< Classical Mechanics

 Lecture 4 Part 1 Part 2 Part 3 Part 4

Forces, Density and Pressure

Classical Mechanics

o The same volume of two different fluids can have very different masses because they have different densities.
- Density is the mass (in kilograms) of an object per volume (in metres cubed), denoted by $\rho=\frac{m}{V}$ where $\rho$ = density $(kg/m^3)$, m = mass (kg), V = volume $(m^3)$
o Archimedes was asked to find a way to measure the volume of the king's crown without damaging it in any way (King Hierro II suspected that the crown made by his goldsmith was not purely gold).
- The solution came to him as he was taking a bath. He noticed that as he got in to the tub, the water level would rise. He figured out that the amount of volume of his body submerged displaced an equal volume of water.
- All he needed to do was submerge the crown in water, and measure the volume of water it pushed out of the way. This would be equal to the volume of the crown.
- As the story goes, he was so excited by his discovery that he jumped out of the tub and ran naked through the streets yelling "Eureka!" which translates from Greek as "I have found it."

o This is known as the Archimedes' Principle, which says that the force of buoyancy that an object will feel when it is immersed in a fluid is equal to the weight of the of the fluid displaced by the object.
- This just means that if you place an object with a volume of (for example) 3 $m^3$ fully into some water, it will displace 3 $m^3$ of water. The buoyant force the object will feel is equal to the weight of that 3 $m^3$ of water.
- It is important to note that only the volume of the object that is submerged into the fluid can displace any volume, so that is the only amount considered when you figure out the buoyant force.
We can figure out a formula for buoyancy based on the formula we have for weight.
- Keep in mind that the buoyancy depends on the mass of the fluid displaced.
- Density $\rho\ = \frac{m}{V} (\frac{mass}{volume}) \quad in \quad (kg/m^3) \rightarrow m = \rho\ V$. This is handy, since we often talk about fluids in terms of density and volume, instead of mass.
$F_g = mg \rightarrow F_B = g m_{fluid} \rightarrow F_B = g \rho_{fluid}\ V_{sub}$ A motor boat is being launched at a lake. When it is placed in the water, it sinks into the water enough to displace 4300000 $cm^3$ of water. Assuming that this is enough for the boat to float, determine the mass of the motor boat.

1. 430000 $g$
2. 4300 $kg$
3. 43000 $kg$
o Imagine that you have a container of fluid.
- From what we've learned so far, we know that the pressure the fluid exerts on the sides of the container is the same everywhere. If they were not, the fluid would no longer be static.
- Now we exert a force somewhere on the outside of the container. The pressure inside the container will increase.
- The important part is that, according to Pascal's Principle, the pressure will increase everywhere in the fluid, not just where you are applying the force.
o Assume a hydraulic lift that lifts a large mass as an example of Pascal's principle application.
- On one side we have a small cylinder (filled with an incompressible liquid) with a piston. This is where we will exert a force downwards $(\Delta\ P = \frac{F_1}{A_1})$.
- This is connected by a pipe to another, larger cylinder, with a large piston, where the large mass will be lifted by a force upwards $(\Delta\ P = \frac{F_2}{A_2})$.
- Based on Pascal's principle the change in pressure must be the same at both sides i.e $\frac{F_1}{A_1} = \frac{F_2}{A_2}$ and since $A_2 > A_1$ then $F_2 > F_1$ for the sides to be equal. This just means that we can exert a small force on the small piston and get a bigger force on the big piston.
- The volume of the fluid being moved is $V = A_1 d_1$ and $V = A_2 d_2$ but $d_1 > d_2$ and substituting this in the above equality: $\frac{F_1}{A_1} (A_1 d_1) = \frac{F_2}{A_2} (A_2 d_2) \rightarrow F_1 d_1 = F_2 d_2$

I want to build a hydraulic press to be able to squeeze all my gold bars down to thin gold disks. The small piston has an radius of 1.0 $cm$ and I will be able to exert a force of 230 $N$ on it. If the large piston has a radius of 30 $cm$ determine the force against the gold bars.

1. 0.256 $N$
2. 155,000 $N$
3. 207,000 $N$
 Part 1 Part 2 Part 3 Part 4 Lecture 5 