How much force must be exerted at the right end to pick up the lever and load if it pivots at the left end? The 2.00 meter long lever weighs 1.50 N and is uniform. There is an 8.00 N weight resting on the 40.0 cm mark.
so would it take .4m (40cm to m)
.4m*8N=32Nm=?1.6M
because 1.6 is the distance left over 2.0-.4=1.6??
so 3.2/1.6=2N?
To solve this problem, we need to find the force required at the right end of the lever to lift the lever and the load.
First, convert the 40 cm mark to meters:
40 cm is equal to 0.4 meters.
Next, calculate the torque exerted by the weight resting on the 40 cm mark:
Torque is calculated by multiplying the force (8 N) by the distance from the pivot point (0.4 m):
Torque = Force × Distance = 8 N × 0.4 m = 3.2 N⋅m.
Since the lever is in equilibrium, the torque exerted by the weight on the left side must be balanced by the torque exerted by the force at the right end of the lever. The torque exerted by the lever itself can be calculated by multiplying its weight (1.50 N) by its center of mass distance from the pivot point (1.00 m):
Torque of the lever = 1.50 N × 1.00 m = 1.50 N⋅m.
To find the force required at the right end of the lever, divide the total torque needed (3.2 N⋅m) by the remaining distance from the 40 cm mark to the right end (1.6 m):
Force = Torque / Distance = 3.2 N⋅m / 1.6 m = 2 N.
Therefore, a force of 2 N must be exerted at the right end of the lever to lift the lever and the load.