solution:
Weight of bucket = 10kg
Length or distance =14m
Weight of rope=0.5kg/m
At any point x of the rope,
=(0.5)(14-x)
=(7-0.5x)
Since the water finishes draining at 14m level and total weight of water is 42kg
Total mass=(7-0.5x)+(42-3x)+10=(59-3.5x)kg
Force=(9.8)(59-3.5x)
Explanation:
Given that,
Mass if the rock, m = 1 kg
It is suspended from the tip of a horizontal meter stick at the 0-cm mark so that the meter stick barely balances like a seesaw when its fulcrum is at the 12.5-cm mark.
We need to find the mass of the meter stick. The force acting by the stone is
F = 1 × 9.8 = 9.8 N
Let W be the weight of the meter stick. If the net torque is zero on the stick then the stick does not move and it remains in equilibrium condition. So, taking torque about the pivot.
W = 3.266 N
The mass of the meters stick is :
So, the mass of the meter stick is 0.333 kg.