Chapter 10, Problem 74A

To determine

The constant rate at which the crane should lift a $410\text{kg}$ crate.

Answer to Problem 74A

  $0.506\text{m/s}$

Explanation of Solution

Given:

The efficiency of the motor, $e_m=88\%$

The efficiency of the crane, $e_c=42\%$

The power supplied to the motor, $P_m=5.5\text{kW}$

The mass of the crate, $m=410\text{kg}$

Formula used:

Power is given by the work done divided by the taken to do the work,

  $P=\frac{W}{t}$

Where $W$ is the work done, and $t$ is the time taken.

Also, work done is given by,

  $W=Fd\cos \theta$

Where $F$ is the force done, $d$ is the distance, and $\theta$ is the angle between the distance and force.

Calculation:

The total efficiency of the system will be,

  $\begin{array}{c}e=e_m\times e_c\\ =\left(88\%\right)\times \left(42\%\right)\\ =36.96\%\end{array}$

Then, the useful power will be,

  $\begin{array}{c}P=36.96\%\left(P_m\right)\\ =36.96\%\left(5.5\text{kW}\right)\\ =2032.8\text{W}\end{array}$

Now, using the known formula of power,

  $\begin{array}{c}P=\frac{W}{t}\\ =\frac{Fd\cos 0°}{t}\\ =F\left(\frac{d}{t}\right)\\ =Fv\end{array}$

Where $F$ will be the weight of the crate and the $v$ be the velocity with which it should be lifted. Then,

  $\begin{array}{l}P=mgv\\ 2032.8=410\times 9.8\left(v\right)\\ v=0.506\text{m/s}\end{array}$

Conclusion:

Thus, the constant rate at which the crane should lift a $410\text{kg}$ crate is $0.506\text{m/s}$ .