Question

Mass of a block of unknown metal alloy is m = 320 g when measured in air. When fully submerged in

water, the apparent mass of the metal block is measured to be m' = 270 g.
a) Draw two free-body force diagrams showing all forces acting on the metal block when its mass is measured in air, and when its mass is measured in water.
b) Write down the force equation at the equilibrium when the metal block is being measured in
water.
c) Start with Archimedes' principle to calculate the volume of the metal block.
D) Determine the density of the metal block.

Answer 1

I’m not sure but I think it could be B??

Answer 2

When a metal block is weighed in water, the buoyant force equals its weight at equilibrium. Using Archimedes' principle, the volume of the metal block is calculated, and its density is determined as the ratio of its mass to the calculated volume.

a) Free-Body Force Diagrams:

1. In Air:

- Weight
`(\(W\))` downward.

- Tension
`(\(T\))` upward (assuming it's held in place).

- No buoyant force in air.

2. In Water:

- Weight
`(\(W\))` downward.

- Buoyant force
`(\(B\))` upward.

- Tension
`(\(T\))` upward (assuming it's held in place).

b) Force Equation at Equilibrium in Water:

- At equilibrium, the block is not accelerating, so the net force is zero.

-
`\( \Sigma F = 0 \)`

-
`\( B - W = 0 \)` (since
`\( T \)` is canceled out by
`\( T \)` in the opposite direction).

-
`\( B = W \)`

-
`\( B = mg \)` (where
`\( g \)` is the acceleration due to gravity, approximately
`(\( 9.8 \ m/s^2 \))`.

c) Archimedes' Principle:
`\( B = \rho_{\text{water}} \cdot V_{\text{block}} \cdot g \)`

- Archimedes' principle states that the buoyant force
`(\( B \))` is equal to the weight of the fluid displaced by the object.

-
`\( B = \rho_{\text{water}} \cdot V_{\text{block}} \cdot g \)`

- Since B=W (at equilibrium),
`\( W = \rho_{\text{water}} \cdot V_{\text{block}} \cdot g \)`.

- Solving for
`\( V_{\text{block}} \): \( V_{\text{block}} = \frac{W}{\rho_{\text{water}} \cdot g} \)`.

d) Determine Density of the Metal Block:

- Density
`(\( \rho_{\text{block}} \))` is given by
`\( \rho_{\text{block}} = \frac{m}{V_{\text{block}}} \)`.

- Substituting the expression for
`\( V_{\text{block}} \)` from part (c):
`\( \rho_{\text{block}} = \frac{m}{\frac{W}{\rho_{\text{water}} \cdot g}} \)`.

- Simplifying:
`\( \rho_{\text{block}}` =
`\frac{m \cdot \rho_{\text{water}} \cdot g}{W} \).`

Given the values of
`\( m \)` (mass in air) and
`\( m' \)` (apparent mass in water), and knowing
`\( g \)` and the density of water
`(\( \rho_{\text{water}} \))`, you can calculate the density of the metal block using the final expression obtained in part (d).