A sailboat at rest on a calm lake has its anchor dropped a distance of 4.0 m bel

Question

A sailboat at rest on a calm lake has its anchor dropped a distance of 4.0 m below the surface of the water. The anchor is suspended by a rope of negligible mass and volume. Depth from suface of water to the seafloor is 7.0 m. The mass of the anchor is 50 kg, and its volume is

6.25 *10^-3 m^3 . The density of water is 1000 kg/m^3 .

(a) On the dot below that represents the anchor, draw and label the forces (not components) 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 int

the boat so that the bottom of the boat is at a new depth d' below the surface of the water. How does d'

compare to d ?

____ d'< d           ____ d' = d           ____ d'>d

Justify your answer.

Explanation / Answer

A) weight acts downward, buoynat forcw and tension in the rope acts upward

b) buoyant force, B = weight of the displaced water

= rho*V*g

= 1000*6.25*10^-3*9.8

= 61.25 N


c)

as the anchor is in equilibrium,

buoyant force + tension = gravitational force

B + T = F

T = Fg - B
= m*g - 61.25
= 50*9.8 - 61.25

= 428.75 N

d)

d' > d

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.