# Tutor profile: Pratik B.

## Questions

### Subject: Physics (Fluid Mechanics)

There is a vertically symmetric glass with curvy walls and flat base, 10 cm in height. It can hold $$ V $$ volume of a fluid. The base of the glass is octagonal in shape. Area of the base is $$ A_b $$ and area of the top surface enclosed by a rim of the glass is $$ A_t $$. The glass is filled fully with water. Find the net force applied by walls (excluding base) of the glass on the water for following cases: (a) Atmospheric pressure is zero (b) Atmospheric pressure is non Zero and equal to $$P_{atm}$$ (c) When will both the cases give the same answer?

This is a problem from fluid statics. There is always a way to calculate force applied by walls on water by using integration of small forces at each of the small elements of walls. But for that, we must know the profile of the walls. Here, we know very little about walls and have to calculate total (net) force by walls on water. So, we can try to find all forces on the water in the glass and equate their sum to zero as water is not accelerating. Forces on water inside glass are: 1. Gravity causes weight (Downward) ($$W$$) 2. Force by glass base (Upward) $$F_b$$ 3. Force by walls (Upward because walls of glass are vertically symmetric) $$F_w$$ 4. Atmospheric pressure at top (Downward) $$F_a$$ Net force in vertical should be zero. So, $$W+F_a=F_w+F_b$$ So, $$F_w=W+F_a -F_b$$ $$W=\rho \times V \times g$$ $$F=P \times Area$$, So, $$F_a= P_{atm} \times A_t$$ and for the base,, the magnitude of force applied by base on water is equal to the force applied by water on base. So, $$F_b= P_{base} \times A_b$$ Using simple statics equation, $$P_{base}=P_{top} + \rho \times g \times height$$. So, $$F_w= 1000 \times V \times 10 + P_{atm} \times A_t - (P_{atm}+ 1000 \times 10 \times 0.1) \times A_b$$ $$F_w= 10000V + P_{atm} (A_t-A_b) - 1000A_b$$ So this is the $$F_w$$ when $$P_{atm}$$ is not zero. When it is zero, the middle term disappears, and $$F_w$$ becomes, $$F_w= 10000V - 1000A_b$$ For (c), we want the middle term to go to zero all the time irrespective of the value of $$P_{atm}$$. So, $$0=P_{atm} (A_t-A_b)$$ $$A_t=A_b$$ This means that area of base and top should be same for the force to be independent of atmospheric pressure.

### Subject: Mechanical Engineering

There is a container holding water at rest at STP. A small blade-fan is put into the water and operated for 1 min. The temperature of the water remains same throughout the process and there is no heat loss to the surrounding or to the fan. Explain the energy balance and KE of the water if the motor driving the fan is operated at 1 kW average. You may ignore the losses within the motor.

This is a simple problem based on the first law of thermodynamics. We can define our thermodynamic system first. The system here can be taken as the water excluding fan, blades, and motor. So, $$Q=W+ \Delta E$$ Here, $$Q$$ represents the heat taken by the system from surrounding $$W$$ represents the work done by the system on surrounding $$\Delta E$ represents the change in energy of the system It is given here that $$Q=0$$ as there is no heat loss to blades or surrounding. $$W$$ is the work done. Here, work is done by fan blades on the water in our container. So, whatever amount of work is done is shown negative in the equation as the definition of W in the equation is work done by the system. So here, $$W=-1kW \times 60 sec=-60 kJ$$ There is no other way of doing work here. $$ \Delta E$$ consists of various ways in which energy of the system may increase. It may include KE change, PE change, internal energy change. Internal energy depends on temperature and there is no change in temperature. And other changes of energy (magnetic, electric, potential etc) are not possible here. So, $$\Delta E= \Delta KE= KE_{final}-KE_{initial}=KE_{final}$$ So, putting in equation of the first law, $$0=-60kJ+KE_{final}$$ $$KE_{final}=60kJ$$ Thus, energy balance here is, all the energy provided by a fan and apparently by the motor driving the fan is transferred to the water. In water, it shows up as KE only.

### Subject: Physics

There is a block lying on an inclined plank. The plank is resting on an airplane floor with inclination angle $$30^o$$ with horizontal. The block is moving slowly downwards with very small velocity. Suddenly, the pilot throttles the airplane to ensure acceleration of the plane as $$a=2m/s^2$$ in the horizontal direction. The plank is still held tightly in place with same inclination and it is observed that the block stops sliding down the plank. Mass of the block is $$m=10kg$$ and you can take acceleration due to gravity, $$g=10 m/s^2$$. What is the friction acting between block and plank when pilot applies throttle?

This is the situation where the plank is accelerating horizontally with $$2m/s^2$$. And a block of mass $$10kg$$ is lying on it at rest. Drawing free body diagram of the block, we can show three forces acting on the block. 1. Friction f 2. Weight mg 3. Normal reaction N If we resolve all the components into horizontal and vertical directions and sum them up in respective directions, we can write, $$\sum F_{vertical}=m \times a_{vertical}=0$$ $$\sum F_{horizontal}=m \times a_{horizontal}=10 \times 2=20$$ Rewriting above equations by modifying left hand side (Putting values of forces), $$f sin(30^o) + N sin(60^o)-mg=0$$ $$ Ncos(60^o)-fcos(30^o)=20$$ Again, putting values of sine and cosine, $$ f+N \sqrt{3}=200$$ $$ N- f sqrt{3}=40$$ We want f. So, eliminate N. Therefore, putting value of N from second equation into first, we get, $$ f+(40+f \sqrt{3}) \sqrt{3}=200$$ $$4f=200-40 \sqrt{3}$$ $$f=32.67 N$$ force of friction.

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