Question

A system of pulleys is used to lift a load of 1000 kg wt. If the velocity ratio efficiencies and the efficiency are $20$and $90\%$ respectively, then the effort required to lift the load is ____ N

• A. $5.6\times {10}^7$
• B. $5.6\times {10}^{-3}$
• C. $5.6\times {10}^5$
• D. $560$

Answer 1

The correct option is D

$560$

Step 1: Given data

Load=$1000kgwt$

Efficiency =$90\%$

Velocity ratio = 20

Effort=?

Step 2: Calculations

Efficiency is given by the equation

$$\begin{gathered} Efficiency=\frac{Mechanicaladvantage}{velocityratio} \\ 0.9=\frac{Mechanicaladvantage}{20} \\ Mechanicaladvantage=0.9\times 20 \\ Mechanicaladvantage=18 \end{gathered}$$

Where mechanical advantage is defined as the ratio of load to the effort.

$$\begin{gathered} Mechanicaladvantage=\frac{load}{effort} \\ So, \\ Effort=\frac{Load}{mechanicaladvantage} \\ Effort=\frac{1000kgwt}{18} \\ Effort=55.55kgwt \end{gathered}$$

Converting to Newton,

$55.5Kgwt=55.5\times 9.8=544N$

Which is nearly equal to $560$ N

Hence, option(d) is correct