A system of pulleys is used to lift a load of 1000 kgwt.If the efficie

1.	A system of pulleys is used to lift a load of 1000 kgwt.If the efficiency of the pulley system is 80% andvelocity ratio is 20, find the effort required to lift theload. explain
Answer» Load ⇒ 1000kgwt Efficiency ⇒ 80% VELOCITY Ratio ⇒ 20 Effort ⇒ ? ⇒ $\sf \ Efficiency = \frac{<klux>MECHANICAL</klux> \ Advantage }{Velocity \ Ratio }$ ⇒ $\sf 0.8 = \frac{Mechanical \ Advantage }{20}$ ⇒ $\sf Mechanical \ Advantage = 0.8 \times 20$ ⇒ $\sf Mechanical \ Advantage = 16$ 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? $MA = \dfrac{F_B}{F_A}$