Question

A solid, aluminum ingot weighs 90 N in air

Part A

What is its volume?

Express your answer using two significant figures.

Part B

The ingot is suspended from a rope and totally immersed in water. What is the tension in the rope (the apparent weight of the ingot in water)?

Express your answer using two significant figures.

Answer 1

Gravitational acceleration = g = 9.81 m/s²

Density of aluminium = = 2700 kg/m³

Weight of the aluminium ingot = W = 90 N

Volume of the aluminium ingot = V

W = Vg

90 = (2700)(9.81)V

V = 3.4 x 10^-3 m³

Density of water = _w = 1000 kg/m³

Buoyancy force of water on the aluminium ingot = F_b = _wVg

Tension in the rope = T

The buoyancy force and the tension in the rope supports the weight of the ingot.

W = T + F_b

W = T + _wVg

90 = T + (1000)(3.4x10^-3)(9.81)

T = 56.65 N

Rounding off to two significant figures,

T = 57 N

A) Volume of the aluminium ingot = 3.4 x 10^-3 m³

B) Tension in the rope (Apparent weight of the ingot in water) = 57 N