\\ {\theta }_1=\text{ 60}{\text{.0}}^{\text{o}}\text{ with respect to +}x\\ {\theta }_2=\text{ 37}{\text{.0}}^{\text{o}}\text{ with respect to +}x\end{array}$

Formula used:

Newton's first law ; In equilibrium ${\displaystyle \sum F_{ext}}=0$

Calculation:

Considering the equilibrium of forces in x direction, $\stackrel{\rightharpoonup }{F_3}$ in positive x direction $\stackrel{\rightharpoonup }{F_{3_x}}=-209.73\text{ N}$

Considering the equilibrium of forces in y direction (perpendicular to x direction), $\stackrel{\rightharpoonup }{F_3}$ force in perpendicular to positive x direction= $\stackrel{\rightharpoonup }{F_{3_y}}=-33.76\text{N}$

  $\begin{array}{l}\stackrel{\rightharpoonup }{F_3}\cos \left({\theta }_3\right)=-209.73\text{N}\Rightarrow \left(p\right)\\ \stackrel{\rightharpoonup }{F_3}\sin \left({\theta }_3\right)=-33.76\text{N}\Rightarrow \left(q\right)\\ \raisebox{1ex}{$\left(q\right)$}\!\left/ \!\raisebox{-1ex}{$\left(p\right)$}\right.;\frac{\stackrel{\rightharpoonup }{\bcancel{F_3}}\sin \left({\theta }_3\right)}{\bcancel{\stackrel{\rightharpoonup }{F_3}}\cos \left({\theta }_3\right)}=\frac{-33.76\text{N}}{-209.73\text{N}}=0.161\\ \tan \left({\theta }_3\right)=0.161\\ {\theta }_3={\tan }^{-}\left(0.161\right)\\ \underset{\_}{\underset{\_}{{\theta }_3={9.146}^{\text{o}}}}\end{array}$

Conclusion:

Thus, the angle subtended by force $\stackrel{\rightharpoonup }{F_3}$ in positive direction of x is ${\theta }_3={9.146}^{\text{o}}$

To determine

(c)

Direction of acceleration of the box

Answer to Problem 50QAP

The direction of acceleration relative to positive direction of x is $\theta ={1.238}^{\text{o}}$.

Explanation of Solution

Given:

  $\begin{array}{l}\stackrel{\rightharpoonup }{F_1}=100\text{ N}\\ \stackrel{\rightharpoonup }{F{\text{'}}_2}=150\text{ N}\\ {\theta }_1=\text{ 60}{\text{.0}}^{\text{o}}\text{ with respect to +}x\\ {\theta }_2=\text{ 37}{\text{.0}}^{\text{o}}\text{ with respect to +}x\end{array}$

Formula used:

Newton's second law ; $F=ma$

Calculation:

Considering forces in + x direction;

$\begin{array}{l}{\displaystyle \sum {\stackrel{\rightharpoonup }{F}}_x}=ma\\ \stackrel{\rightharpoonup }{F_{1_x}}+\stackrel{\rightharpoonup }{F_{2_x}}=m{a}_x\\ \stackrel{\rightharpoonup }{F_1}\cos \left({\theta }_1\right)+\stackrel{\rightharpoonup }{F_2}\text{'}\cos \left({\theta }_2\right)=m{a}_x\\ 100\cos \left({60.0}^{\text{o}}\right)+150\cos \left({37.0}^{\text{o}}\right)=m{a}_x={\stackrel{\rightharpoonup }{F}}_{x\text{ total}}\\ {\stackrel{\rightharpoonup }{F}}_{x\text{ total}}=\left(50\text{N}\right)+\left(119.78\text{N}\right)\\ \underset{\_}{{\stackrel{\rightharpoonup }{F}}_{x\text{ total}}=169.78\text{N}}\\ \underset{\_}{\underset{\_}{{\stackrel{\rightharpoonup }{F}}_{\text{total}}\cos \left(\theta \right)=169.78\text{ N}}}\end{array}$

Considering the equilibrium of forces in + y direction (perpendicular to x direction);

Therefor the direction of the force $\theta$ ;

  $\begin{array}{l}{\stackrel{\rightharpoonup }{F}}_{total}\sin \left(\theta \right)=-3.67\text{N }\Rightarrow \left(r\right)\\ {\stackrel{\rightharpoonup }{F}}_{\text{total}}\cos \left(\theta \right)=-169.78\text{ N}\Rightarrow \left(s\right)\\ \raisebox{1ex}{$\left(r\right)$}\!\left/ \!\raisebox{-1ex}{$\left(s\right)$}\right.;\frac{{\stackrel{\rightharpoonup }{F}}_{total}\sin \left(\theta \right)}{{\stackrel{\rightharpoonup }{F}}_{\text{total}}\cos \left(\theta \right)}=\frac{3.67\text{N }}{169.78\text{ N}}=0.022\\ \tan \left(\theta \right)=0.022\\ \underset{\_}{\underset{\_}{\theta ={1.238}^{\text{o}}}}\end{array}$

Acceleration is in the direction of the force. Therefor the direction of acceleration relative to positive direction of x is $\theta ={1.238}^{\text{o}}$.

Conclusion:

Thus, the direction of acceleration relative to positive direction of x is $\theta ={1.238}^{\text{o}}$.

To determine

(d)

Mass of the box

Answer to Problem 50QAP

Mass of the box is $m=16.981\text{ kg}$.

Explanation of Solution

Given:

  $\begin{array}{l}\stackrel{\rightharpoonup }{F_1}=100\text{ N}\\ \stackrel{\rightharpoonup }{F{\text{'}}_2}=150\text{ N}\\ {\theta }_1=\text{ 60}{\text{.0}}^{\text{o}}\text{ with respect to +}x\\ {\theta }_2=\text{ 37}{\text{.0}}^{\text{o}}\text{ with respect to +}x\end{array}$

Acceleration of the box= 10 ms-2

Formula used:

Newton's second law ; $F=ma$

Calculation:

Conclusion:

Thus, mass of the box is $m=16.981\text{ kg}$.