At one end of a lever is a 12-pound weight which is 8 inches from the fulcrum. How much weight should be placed on the other end 10 inches from the fulcrum to balance the lever?
At one end of a lever is a 12-pound weight which is 8 inches from the fulcrum. How much weight should be placed on the other end 10 inches from the fulcrum to balance the lever?
At one end of a lever is a 12-pound weight which is 8 inches from the fulcrum. How much weight should be placed on the other end 10 inches from the fulcrum to balance the lever?
yea this is a physics problem, definitely dont think this would be on the DAT. But ill give it a shot anyways just for kicks. you need the torque on both ends to equal each other which is RFsin(theta). since the torques are perpendicular to the lever arm, it sin (90) = 1.
radius X F = radius X force
8 X 12 = 10 X weight
weight = 9.6 lbs