Chapter 10, Problem 77A

(a)

To determine

To prove: TheIMA of the compound machine is 6.0.

Answer to Problem 77A

IMA of compound machine is 6.0.

Explanation of Solution

Given:

IMA of lever is 3.0

IMA of pulley system is 2.0

Formula used:

For a compound machine, $IM{A}_c=\frac{d_{e1}}{d_{r2}}$

Where, $d_{e1}$ =distance over which the effort is applied.

  $d_{r2}$ = distance that the load travels.

Calculation:

Due to conservation of energy, the input work of the machine, $W_{i1},W_{i2}$ , must be equal to the output work, $W_{o1},W_{o2}$ . So, mathematically,

  $\begin{array}{l}{W}_{i1}=W_{o1}=W_{i2}=W_{o2}\\ W_{i1}=W_{o2}\\ F_{e1}{d}_{e1}=F_{r2}{d}_{r2}\end{array}$

Now, the IMA for lever and pulley system is,

$IM{A}_1=\frac{d_{e1}}{d_{r1}}$ and $IM{A}_2=\frac{d_{e2}}{d_{r2}}$

Here $d_{e2}=d_{r1}$ , so the ratio will be

  $\begin{array}{c}\frac{d_{e1}}{d_{r2}}=IM{A}_c\\ IM{A}_c=IM{A}_1\times IM{A}_2\\ IM{A}_c=3.0\times 2.0=6.0\end{array}$

Conclusion:

Therefore, the IMA of compound machine is 6.0.

(b)

To determine

Effort force applied to the lever.

Answer to Problem 77A

The effort force, $F_e=150\text{ N}$ .

Explanation of Solution

Given:

Weight of box is, $F_r=540\text{ N}$

Efficiency, $e=60\%$

  $IM{A}_c=6.0$

Formula used:

The efficiency is given by: $e=\frac{F_r}{F_e\times IM{A}_c}\times 100$

Calculation:

Substituting all the values in the above formula,

  $\begin{array}{l}60=\frac{540}{F_e\times 6.0}\times 100\\ F_e=150\text{ N}\end{array}$

Conclusion:

The effort force, $F_e=150\text{ N}$ .

(c)

To determine

The distance travelled by box, $d_{r2}$

Answer to Problem 77A

The box is lifted at a distance of 2.0 cm.

Explanation of Solution

Given:

Change in the distance of effort side of lever, $d_{e1}=12.0\text{ cm}$

Formula used:

  $\frac{d_{e1}}{d_{r2}}=IM{A}_c$

Calculation:

Substituting all the values in the formula,

  $\begin{array}{l}\frac{12.0}{d_{r2}}=6.0\\ d_{r2}=2.0\text{ cm}\end{array}$

Conclusion:

Change in distance, $d_{r2}=2.0\text{ cm}$ .