A balanced lever has two weights on it, the first with mass #2 kg # and the second with mass #38 kg#. If the first weight is # 4 m# from the fulcrum, how far is the second weight from the fulcrum?
The second weight roughly 0.21 meters from the fulcrum.
In order for the lever to balanced the net torque on it must be zero. Before we continue we must be familiar with a few concepts.
A. Torque is equal to the force applied in the perpendicular direction of the lever multiplied by the distance of of the force from the center of rotation.
B. The force of gravity that an object applies is equal to it's mass multiplied by Earth's gravitational Constant which is 9.81 m/
Now for the math:
Torque = Force * Lever Arm(distance from fulcrum)
In order for the two weights to balance we know that the net torque must equal zero.
The symbol for torque is
We can now plug in the equation for Torque that we stated earlier.
We can continue further by plugging in the equation used to solve the force applied by gravity.
But because each weight is multiplied by 9.81 we can cancel it out by dividing each side of the equation by 9.81. Zero divided by 9.81 is equal to zero. and the other side of the equation simply the former equation without the two 9.81 terms.
Now we'll plug in our values. The first weight is 2kg and 4m from the fulcrum. The second weight is 38kg, but we don't know how far it is from the fulcrum.
Because the weights are applying torque in opposite directions we'll assume that weight two is the weight applying the negative torque. Thus our equation must be:
2 kg * 4 M + (-38 kg * Lever Arm) = 0
2 kg * 4M - (38 kg * Lever Arm) = 0
+ (38 kg * Lever Arm) +(38 kg * Lever Arm)
2 kg * 4M = 38 kg * Lever Arm
8 kg/m = 38 * Lever Arm
/38 kg /38
.2105 m = Lever Arm
Thus we know that if you multiply