A balanced lever has two weights on it, the first with mass 3 kg 3kg and the second with mass 4 kg4kg. If the first weight is 8 m8m from the fulcrum, how far is the second weight from the fulcrum?

1 Answer
Jan 23, 2017

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"=0τnet=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.