Chapter 6, Problem 70P

To determine

Minimum horse power must a motor have to be able to drag a $\text{370}\text{kg}$ box along a level floor.

Answer to Problem 70P

Solution:$\text{2}\text{.62 hp}$

Explanation of Solution

Given:

Mass of box $=370\text{kg}$

Speed of box $=1.20\text{m/s}$

  $\mu =0.45$

Concept used:

As per Newton’s second law of motion:

  ${\displaystyle \sum F=ma}$

Where, m is the mass and a is the acceleration.

Power $P=\overrightarrow{F}.\overrightarrow{V}$

  $=FV \cos \theta$

F : force

v : velocity

  $\theta$: Angle between force and velocity.

P : power supply by force F

Calculation:

Consider free body diagram:

  

  $F_p$: pulling force

f : friction force

N : normal force

Mg : weight of box

Apply Newton’s first law $\left(\text{a}=0\right)$

  $\text{N=mg}$

  $\begin{array}{c}{F}_p=f\\ =\mu N\\ =\mu mg\end{array}$

  $P=FV \cos \theta$

Here $\theta$

  $=0$

  $\begin{array}{c}P=\left(\mu mg\right)\left(V\right)\cos 0\\ =\mu mg V\end{array}$

Putting the value

  $\mu =0.45$

  $m=3\text{70}\text{kg}$

  $\text{V=1}\text{.20}\text{m/s}$

  $\text{g = 9}\text{.80}{\text{m/s}}^{\text{2}}$

Power supply by pulling force

  $\begin{array}{c}P=\mu mg V\\ =0.45\times 370\times 1.20\times 9.80\\ \text{=1958}\text{.04 watt}\\ \because \text{746 W = 1hp}\\ \text{=1958}\text{.04}\text{watt×}\frac{\text{1 hp}}{\text{746 watt}}\\ =2.62 \text{hp}\end{array}$

Conclusion: Hence, the power that motor must be $\text{2}\text{.62 hp}$ .