Question

In a lifting machine an effort of \( 150 N \) raised a load of \( 7700 N \). What is the mechanical advantage? Find the velocity ratio if the efficiency at this load is \( 60 \% \). If by the same machine, a load of \( 13200 N \) is raised by an effort of \( 250 N \). what is the efficiency? Calculate the maximum mechanical advantage and the maximum efficiency.

Answer 1

Effort, P = 150 N

Load, W = 7700 N

∴ Mechanical advantage:

\(MA = \frac WP = \frac{7700}{150} = 51.33\)

If the efficiency is 60%

\(\eta = 0.6\)

\(\eta = \frac {MA}{VR}\)

\(\therefore 0.6 = \frac{51.33}{VR}\)

or \(VR = \frac{51.33}{0.6}\) 

i.e, VR = 85.55

Let the law of machine be  P = mW + C

In the first case, 150 = 7700 m + C     ...(1)

In the second case, 250 = 13,200 m + C     ...(2)

Subtracting eqn. (1) from eqn. (2), we get

100 = 5500 m

m = 0.01818

∴ Maximum mechanical advantage

\(= \frac 1m\)

\(= \frac 1{0.01818}\)

\( = 55\)

Maximum efficiency \(= \frac 1m \times \frac 1{VR}\) 

\(= \frac 1{0.01818} \times\frac1{85.55}\)

\(= 0.6429\)

\(= 64.29\%\)