In this problem, take atmospheric pressure to be 1.0 x 10 5 Pa, and the density

Question

In this problem, take atmospheric pressure to be 1.0 x 105 Pa, and the density of water to be 1000 kg/m3. Use g = 10 N/kg.

In 1690, Sir Edmund Halley (of comet fame) invented the diving bell. It was made of wood and roughly cylindrical in shape. We will treat the bell as if it were a cylinder 3 m high and 1 m in diameter.

The bell is fully submerged underwater so that the bottom face of the bell is at a depth of 24 m.

a. What is the water pressure at the bottom of the bell? _______ Pa   What is the water pressure at the top of the bell? _______ Pa

b. What is the force of the water on the bottom face of the bell? _______N   What is the force of the water on the top face of the bell? _______N

c. What is the net force of the water pushing on the top and bottom of the bell? _______N , and what direction is it, Upward or Downward? (Verify that this is the same as the buoyant force calculated using the equation B = RhofluidVsubmergedg.)

d. If the bell and its contents have a mass of 790 kg, what is the net force on the bell? _______N, and what direction is it, Upward or Downward?

Explanation / Answer

Given

atmospheric pressure is P = 1.0*10^5 Pa = 1 atm

density of water rho = 1000 kg/m3 , g = 10 N/kg

bell of cylinder shape with height 3m, diamter 1 m

the bell was fully submerged and at a depth of h = 24 m  

a.

What is the water pressure at the bottom of the bell? _______ Pa What is the water pressure at the top of the bell? _______ Pa

we know that the pressure is P = rho*g*h  

P_b = rho*g*h2 = 1000*10*24 Pa = 240000 Pa = 2.4*10^5 Pa

P_t = rho*g(h2-height of cylinder)

= 1000*10(24-3) Pa

= 210000 Pa

= 2.1*10^5 Pa

b. What is the force of the water on the bottom face of the bell? _______N What is the force of the water on the top face of the bell? _______N

we know that the Buoyance force is F = rho*v*g , v = h*A  

at bottom the buoyance force is F_b = rho*h2*A*g = 1000*24*pi*0.5^2*10 N = 188495.56 N

and the buoyance force at the top is F_t = rho*h2*A*g = 1000*(24-3)*pi*0.5^2*10 N = 164933.614 N

F_b > F_t so

c. net force of the water direction is upward

d. mass of bell is 790 kg

mg (h2-h1) = 790*10(24-3) = 165900 N and due to buoyance force net F_B = 23561.945

so total force acting is 165900 -23561.945b= 142338 downward  

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