Question

A system of pulleys is used to lift a load of 1000 kgwt.
If the efficiency of the pulley system is 80% and
velocity ratio is 20, find the effort required to lift the
load. explain​

Answer 1

Load ⇒ 1000kgwt

Efficiency ⇒ 80%

Velocity Ratio ⇒ 20

Effort ⇒ ?

⇒  $\sf \ Efficiency = \frac{Mechanical \ Advantage }{Velocity \ Ratio }$

⇒ $\sf 0.8 = \frac{Mechanical \ Advantage }{20}$

⇒ $\sf Mechanical \ Advantage = 0.8 \times 20$

⇒ $\sf Mechanical \ Advantage = 16$

So now the question comes up that what is Mechanical Advantage?

⇒ $\sf Mechanical \ Advantage = \frac{Load}{Effort}$

⇒ $\sf Effort = \frac{Load}{Mechanical \ Advantage }$

⇒ $\sf Effort = \frac{1000kgwt}{16}$

⇒ $\sf Effort = \frac{125}{2}$

⇒ $\sf Effort = 62.5kgwt$

∴The effort required to lift the load would be 62.5kgwt.

Additional Information :)

What is velocity?

The rate of change of an object’s position with respect to a frame of reference and time is known as velocity.

What is the S.I unit of velocity?

S.I unit of velocity is m/s

What is mechanical advantage?

A calculation that measures the amplified force used up by a mechanical system is known as mechanical advantage.

What is the formula for mechanical advantage?

$MA = \dfrac{F_B}{F_A}$