Chapter 4, Problem 62QAP

To determine

(a)

The magnitude of ${\overrightarrow{F}}_2$

Answer to Problem 62QAP

The magnitude of ${\overrightarrow{F}}_2=\left|{\overrightarrow{F}}_2\right|=10.7\text{N}$

Explanation of Solution

Given info:

  

Mass, $M=3.00\text{kg}$

Acceleration, $a=7.00{\text{m/s}}^{\text{2}}$ in $+x$ direction

And ${\theta }_1={30.0}^{\circ }$

Formula Used:

  $\sum F_{\text{ext,x}}=F_{1x}+F_{2x}$

  $\sum F_{\text{ext,y}}=F_{1y}+F_{2y}$

  $F_2=\sqrt{F_{2x}{}^2+F_{2y}{}^2}$

  $F=Ma$

Calculation:

Consider as two forces are attached to an object of mass M.

Let rope $1$ applies a force with magnitude $F_1$ at an angle of ${30.0}^{\circ }$.

We need to determine the magnitude and direction of rope $2$ 's force, ${\overrightarrow{F}}_2$.

We are told that the box accelerates only in the x-direction at a rate of $7.00{\text{m/s}}^{\text{2}}$, which means the acceleration in the y-direction is equal to $0$.

We can use Newton's second law in component form in order to calculate the components and, therefore, the magnitude and direction of force $2$.

  $\begin{array}{l}{\displaystyle \sum F_{\text{ext,x}}=F_{1x}+F_{2x}\text{=}}{F}_1\cos ({30.0}^{\circ })+F_{2x}=M(7.00{\text{m/s}}^{\text{2}})\\ =>F_{2x}=M(7.00{\text{m/s}}^{\text{2}})-F_1\cos ({30.0}^{\circ })\end{array}$

  $\begin{array}{l}{\displaystyle \sum F_{\text{ext,y}}=F_{1y}+F_{2y}\text{=}}{F}_1\sin ({30.0}^{\circ })+F_{2y}=0\\ =>F_{2y}=-F_1\sin ({30.0}^{\circ })\end{array}$

  $\begin{array}{l}{F}_2=\sqrt{F_{2x}{}^2+F_{2y}{}^2}=\sqrt{{(M(7.00{\text{m/s}}^{\text{2}})-F_1\cos ({30.0}^{\circ }))}^2+{(-F_1\sin ({30.0}^{\circ }))}^2}\\ =\sqrt{M^2(49.0{\text{m}}^2{\text{/s}}^4)-(14.0{\text{m/s}}^{\text{2}})M{F}_1\cos ({30.0}^{\circ })+F_1{}^2{\cos }^2({30.0}^{\circ })+F_1{}^2{\sin }^2({30.0}^{\circ })}\\ =\sqrt{M^2(49.0{\text{m}}^2{\text{/s}}^4)-(14.0{\text{m/s}}^{\text{2}})M{F}_1\cos ({30.0}^{\circ })+F_1{}^2}\end{array}$

  $\begin{array}{l}{F}_2=\sqrt{{(3.00\text{kg)}}^2(49.0{\text{m}}^2{\text{/s}}^4)-(14.0{\text{m/s}}^{\text{2}})(3.00\text{kg)}(20.0\text{N})\cos ({30.0}^{\circ })+{(20.0\text{N})}^2}\\ F_2=10.7\text{N}\end{array}$

Conclusion:

The magnitude of ${\overrightarrow{F}}_2$ is $10.7\text{N}$

To determine

(b)

A careful drawing to show direction of $F_2$.

Answer to Problem 62QAP

The below diagram shows the direction of $F_2$.

  

Explanation of Solution

Given info:

  

Mass, $M=3.00\text{kg}$

Acceleration, $a=7.00{\text{m/s}}^{\text{2}}$ in $+x$ direction

And ${\theta }_1={30.0}^{\circ }$

Formula Used:

  $\sum F_{\text{ext,x}}=F_{1x}+F_{2x}$

  $\sum F_{\text{ext,y}}=F_{1y}+F_{2y}$

  $F_2=\sqrt{F_{2x}{}^2+F_{2y}{}^2}$

  $F=Ma$

  ${\theta }_2={\tan }^{-1}\left(\frac{F_{2y}}{F_{2x}}\right)$

Calculation:

Consider as two forces are attached to an object of mass M.

Let rope $1$ applies a force with magnitude $F_1$ at an angle of ${30.0}^{\circ }$.

We need to determine the magnitude and direction of rope $2$ 's force, ${\overrightarrow{F}}_2$.

We are told that the box accelerates only in the x-direction at a rate of $7.00{\text{m/s}}^{\text{2}}$, which means the acceleration in the y-direction is equal to $0$.

We can use Newton's second law in component form in order to calculate the components and, therefore, the magnitude and direction of force $2$.

  $\begin{array}{l}{\displaystyle \sum F_{\text{ext,x}}=F_{1x}+F_{2x}\text{=}}{F}_1\cos ({30.0}^{\circ })+F_{2x}=M(7.00{\text{m/s}}^{\text{2}})\\ =>F_{2x}=M(7.00{\text{m/s}}^{\text{2}})-F_1\cos ({30.0}^{\circ })\end{array}$

  $\begin{array}{l}{\displaystyle \sum F_{\text{ext,y}}=F_{1y}+F_{2y}\text{=}}{F}_1\sin ({30.0}^{\circ })+F_{2y}=0\\ =>F_{2y}=-F_1\sin ({30.0}^{\circ })\end{array}$

On substituting given values we have: -

  $\begin{array}{l}{F}_{2x}=(M(7.00{\text{m/s}}^{\text{2}})-F_1\cos ({30.0}^{\circ }))=((3.00\text{kg)}(7.00{\text{m/s}}^{\text{2}})-(20.0\text{N})\cos ({30.0}^{\circ }))\\ F_{2x}=3.68\text{N}\\ \\ F_{2y}=-F_1\sin ({30.0}^{\circ })=-(20.0\text{N})\sin ({30.0}^{\circ })\\ F_{2y}=-(10.0\text{N})\\ \\ {\theta }_2={\tan }^{-1}\left(\frac{-10.0\text{N}}{3.68\text{N}}\right)=-{69.8}^{\circ }={290.2}^{\circ }\end{array}$

290.2⁰

Conclusion:

Thus, we obtained the direction of $F_2$ as below: -