Chapter 10, Problem 68A

To determine

The distance by which the crate is lifted if the efficiency of the lever is $90.0$ percent.

Answer to Problem 68A

  $0.245\text{m}$

Explanation of Solution

Given:

The force exerted on a lever, $F_e=250\text{N}$

The mass of the crate, $m_r=150\text{kg}$

The distance upto which the force is exerted by the student, $d_e=\text{1}\text{.6 m}$

The efficiency, $e=90\%$

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, $d_r$ is the displacement of the load, and $d_e$ is the displacement of the effort force.

Calculation:

The weight of the crate is,

  $\begin{array}{c}{F}_r=m_{r}g\\ =\left(150\right)\left(9.8\right)\\ =1470\text{N}\end{array}$

Using the given values in above formula, the distance by which the crate is lifted ( $d_r$ ) will be,

  $\begin{array}{c}90=\frac{\left(1470\right)d_r}{\left(250\right)\left(1.6\right)}\times 100\\ d_r=0.245\text{m}\end{array}$

Conclusion:

Thus, the distance by which the crate is lifted is $0.245\text{m}$ .