Chapter 6, Problem 69P

To determine

The horsepower engine which is required if $30$ such skiers (max)are on the rope at one time.

Answer to Problem 69P

Solution:$\text{33 hp}$

Explanation of Solution

Given:

Weight of skier $\text{=65}\text{kg}$

Angle made by hill with horizontal $\text{=23}°$

Distance moved by skier $\text{=320}\text{m}$ (along the inclined plane)

Time take taken $\text{=2}\text{.0}\text{min}$

Co efficient of kinetic friction $\left(\text{μk}\right)=0.10$

Concept used:

Work $\left(W\right)=\overrightarrow{F}.\overrightarrow{S}$

  $\overrightarrow{F}$ : Force vector

  $\overrightarrow{S}$ : Displacement vector

Power, $P=\frac{W}{t}$

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

Calculation:

Work done by force of rope to move a skier from bottom to top of hill

  $={\overrightarrow{F}}_p.\overrightarrow{S}=F_{p}S \cos 0$

  ${\overrightarrow{F}}_p$ : Force by rope

  $\overrightarrow{S}$ : Displacement vector

  

  

N : normal force

F : friction force

Mg : weight of one skier

  $F_p$: pull by rope

Skier moves with constant speed, so net force acting on skier is zero.

For equilibrium perpendicular to plane of hill

  $N=mg\text{cos}\theta$

For equilibrium along the plane of hill,

  $\begin{array}{c}{F}_p=f+mg \sin \theta \\ =\mu N+mg\sin \theta \\ =\mu (mg)\cos \theta +mg\sin \theta \\ =mg(\mu \cos \theta +\sin \theta )\end{array}$

  $\begin{array}{l}{W}_p=mg(\cos \theta +\sin \theta )S\\ m=65\text{kg}\\ \theta =23°\\ \mu =0.10\\ g=9.80\\ S=320\text{m}\end{array}$

  $\begin{array}{l}{W}_p=65\times 9.8\times \left(0.10\cos 23°+\sin 23°\right)320\\ =98410.20\text{J}\end{array}$

Work done by pulling force to move one skier from bottom of hill to top of hill

  $=98410.20 \text{J}$

So, the work done by pulling force to move 30 skiers (maximum):

  $=98410.20\times 30 \text{J}$

Rope is moved by engine so work done by engine $=98410.20\times 30 \text{J}$

Power of engine $\text{=}\frac{\text{work}}{\text{time}}$

  $\begin{array}{l}\text{=}\frac{\text{98410}\text{.20×30 J}}{\text{2}\text{.0×60 s}}\\ \text{=24602}\text{.55}\text{watt}\end{array}$

Because $\text{746 watt}=\text{1 hp}$

  $\begin{array}{l}\text{P=}\frac{\text{24602}\text{.55 watt}}{\text{746 watt}}\times 1\text{hp}\\ =33 \text{hp}\end{array}$

Conclusion: Power of engine to work with maximum capacity to move 30 skiersshould be equal to $33\text{ hp}$ .