Help with calcuating force constant ?

[nukeman]

Homework Statement

A 320 kg wooden raft floats on a lake in equilibrium. When a 75 kg man steps carefully onto the raft, it floats 3.5 cm deeper into the water. When he steps off, the raft oscillates for a while.

(a) The buoyant force of the water on the raft is a restoring force that is directly proportional to how deep the raft sits in the water. In this way, the water acts like a mechanical spring. Calculate the force constant of this spring.

Homework Equations

The Attempt at a Solution

I am not sure how to get the force constant out of this, as if it were a spring.

k = f/displacement ?

 

[Curious3141]

Homework Statement

A 320 kg wooden raft floats on a lake in equilibrium. When a 75 kg man steps carefully onto the raft, it floats 3.5 cm deeper into the water. When he steps off, the raft oscillates for a while.

(a) The buoyant force of the water on the raft is a restoring force that is directly proportional to how deep the raft sits in the water. In this way, the water acts like a mechanical spring. Calculate the force constant of this spring.

Homework Equations

The Attempt at a Solution

I am not sure how to get the force constant out of this, as if it were a spring.

k = f/displacement ?

So what is the force the man exerts on the raft (and hence the water) by stepping onto it? This is exactly balanced by the buoyant force. Express this in N (Newton).

And what is the displacement of the raft in the water caused by that force? Express this in m (metre).

After you work out what those two are, just divide as per the equation you quoted. The units are $Nm^{-1}$

 

[nukeman]

so, 75kg * 9.8 = 735 J

So,

735j = (x)(0.035) solve for x gives me 21,000 NM

is that correct?

 

[Curious3141]

so, 75kg * 9.8 = 735 J

So,

735j = (x)(0.035) solve for x gives me 21,000 NM

is that correct?

Why J (Joule) as the unit for force? Remember, weight = mg, and this is a force measured in Newton. Potential energy due to Earth's gravitation = mgh, and this is energy measured in Joule. Big difference.

So the weight of the man is mg = 75*9.8 = 735N (Newton).

The displacement x is 3.5cm = 3.5/100m = 0.035m.

Hence the force constant k = F/x = 735/0.035 = 21000N/m (Newton-per-metre, not the slash signifying division between Newton and metre. Instead of the slash one can also write $Nm^{-1}$ (and many modern texts prefer this notation). This is NOT the same as Nm (Newton-metre, obtained by multiplying Newton and metre, and which is a unit of energy equivalent to Joule).

Please be careful with your units. Wrong unit = wrong answer. Other than that, your figures are fine.

 

[asteriatic]

To calculate the force constant, we first need to understand the concept of Hooke's law. Hooke's law states that the force exerted by a spring is directly proportional to the distance it is stretched or compressed from its equilibrium position. In this case, the water is acting like a spring, with the buoyant force as the restoring force and the displacement being the change in depth of the raft.

To calculate the force constant, we can use the equation k = F/x, where k is the force constant, F is the restoring force, and x is the displacement. In this case, the restoring force is the buoyant force, which can be calculated using the formula F = ρVg, where ρ is the density of water, V is the volume of the displaced water, and g is the acceleration due to gravity.

We know that the displacement of the raft is 3.5 cm, which is equal to 0.035 m. We also know that the man's weight is 75 kg and the total weight of the raft and the man is 395 kg (320 kg raft + 75 kg man). Therefore, the buoyant force when the man is on the raft is:

F = ρVg
= (1000 kg/m^3)(0.035 m)(9.8 m/s^2)
= 343 N

Now, we can plug in these values into the equation for the force constant:

k = F/x
= (343 N)/(0.035 m)
= 9800 N/m

Therefore, the force constant of the water acting like a spring on the raft is 9800 N/m. This means that for every meter the raft sinks into the water, the buoyant force will increase by 9800 N.