A balanced lever has two weights on it, one with mass #1 kg# and one with mass #4 kg#. If the first weight is # 9 m# from the fulcrum, how far is the second weight from the fulcrum?

Question

A balanced lever has two weights on it, one with mass #1 kg# and one with mass #4 kg#.  If the first weight is # 9 m# from the fulcrum, how far is the second weight from the fulcrum?

Olivia Askew · Accepted Answer

#2.25 \ "m"#

The relationship for a balanced lever is as follows:

#m_1d_1=m_2d_2#

#m_1,m_2# are the masses of the two weights

#d_1,d_2# are their distances from the fulcrum

Thus, we obtain:

#1 \ "kg"*9 \ "m"=4 \ "kg"*d_2#

#d_2=(9color(red)cancelcolor(black)"kg""m")/(4color(red)cancelcolor(black)"kg")#

#=2.25 \ "m"#

Xavier Adame · Answer

undefined To find the distance of the second weight from the fulcrum, you can use the principle of moments:

$ \text{Moment}_1 = \text{Moment}_2 $

$ \text{Moment}_1 = \text{Force}_1 \times \text{Distance}_1 $

$ \text{Moment}_2 = \text{Force}_2 \times \text{Distance}_2 $

Given:
$ \text{Force}_1 = 1 \, \text{kg} $
$ \text{Distance}_1 = 9 \, \text{m} $
$ \text{Force}_2 = 4 \, \text{kg} $
$ \text{Distance}_2 = ? $

Using the principle of moments:
$ \text{Force}_1 \times \text{Distance}_1 = \text{Force}_2 \times \text{Distance}_2 $

$ 1 \, \text{kg} \times 9 \, \text{m} = 4 \, \text{kg} \times \text{Distance}_2 $

Solve for $ \text{Distance}_2 $:
$ \text{Distance}_2 = \frac{1 \, \text{kg} \times 9 \, \text{m}}{4 \, \text{kg}} $

$ \text{Distance}_2 = \frac{9}{4} \, \text{m} $

$ \text{Distance}_2 = 2.25 \, \text{m} $

Therefore, the second weight is 2.25 meters from the fulcrum.