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(III) A scuba tank, when fully submerged, displaces 15.7L of seawater. The tank itself has a mass of 14.0kg and, when โ€œfull,โ€ contains 3.00kg of air. Assuming only its weight and the buoyant force act on the tank, determine the net force (magnitude and direction) on the fully submerged tank at the beginning of a dive (when it is full of air) and at the end of a dive (when it no longer contains any air).

Short Answer

Expert verified

The net force on the fully submerged tank at the beginning of a dive is 9N(downwards) and the net force on the fully submerged tank at the end of a dive is 21N(upwards).

Step by step solution

01

Concept of net force

Net force is the reduced forces applied on a system where major forces take over the minor forces. The direction of major forces shows the direction of acceleration of the system as well.

02

Given data

The volume of the tank is Vt=15.7L.

The mass of the tank is mt=14kg.

The mass of the air is ma=3kg.

03

Calculation of net force acting on the tank in the beginning

The buoyant force is calculated as:

FBouyant=ฯVg

Here,ฯis density of fluid, V is volume occupied and g is acceleration due to gravity.

The density of a fluid is calculated as:

ฯ=mV

Here,mis the mass of the fluid.

Using the law of equilibrium of forces, calculate the net forces.

Ffull=FBโˆ’mTg=ฯwVtgโˆ’(mt+ma)g

Here, ฯwis the density of water and mT is the total mass.

Substitute the known values in the above equation to find the net force.

Ffull=(ฯwaterVtankโˆ’(mtank+mair))g=((1025kg/m3)(15.7Lร—1m31000L)โˆ’(14kg+3kg))(9.8m/s2)=โˆ’8.89Nโ‰ˆโˆ’9N

Hence, the net force is 9N downwards.

04

Calculation of net force acting on the tank in the end

Using law of equilibrium of forces,

Fempty=FBโˆ’mTg=ฯwVtgโˆ’(mt)g

Substitute the known valuesin the above equation to find the net force.

Ffull=(ฯwVtโˆ’(mt))g=((1025kg/m3)(15.7Lร—1m31000L)โˆ’14kg)(9.8m/s2)=20.51Nโ‰ˆ21N

Hence, the net force is 21N upwards.

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