Buoyany and change in volume underwater

[sweetpete28]

By breathing in deeply, you can change volume of your body. Suppose a swimmer underwater weighs wf = 23.4 N with his lungs full, and we = 48.0 N with lungs empty. Find change in body volume when swimmer inhales.

I know the apparent weight = weight in air - buoyant force and buoyant force = weight of displaced fluid = desnity of liquid x g x volume but how do I solve for the change in volume without weight in air??

 

[Doc Al]

Just set up the equations and solve for the change in volume. You won't need the weight in air. (Note that you are finding the change in volume, not the volume.)

 

[sweetpete28]

Yup! Just got that one...here is last one I need to finish in 20 min...

A rectangular block of wood (M = 260 kg) floats on a calm fresh water lake with do = 12.6 cm below the water. When a dog steps on the block, the block is pushed downward so that now it floads with d = 15.9 cm beneath the water. Find the mass of the dog.

any suggestions?

 

[Doc Al]

Just set up your force equations. Assume that the block remains upright and has some cross-sectional area A. (You won't need a numerical value.)

 

[asteriatic]

I would approach this problem by first considering the principles of buoyancy and the effects of volume on weight. When a swimmer is underwater, their weight is affected by the buoyant force exerted by the surrounding water. This buoyant force is equal to the weight of the water that is displaced by the swimmer's body.

In this case, we are given the weight of the swimmer when their lungs are full (wf = 23.4 N) and when their lungs are empty (we = 48.0 N). This difference in weight is due to the change in volume of the swimmer's body when they breathe in and out.

To find the change in volume, we can use the formula for buoyant force: buoyant force = density of liquid x g x volume. We know that the density of water is approximately 1000 kg/m^3 and the acceleration due to gravity is 9.8 m/s^2. We can also assume that the swimmer's body is mostly water, so we can use the density of water to represent the density of the swimmer's body.

Using this information, we can set up the following equation:

wf = 1000 kg/m^3 x 9.8 m/s^2 x Vf
we = 1000 kg/m^3 x 9.8 m/s^2 x Ve

where Vf is the final volume of the swimmer's body when their lungs are full and Ve is the initial volume of their body when their lungs are empty.

We can then solve for the change in volume (Vf - Ve) by rearranging the equations and substituting in the given values:

Vf - Ve = wf / (1000 kg/m^3 x 9.8 m/s^2) - we / (1000 kg/m^3 x 9.8 m/s^2)
= 23.4 N / (1000 kg/m^3 x 9.8 m/s^2) - 48.0 N / (1000 kg/m^3 x 9.8 m/s^2)
= 0.0024 m^3 - 0.0049 m^3
= 0.0025 m^3

Therefore, the change in volume when the swimmer inhales is approximately 0.0025 m^3. This change in volume is a result of