Question

A pulley system with $V.R=4$ is used to lift a load of $175kgf$through a vertical height of $15m$. The effort required is $50kgf$in the downward direction.$(g=10N/kg)$.

Calculate:

(I) Distance moved by the effort.

(ii) Work done by the effort.

(iii) M.A. of the pulley system.

(iv) Efficiency of the pulley system

Answer 1

Step 1: Defining levers: A lever is a straightforward machine comprising of a pillar or inflexible bar turned at a decent pivot or fulcrum. A lever is an unbending body equipped for turning a point on itself. Based on the areas of fulcrum, load, and effort, the lever is separated into three sorts.

Step 2: Finding the distance moved by the effort:

According to the problem,

We know that,

$V.R=\frac{Dis\tan cethrougheffortapplied}{Dis\tan cethroughwhichloadisraised}$

$$\begin{gathered} \Rightarrow 4=\frac{d}{15} \\ \Rightarrow d=60m \end{gathered}$$

Step 3: Finding the work done by the effort:

According to the problem,

We know that,
Effort in newton$=50\times 10=500N$

$$\begin{gathered} Workdonebyeffort=d\times EffortArm \\ =60m\times 500N \\ =30000J \end{gathered}$$

Step 4: Finding the M.A.:

According to the problem,

We know that,

$$\begin{gathered} M.A=\frac{Load}{Effort} \\ \Rightarrow M.A=\frac{175}{50} \\ \Rightarrow M.A=3.5 \end{gathered}$$

Step 5: Finding the efficiency$\left(\eta \right)$ :

According to the problem,

We know that,

$$\begin{gathered} Efficiency\left(\eta \right)=\frac{M.A.}{V.R}\times 100 \\ \Rightarrow \eta =\frac{3.5}{4}\times 100 \\ \Rightarrow \eta =87.5\% \end{gathered}$$