2. A sailboat at rest on a calm lake has its anchor dropped a distance of4.0 m b

Question

2. A sailboat at rest on a calm lake has its anchor dropped a distance of4.0 m below the surface of the water. The anchor is suspended by a rope of negligible mass and volume. The mass of the anchor is 60 kg and its volume is 7.25 x 10-3m3. The density of water is 4.0 m 7.0 m 1000 kg/m3. a. Draw and label the forces that act on the anchor. b. Calculate the magnitude of the buoyant force acting on the anchor. c. Calculate the tension in the rope. d. The bottom of the boat is at a depth d below the surface of the water. Suppose the anchor is lifted back into the boat so that the bottom of the boat is at a new depth of d' below the surface off the water. How does d' compare to d. Explain your reasoning.

Explanation / Answer

2.(a)

weight (W) acts downward, buoynat force (B) and tension (T) in the rope acts upward

(b)

buoyant force is,

B = rho*V*g

B = 1000*7.25*10^(-3)*9.8

B = 71.05 N

(c)

as the anchor is in equilibrium,

gravitational force Fg = B + T

T = Fg - B

T = m*g - B = 60*9.8 - 71.05

T = 516.95 N

(d)

when the anchor is lifted back to the ship,

To balance the total weight more water should be displaced. it means the bottom of boat lowers little more depth.

so, d' > d

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