Question

An anchor with a density of 700 lb/ft tied to a rope is submerged in water (but is not touching the bottom). It has a volume of 3.o ft3. What is the buoyancy force on that object? 1. (700 lb/ft3)(3 ft3) 2. (700 lb/ft3)(32.2 ft/s2)(3 ft 3. (62.4 lb/ft3(3 ft3 4. (62.4 lb/ft3)(32.2 ft/s2(3 ft What is the weight of the anchor? 1. (700 lb/ft(3 ft3) 2. (700 lb/ft)(32.2 ft/s2)(3 3. (62.4 lb/ft)(3 ftS) 4. (62.4 lb/ft)(32.2 ft/s )(3 What is the relative density (specific gravity) of the anchor? 1. (700 lb/ft/(62.4 Ib/ft3) 2. (62.4 lb/ft)/(700 lb/ft3) 3. (62.4 lb/ft)(32.2 ft/s)/(700 lb/ft3) Draw a FBD for this scenario The buoyancy force, FB, points up. The weight of the anchor, w, points down. The tension in the rope, T, points up. Which equation will correctly determine Tension in the rope? The rope is cut. Which equation will correctly determine its acceleration as it sinks? Draw a FBD KD 2. FB W-ma

Answer 1

DATA REQUIRED:

Density of anchor, ρ = 700 lb/ft³

Density of water, ρ_w = 62.4 lb/ft³

Volume of anchor, V = 3 ft³

Acceleration due to gravity, g = 32.2 ft/s²

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PART A:

According to Archimedes principle, buoyant force is given by the weight of the fluid displaced.

Hence, Buoyant force is given by:

FB = mg

FB = (V.Pw).

:: F3 = (3 ft) x (62.4 lb/ft3) x (32.2 ft/s)

So the correct option is (4).

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PART B:

Weight of the anchor is given by:

W = mass xg

W = Density x volume xg

W = pVg

W = (3 ft) x (700 lb/ft) x (32.2 ft/s)

or

W = (700 lb/ft3) x (32.2 ft/s) x (3 ft)

So the correct option is (2).

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PART C:

Relative density is given by:

density of anchor R.D= density of water

R.D=

700 lb/ft3 R.D= 62.4 lb/ft?

So the correct option is (1).

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PART D:

Free body diagram:

From FBD, we have:

FB-W+T=0

So the correct option is (2).

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PART E:

Let a be the acceleration downwards.

LW-F8 = ma

- (W-FB) = -ma

FR - W = -ma

So the correct option is (2).