Question

If the load of 600N is lifted using a first class lever applying an effort of 350N. If the distance between the fulcrum and the effort is 60cm and the distance between the load and fulcrum is 30cm , calculate its efficiency .

Answer 1

Answer:

85.5%

Explanation:

Mechanical advantage = Load/Effort = 600/350 = 1.71 Velocity ratio = distance moved by effort/distance moved by load = 60/30 = 2 Efficiency = (M.A/V.R) × 100% = (1.71/2) × 100 = 0.855 × 100 = 85.5%

Answer 2

Answer:

If the load of 600N is lifted using a first-class lever applying an effort of 350N. If the distance between the fulcrum and the effort is 60cm and the distance between the load and fulcrum is 30cm

Efficiency = $85.71\%$

Explanation:

• For a lever, the efficiency can be expressed as,

                  $\text { Efficiency}\ (\eta) = \frac{\text { M.A }}{\text { V.R } }$                  ...(1)

• M.A is the mechanical advantage and V.R is the velocity ratio. It can be defined as,

                  $\text{ { Mechanical\ advantage }(M . A)}=\frac{\text { Load }}{\text { Effort }} = \frac{L}{E}$    ...(2)

                  $\text { Velocity ratio }(\text { V.R })=\frac{E_{d}}{L_{d}}$         ...(3)        

         Where,

         L   -    Load

         E   -   Effort

        $L_d$  -   Load distance

        $E_d$  -   Effort distance

In the question, it is given that,

$L = 600\ N$                                 

$E = 350\ N$            

$L_d = 30\ cm=0.30\ m$                  

$E_d =60\ cm = 0.60\ m$  

Substitute the above values into equation (1)

$\text M \cdot \text A=\frac{600}{350}=\frac{40}{15} =1.71428$

V.R  can be obtained by substituting the values of $E_d$ and $L_d$ into equation (2)

$\text V \cdot \text R =\frac{0.60}{0.30}=2$

Using equation (3), Efficiency is,

$\eta = \frac{1.71428}{2} \times 100 = 85.71 \%$