What is the resultant force exerted by the sides of the glass on the water?

[lavankohsa]

Homework Statement

A glass full of water has a bottom of area 20 cm 2 , top of area 20 cm 2 , height 20 cm and volume half a litre.
(a) Find the force exerted by the water on the bottom. (b) Considering the equilibrium of the water, find the
resultant force exerted by the sides of the glass on the water. Atmospheric pressure = 1'0 x 10 N/m 2 . Density
of water = 1000 kg/m 3 and g = 10 m/s 2 . Take all numbers to be exact

Homework Equations

The Attempt at a Solution

I calculated the first part by pressure at bottom*area of bottom. But i am not able to solve the 2nd part. Please help.

 

[Chestermiller]

It looks like you did the first part correctly. Now, to do the second part, start by drawing a free body diagram on the water in the container (which is in equilibrium). What are the forces acting on the water?

Chet

 

[lavankohsa]

For first part i found out the pressure at the bottom which is P_atm+density*g*h and multiply it by base area. The answer i got 204 N.
Now for 2nd part i calculated the weight of water (vol*density) which is 5N. I drew below diagram,

 

[Chestermiller]

You are very close to having the answer. What is the downward force exerted by the air on the top of the water? What is the upward force exerted by the bottom of the vessel on the water? You already got 5 N for the weight of the water (the downward force of gravity on the water). What force does the side walls of the vessel have to exert on the water (and in what direction) to keep the water in equilibrium? This is determined by your overall force balance in the vertical direction.

As an aside, your diagram looks very nice (especially the directions of the arrows), but the lengths of the arrows should be getting longer as you go from the top of the vessel to the bottom of the vessel. Do you know why?

Chet

 

[lavankohsa]

Will the normal force of base will be 204 N or weight of water which is 5 N ?? I know the arrow should be longer as we move down because pressure increases as we go down.

 

[lavankohsa]

Ok i solved it i guess. The normal force will be 204 N. Suppose the force by the side wall is F(upward). So for equilibirium,

P_atm*A+weight of water=N+F.

1*10⁵*20*10^-4+5=204+F

205=204+F

so F=1N

 

[Chestermiller]

Ok i solved it i guess. The normal force will be 204 N. Suppose the force by the side wall is F(upward). So for equilibirium,

P_atm*A+weight of water=N+F.

1*10⁵*20*10^-4+5=204+F

205=204+F

so F=1N

Why do you say "I guess." Do you not feel like you understand it to the extent that you would like, or do you feel that I helped too much?

Chet