Question

A balanced lever has two weights on it, the first with mass `3 kg` and the second with mass `4 kg`. If the first weight is `8 m` from the fulcrum, how far is the second weight from the fulcrum?

Answer 1

6 m away from the fulcrum, opposite from the first weight.

Explanation:

For the lever to be "balanced", there must be no net torque on the system, or

`tau_"net"=0`

or

`Sigma tau = 0`

Assuming the fulcrum is at the center of mass of the lever (or the lever is massless), the torques created by each of the weights must be equal and opposite:

`tau_1 + tau_2 = 0`
`tau_1 = - tau_2`

Torque is the force perpendicular to a point of rotation multiplied by the distance from that pivot point. Assuming the lever to be horizontal, the torque equation becomes:

`F_1xxr_1 = -(F_2xxr_2)`
`m_1gxxr_1=-(m_2gxxr_2)`

Solving for `r_2` gives:

`r_2=((m_1cancel(g))r_1)/(-m_2cancel(g))=-m_1/m_2r_1`

Applying the given information, we get:

`r_2=-(3"kg")/(4"kg")(8"m")=-6"m"`

The fact that `r_2` is a negative value tells us that it is on the opposite side of the fulcrum from the first weight.

The second weight must be positioned 6 m away from the fulcrum on the other side of the fulcrum.