Question

a large cyllindrical vessel contains water to a height of 10m. it is found that the thrust acting on the curved surface is equal to that at the bottom. if atmospheric pressure can support a water column of 10m, the radius of the vessel is

Answer 1

15m

Answer 2

To solve this problem, we need to calculate the hydrostatic thrust (force) on the bottom surface and the curved surface of the cylindrical vessel and then equate them.

1. Given Information:

• Height of water in the vessel, $H = 10 \text{ m}$.
• Atmospheric pressure, $P_{atm}$, can support a water column of $10 \text{ m}$. This means $P_{atm} = \rho g (10 \text{ m})$, where $\rho$ is the density of water and $g$ is the acceleration due to gravity.
• Let the radius of the cylindrical vessel be $R$.

2. Thrust on the Bottom Surface ($F_B$):

The pressure at the bottom of the vessel is the sum of the atmospheric pressure and the pressure due to the water column.

$P_{bottom} = P_{atm} + \rho g H$

Substitute the given values:

$P_{bottom} = \rho g (10) + \rho g (10) = 20 \rho g$

The area of the bottom surface is $A_B = \pi R^2$.

The thrust on the bottom surface is:

$F_B = P_{bottom} \times A_B = (20 \rho g) \times (\pi R^2) = 20 \pi \rho g R^2$

3. Thrust on the Curved Surface ($F_C$):

The pressure on the curved surface varies with depth. It ranges from $P_{atm}$ at the water surface to $P_{atm} + \rho g H$ at the bottom. Since the pressure varies linearly with depth, the effective average pressure acting on the curved surface is the pressure at the centroid of the vertical side area. The centroid of a vertical rectangular area of height $H$ is at a depth of $H/2$ from the water surface.

So, the average pressure on the curved surface is:

$P_{avg} = P_{atm} + \rho g (H/2)$

Substitute the given values:

$P_{avg} = \rho g (10) + \rho g (10/2) = \rho g (10) + \rho g (5) = 15 \rho g$

The area of the curved surface is $A_C = (2 \pi R) \times H$.

$A_C = 2 \pi R (10) = 20 \pi R$

The thrust on the curved surface is:

$F_C = P_{avg} \times A_C = (15 \rho g) \times (20 \pi R) = 300 \pi \rho g R$

4. Equating the Thrusts:

According to the problem statement, the thrust acting on the curved surface is equal to that at the bottom:

$F_C = F_B$

$300 \pi \rho g R = 20 \pi \rho g R^2$

5. Solve for R:

We can cancel common terms from both sides: $\pi$, $\rho$, $g$, and $R$ (assuming $R \neq 0$).

$300 = 20 R$

$R = \frac{300}{20}$

$R = 15 \text{ m}$

The radius of the vessel is 15 m.

The final answer is $\boxed{\text{15m}}$.

Explanation of the solution:

1. Calculate the force on the bottom: $F_B = (P_{atm} + \rho g H) \times (\pi R^2)$.
2. Calculate the force on the curved surface: $F_C = (P_{atm} + \rho g H/2) \times (2\pi R H)$.
3. Equate $F_B = F_C$ and solve for $R$.

Given $H = 10 \text{ m}$ and $P_{atm} = \rho g (10 \text{ m})$, we have:

$F_B = (\rho g (10) + \rho g (10)) \pi R^2 = 20 \pi \rho g R^2$.

$F_C = (\rho g (10) + \rho g (10/2)) (2\pi R (10)) = (15 \rho g) (20 \pi R) = 300 \pi \rho g R$.

Equating them: $20 \pi \rho g R^2 = 300 \pi \rho g R$.

Dividing by $20 \pi \rho g R$ (assuming $R \neq 0$), we get $R = 15 \text{ m}$.