Question

If a 500kg load is to be lifted with a 100N effort using a first class lever at what distance from the fulcrum must the effort be applied when the load is 20cm away from the fulcrum?(answer is 1m)​

Answer 1

Explanation:

According to the question

$\sf \: Lord (i)=500kg$

$\sf \: Effort (e)= \bold{100N}$

$\sf \: Lord \: distance \: (id)= \bold{20cm}$

To Find

Effect distance (ed)=?

According to the principal of lever

L×LD=E×ED

$\red{ \sf \: 500×20=100×ED}$

$\frac{500 \times 10}{100} = 100cm$

100cm =1m

Hence Proved

Answer 1m

Answer 2

Answer:

If a 500kg load is to be lifted with a 100N effort using a first-class lever. The distance from the fulcrum at which the effort be applied when the load is 20cm away from the fulcrum will be $1m$

Explanation:

• According to the principle of levers,

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

• If L is the load and E is the effort. The distance between the load and fulcrum be the load arm$L_d$ and the distance between effort and fulcrum be the Effort arm$E_d$.

        So,

         $L\times L_d = E\times E_d$    ...(1)

In the question, it is given that,

$L = 500\ N$                   $E = 100\ N$

$L_d = 20\ c m$                 $E_d = ?$

Using equation (1),

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

Substitute the values into the above equation.

$E_d = \frac{500\times 20}{100} = 100 cm = 1m$

Effort distance is $1m$