Chapter 10.2, Problem 27PP

a.

To determine

The MA of the pulley.

Answer to Problem 27PP

  $1.82$

Explanation of Solution

Given:

The mass of the carton, $m=24\text{}\text{kg}$

The height by which carton is raised, $d_r=16.5\text{m}$

The force exerted, $F_e=129\text{N}$

The distance by which the rope is pulled, $d_e=33.0\text{m}$

Formula used:

The mechanical advantage of a system is given by,

  $MA=\frac{F_r}{F_e}$

Where $F_r$ is the resistance force, and $F_e$ is the effort force.

Calculation:

The weight of the carton will act as the resistance force, thus $F_r=mg$ .

Then, the MA of the pulley will be,

  $\begin{array}{c}MA=\frac{F_r}{F_e}\\ =\frac{mg}{129}\\ =\frac{\left(24\right)\left(9.8\right)}{129}\\ =1.82\end{array}$

Conclusion:

Thus, the MA of the pulley is $1.82$ .

b.

To determine

The efficiency of the pulley.

Answer to Problem 27PP

  $91.16\%$

Explanation of Solution

Given:

The mass of the carton, $m=24\text{}\text{kg}$

The height by which carton is raised, $d_r=16.5\text{m}$

The force exerted, $F_e=129\text{N}$

The distance by which the rope is pulled, $d_e=33.0\text{m}$

Formula used:

The efficiency of a machine is the ratio of real and idea mechanical advantages,

  $e=\frac{MA}{IMA}\times 100=\frac{F_r{d}_r}{F_e{d}_e}\times 100$

Where $F_r$ is the resistance force, $F_e$ is the effort force, $d_e$ is the displacement by the effort force, and $d_r$ is the displacement of the load.

Calculation:

The weight of the carton will act as the resistance force, thus $F_r=mg$ .

Using the given values in above formula, the force to be exerted on the handle of the crank will be,

  $\begin{array}{c}e=\frac{F_r{d}_r}{F_e{d}_e}\times 100\\ =\frac{\left(24\right)\left(9.8\right)\left(16.5\right)}{\left(129\right)\left(33.0\right)}\times 100\\ =91.16\%\end{array}$

Conclusion:

Thus, the efficiency of the pulley is $91.16\%$ .