Physics

A lever of length 9 cm has its load arm 5 cm long and the effort arm is 9 cm long.
(a) To which class does it belong?
(b) Draw diagram of the lever showing the position of fulcrum F and directions of both the load L and effort E.
(c) What is the mechanical advantage and velocity ratio if the efficiency is 100%?
(d) What will be the mechanical advantage and velocity ratio if the efficiency becomes 50%?

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(a) It belongs to Class II lever as length of the lever is equal to the effort arm and the effort arm is also more than the load arm.

(b) Below is the diagram of the lever with position of fulcrum F and directions of both the load L and effort E labelled:

$\text{Mechanical advantage M.A.} = \dfrac{\text{Effort arm}}{\text{Load arm}} \\[0.5em]$

Given,

Load arm = 5 cm

Effort arm = 9 cm

Substituting the values in the formula we get,

$M.A. = \dfrac{9}{5} \\[0.5em] \Rightarrow M.A. = 1.8 \\[0.5em]$

Relation between mechanical advantage, efficiency and velocity ratio is

$M.A. = \eta \times V.R. \\[0.5em]$

Given, η = 100% = 1

So,

$M.A. = 1 \times V.R. \\[0.5em]$

Therefore, Mechanical advantage = Velocity ratio = 1.8

(d) When efficiency reduces to 50%, its mechanical advantage reduces however, its velocity ratio remains the same. So,

M.A. = 0.5 x 1.8 = 0.9
V.R. = 1.8

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