Problem 19: Weighing Machine Reading with Pump
Concept: Dynamic Force & Reaction
The weighing machine measures the normal reaction $N$.
$$N = W_{total} + F_{dynamic}$$
The dynamic force arises because the pump accelerates water from rest to a velocity $v$ (suction side) and then the system brings it to rest or redirects it (ejection side). This change in momentum requires force.
Derivation:
Let the rate of mass transfer be $\mu$ (kg/s). The velocity at which level of water changes is $v$. $$\mu = \rho S v \implies v = \frac{\mu}{\rho S}$$ Consider the transfer from the lower vessel to the upper vessel. The total additional force is the sum of forces at the suction and ejection points:
1. Suction (Lower Vessel):
Water is accelerated from rest ($0$) to velocity $v$.
2. Ejection (Upper Vessel):
The water jet enters the upper tank and comes to a halt (or hits the water surface).
Let the rate of mass transfer be $\mu$ (kg/s). The velocity at which level of water changes is $v$. $$\mu = \rho S v \implies v = \frac{\mu}{\rho S}$$ Consider the transfer from the lower vessel to the upper vessel. The total additional force is the sum of forces at the suction and ejection points:
1. Suction (Lower Vessel):
Water is accelerated from rest ($0$) to velocity $v$.
- The force required on the water is upward ($F = \frac{dp}{dt} = \mu v$).
- By Newton’s 3rd law, the water exerts an equal and opposite force on the pump mechanism (which is attached to the stand).
- Thus, the pump pushes the stand downwards with force $F_1 = \mu v$.
Insight: If the pump were fixed on an external stand (not on the weighing pan), this recoil force would not be measured. The reading would only increase due to the impact on the upper vessel (half the present value).
2. Ejection (Upper Vessel):
The water jet enters the upper tank and comes to a halt (or hits the water surface).
- We assume the water loses its vertical momentum upon impact.
- The rate of momentum destruction exerts a force on the tank (and thus the scale) downwards.
- Magnitude of force $F_2 = \mu v$.
Total Force Calculation:
The total additional downward force $F_{total}$ is the sum of the suction recoil and the ejection impact:
$$F_{total} = F_1 + F_2 = \mu v + \mu v = 2\mu v$$
Substituting $v = \frac{\mu}{\rho S}$:
$$F_{total} = 2\mu \left( \frac{\mu}{\rho S} \right) = \frac{2\mu^2}{\rho S}$$
Final Answer:
The reading of the weighing machine is: $$R = W + \frac{2\mu^2}{\rho S}$$
Alternate solution contributed by Sayan Chatterjee.
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The reading of the weighing machine is: $$R = W + \frac{2\mu^2}{\rho S}$$