# A balanced lever has two weights on it, one with mass 2 kg and one with mass 8 kg. If the first weight is  4 m from the fulcrum, how far is the second weight from the fulcrum?

Mar 30, 2018

$1 m$

#### Explanation:

The concept that comes into use here is torque. For the lever to not tip over or rotate, it must have a net torque of zero.

Now, the formula of torque is $T = F \cdot d$.

Take an example to understand, if we hold a stick and attach a weight at the front of the stick, it doesn't seem too heavy but if we move the weight to the end of the stick, it seems a lot heavier. This is because the torque increases.

Now for the torque to be same,
${T}_{1} = {T}_{2}$

${F}_{1} \cdot {d}_{1} = {F}_{2} \cdot {d}_{2}$

The first block weighs 2 kg and exerts approximately $20 N$ of force and is at a distance of 4m

The first block weighs 8 kg and exerts approximately $80 N$

Putting this in the formula,

$20 \cdot 4 = 80 \cdot x$

We get that x= 1m and hence it must be placed at a distance of 1m

Mar 30, 2018

The distance is $= 1 m$

#### Explanation: The mass ${M}_{1} = 2 k g$

The mass ${M}_{2} = 8 k g$

The distance $a = 4 m$

${M}_{1} \times a = {M}_{2} \times b$
$b = \frac{{M}_{1} \times a}{{M}_{2}} = \frac{2 \cdot 4}{8} = 1 m$