Chapter 4, Problem 37P

To determine

The magnitude of the acceleration of object for the given figures.

Answer to Problem 37P

The magnitude of the acceleration of object for case (a) is $4.24\text{m}/{\text{s}}^{\text{2}}$ and for case (b) is $8.39\text{m}/{\text{s}}^{\text{2}}$ .

Explanation of Solution

Given:

The given diagram is shown in Figure 1.

  

Figure 1

The acceleration when it acts on object on frictionless surface is, $\overrightarrow{a}=3.0\text{m}/{\text{s}}^2$

Formula used:

The expression for the force is given by,

  ${\overrightarrow{F}}_0=m\overrightarrow{a}$

Here, ${\overrightarrow{F}}_0$ is the required force.

The net force acting on object due to ${\overrightarrow{F}}_0$ and ${\overrightarrow{F}}_0$ which are $90°$ apart is given by,

  $\begin{array}{l}{\overrightarrow{F}}_{\text{net}}={\overrightarrow{F}}_0+{\overrightarrow{F}}_0\\ \left|{\overrightarrow{F}}_{\text{net}}\right|=\sqrt{{\left|{\overrightarrow{F}}_0\right|}^2+{\left|{\overrightarrow{F}}_0\right|}^2+2{\left|{\overrightarrow{F}}_0\right|}^2\cos \theta }\end{array}$

The net force acting on object due to ${\overrightarrow{F}}_0$ and $2{\overrightarrow{F}}_0$ which are $45°$ apart is given by,

  $\begin{array}{l}{\overrightarrow{F}}_{\text{net}}={\overrightarrow{F}}_0+2{\overrightarrow{F}}_0\\ \left|{\overrightarrow{F}}_{\text{net}}\right|=\sqrt{{\left|2{\overrightarrow{F}}_0\right|}^2+{\left|{\overrightarrow{F}}_0\right|}^2+2\times 2\times {\left|{\overrightarrow{F}}_0\right|}^2\cos \theta }\end{array}$

Calculation:

The value of the force is calculated as,

  $\begin{array}{c}{\overrightarrow{F}}_0=m\overrightarrow{a}\\ =m\left(3.0\text{m}/{\text{s}}^2\right)\\ =3m\text{N}\end{array}$

For the case (a),

The net force acting on object due to ${\overrightarrow{F}}_0$ and ${\overrightarrow{F}}_0$ which are $90°$ apart is calculated as,

  $\begin{array}{c}\left|{\overrightarrow{F}}_{\text{net}}\right|=\sqrt{{\left|{\overrightarrow{F}}_0\right|}^2+{\left|{\overrightarrow{F}}_0\right|}^2+2{\left|{\overrightarrow{F}}_0\right|}^2\cos \theta }\\ =\sqrt{{\left(3m\right)}^2+{\left(3m\right)}^2+2{\left(3m\right)}^2\cos 90°}\\ =\sqrt{18m^2}\\ =3\sqrt{2}m\text{N}\end{array}$

The value of the acceleration is calculated as,

  $\begin{array}{c}{\overrightarrow{F}}_{\text{net}}=m\left|{\overrightarrow{a}}_1\right|\\ 3\sqrt{2}m\text{N}=m\left|{\overrightarrow{a}}_1\right|\\ \left|{\overrightarrow{a}}_1\right|=3\sqrt{2}\text{m}/{\text{s}}^{\text{2}}\\ \left|{\overrightarrow{a}}_1\right|=4.24\text{m}/{\text{s}}^{\text{2}}\end{array}$

For the case (b),

The net force acting on object due to ${\overrightarrow{F}}_0$ and $2{\overrightarrow{F}}_0$ which are $45°$ apart is calculated as,

  $\begin{array}{c}\left|{\overrightarrow{F}}_{\text{net}}\right|=\sqrt{{\left|2{\overrightarrow{F}}_0\right|}^2+{\left|{\overrightarrow{F}}_0\right|}^2+2\times 2\times {\left|{\overrightarrow{F}}_0\right|}^2\cos \theta }\\ =\sqrt{{\left(\left(2\right)3m\right)}^2+{\left(3m\right)}^2+2\times 2\times {\left(3m\right)}^2\cos 45°}\\ =\sqrt{\left(36+9+\frac{36}{\sqrt{2}}\right)}m\text{N}\\ =8.39m\text{N}\end{array}$

The value of the acceleration is calculated as,

  $\begin{array}{c}{\overrightarrow{F}}_{\text{net}}=m\left|{\overrightarrow{a}}_2\right|\\ 8.39m\text{N}=m\left|{\overrightarrow{a}}_2\right|\\ \left|{\overrightarrow{a}}_2\right|=8.39\text{m}/{\text{s}}^{\text{2}}\end{array}$

Conclusion:

Therefore, the magnitude of the acceleration of object for case (a) is $4.24\text{m}/{\text{s}}^{\text{2}}$ and for case (b) is $8.39\text{m}/{\text{s}}^{\text{2}}$ .