Chapter 4, Problem 51QAP

To determine

The magnitude of forces $F_2$ and $F_3$

Answer to Problem 51QAP

The magnitude of forces are the magnitudes of forces are $F_2=1.388\text{N}$ and $F_3=1.638\text{N}$

Explanation of Solution

Given:

  $\begin{array}{l}{\theta }_1=\text{ 40}{\text{.0}}^{\text{o}}\text{ with respect to +}x\\ {\theta }_2=\text{ 60}{\text{.0}}^{\text{o}}\text{ with respect to +}x\\ {\theta }_3={20.0}^{\text{o}}\text{with respect to +}x\\ F_1=1.00\text{ N}\end{array}$

Acceleration =1.50 ms-2

Formula used:

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

Newton's second law $F=ma$

Calculation:

In positive $x$direction, from Newton's second law $F=ma$

  $\begin{array}{l}{F}_x=m{a}_x\\ {\stackrel{\rightharpoonup }{F}}_{1_x}+{\stackrel{\rightharpoonup }{F}}_{2_x}+{\stackrel{\rightharpoonup }{F}}_{3_x}=m{a}_x\\ F_1\cos \left({40}^o\right)+F_{2_x}+F_{3_x}=2\text{ kg}\times \text{1}{\text{.50 ms}}^{-2}\\ F_2\cos \left({60}^o\right)+F_3\cos \left({20}^{\text{o}}\right)=\left(3-0.766\right)\\ F_2\times 0.5+F_3\times 0.94=2.234\Rightarrow \left(p\right)\end{array}$

In positive y (perpendicular to $x$ direction) as the acceleration is zero, considering equilibrium in y direction.

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

  $\begin{array}{l}{\displaystyle \sum F_{ext}}=0\\ {\stackrel{\rightharpoonup }{F}}_1{}_y+{\stackrel{\rightharpoonup }{F}}_{2_y}+{\stackrel{\rightharpoonup }{F}}_{3_y}=0\\ F_1\sin \left({40}^{\text{o}}\right)+F_3\sin \left({20}^{\text{o}}\right)=F_2\sin \left({60}^{\text{o}}\right)\\ 1\text{ N}\times \text{0}\text{.642}=F_2\times 0.866-F_3\times 0.342\Rightarrow \left(q\right)\end{array}$

  $\begin{array}{l}{F}_2\times 0.5+F_3\times 0.94=2.234\text{N}\Rightarrow \left(p\right)\\ \text{0}\text{.642 N}=F_2\times 0.866-F_3\times 0.342\Rightarrow \left(q\right)\\ \left(p\right)\times 0.866-\left(q\right)\times 0.5;\\ 2.234\text{N}\times 0.866-\text{0}\text{.642 N}\times 0.5=\left(F_2\times 0.5+F_3\times 0.94\right)\times 0.866-\left(F_2\times 0.866-F_3\times 0.342\right)\times 0.5\\ 1.934-0.321=F_3\times 0.814+F_3\times 0.176\\ \underset{\_}{\underset{\_}{F_3=1.638\text{N}}}\\ \\ \left(p\right)\times 0.342+\left(q\right)\times 0.94;\\ 2.234\text{N}\times 0.342-\text{0}\text{.642 N}\times 0.94=\left(F_2\times 0.5+F_3\times 0.94\right)\times 0.342-\left(F_2\times 0.866-F_3\times 0.342\right)\times 0.94\\ 0.764+0.603=F_2\times 0.176+F_2\times 0.814\\ \underset{\_}{\underset{\_}{F_2=1.388\text{N}}}\end{array}$

Conclusion:

Thus, the magnitudes of forces are $F_2=1.388\text{N}$ and $F_3=1.638\text{N}$