Question

In: Physics

a) A block of Aluminum has a density of 2700 kg/m^3 and a mass of 63...

a) A block of Aluminum has a density of 2700 kg/m^3 and a mass of 63 grams. The block is tied to one end of a string and the other end of the string is tied to a force scale. The block is allowed to hang in the air from the string. Calculate the force which will be registered on the scale.

b) The block remains tied to the string but is now completely submerged in a container of water. The block is not touching the sides or bottom of the container. Calculate the force which will be registered on the scale.

c) The block is raised so that it is only 39% submerged in the water. Calculate the force which will be registered on the scale.

Solutions

Expert Solution

(a)The block is tied to one end of a string and the other end of the string is tied to a force scale.

Given the mass of the aluminum block,

Force on the scale will be equal to the weight of the aluminum block

Force on the scale,

ANSWER:

===========================

(b) The block remains tied to the string but is now completely submerged in a container of water.

Density = Mass/Volume

Volume = Mass/Density

The volume of aluminum block = 0.063kg/(2700kg/m3) = 2.33*10-5m3

Weight of the block will be downwards

The buoyant force will be upwards.

Force on the scale = the Weight of the aluminum block - Buoyant force

F = W-Fb

F = mg - Density of water * volume of water displaced *g

Since the block is completely submerged, the volume of water displaced will be equal to the volume of aluminum block

F = mg - Density of water * Volume of aluminum block *g

F = 0.063kg*9.81m/s2 - 1000kg/m3*2.33*10-5m3 * 9.81m/s2

ANSWER: F = 0.38913N

===========================

(c) The block is raised so that it is only 39% submerged in the water

The volume of aluminum block = 0.063kg/(2700kg/m3) = 2.33*10-5m3

Volume of aluminum block under water = 0.39 * 2.33*10-5 m3

=============

Consider the block

Weight of the block will be downwards

The buoyant force will be upwards.

Force on the scale = the Weight of the aluminum block - Buoyant force

F = W-Fb

F = mg - Density of water * volume of water displaced *g

F = mg - Density of water * Volume of aluminum block under water *g

F = 0.063kg*9.81m/s2 - 1000kg/m3*0.39*2.33*10-5m3 * 9.81m/s2

ANSWER: F = 0.52876N

===========================

genius_generous answered 10 months ago
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