What is the effort required to lift a load of $500 N$ by using a lever with a mechanical advantage of $1.6$ ?

Open in App

Solution

Given,

Mechanical advantage $= \frac{o u t p u t f o r c e}{i n p u t f o r c e} = 1.6$

$i n p u t l o a d = \frac{O u t p u t l o a d}{1.6} = \frac{500}{1.6} = 312.5 N$

Effort of $312.5 N$ is required

Suggest Corrections

Similar questions

A machine has a mechanical advantage of $10$ . It used to lift a load of $200 k g f$ . Find the required effort.

In fig below shows the use of a lever.

(a) State the principle of moments as applied to the above lever.

(b) To which class of lever does it belong? Give an example of this class of lever.

(c) If FA = 10 cm , AB= 490 cm, calculate : (i) the mechanical advantage, and (ii) the minimum effort required to lift the load (= 50 N).

Join BYJU'S Learning Program