Question

If 100 N effort is required to lift a load of 300N in a lever of 1m and the distance between fulcrum to load is 30 cm then calculate effort distance, MA, VR, Efficiency?

Answer 1

Answer:

If the frictional force in the machine increases the efficiency decreases. ... Relation between MA, VR, and η ... A lever raises a load of 300N through a distance of 10 cm by an effort of 100N ... Effort distance (Ed) = 1m.

Answer 2

Answer:

If 100 N effort is required to lift a load of 300N in a lever of 1m and the distance between fulcrum to load is 30 cm then

1. Effort distance = $90\ cm$
2. Mechanical advantage = $3$
3. Velocity ratio = $3$
4. Efficiency =$100\%$

Explanation:

• Consider a lever of Load L, effort E, Load distance $L_d$ and effort distance $E_d$.  Then, the mechanical advantage, velocity ratio, and efficiency of the lever can be expressed as,

        $\text{ { Mechanical\ advantage }(M . A)}=\frac{\text { Load }}{\text { Effort }} = \frac{L}{E}$    ...(1)

        $\text { Velocity ratio }(\text { V.R })=\frac{E_{d}}{L_{d}}$         ...(2)

        $\text { Efficiency}\ (\eta) = \frac{\text { M.A }}{\text { V.R } }$                  ...(3)

• According to lever principle,

       $Load \times Load\ arm = Effort\times Effort\ arm$      

      Or

       $L\times L_d = E\times E_d$                              ...(4)  

In the question, it is given that,

$L = 300\ N$                               $E = 100\ N$            

$L_d = 30\ cm = 0.30\ m$              $E_d = ?$  

By using equation(4),

$E_d = \frac{L\times L_d}{E}$

      $=\frac{300\times 0.30}{100} =0.9 m = 90\ cm$

M.A can be obtained by substituting these values into equation (1)

$\text M \cdot \text A=\frac{300}{100}=3$

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

$\text V \cdot \text R =\frac{0.90}{0.30}=3$

Efficiency,

$\eta = \frac{3}{3} \times 100 = 100 \%$