Category: 04 Forces

# 4.6.1 Why must Three Forces Intersect at the Same Point?

A video explanation of the origin of this “rule”.

A visual “proof” of this rule.

The classic “ladder against wall” problem:

# 4.6 Static Equilibrium

Here is the “proof” for the summation of moments is constant regardless of pivot point when net force is zero. This is not required by H2 syllabus. So watch it only if you’re interested.

Below are a few worked examples

# 4.5.1 Moments

Just a video explanation on how to calculate moments.

# 4.4 Translational Equilibrium

How can a 50 g mass lift a 100 g mass?

A few worked examples on translational equilibrium:

Picture Frame

Block on Slope

Submerged Buoy

# 4.3.4 Upthrust Demonstrations

Cartesian Diver
Watch me use my telekinesis power to push the diver down.

Reverse Cartesian
My telekinesis power works both ways. Impressive sia?

Double Layer Golf Ball Float
A golf ball is barely floating in brine. Cooking oil (which is less dense than brine) is going to be poured over the top. Will the golf ball sink deeper into the brine, float higher, or what?

Water See-Saw
A tub of water is originally perched precariously on a wooden peg. When a copper bob is lowered into the tub, can the see-saw hold its balance?

Water Bridge
A tub of water containing a boat is originally perched precariously on a wooden peg. When a boat embarks on a cruise around the tub, can the tub hold its balance?

# 4.3.2 Law of Floatation

• When a foreign object is submerged in a fluid, the surrounding fluid will exert pressure forces perpendicularly into each point on the surface of the foreign object.
• Because pressure increases with depth, the resultant of these pressure forces is an vertically upward force. This force is called the force of upthrust, U.

• With some simple reasoning, we can deduce that the magnitude of the upthrust must be equal to the weight of the displaced fluid. Hence

$U={{m}_{f}}g={{rho }_{f}}{{V}_{f}}g$

• To generate a larger upthrust, a body must displace more fluid.
• Similar to weight, upthrust appears to act at a single point, the centre of gravity of the displaced fluid.

• Compare the upthrust U

$U={{m}_{f}}g={{rho }_{f}}{{V}_{f}}g$

• with the weight of the object W

$W={{m}_{o}}g={{rho }_{o}}{{V}_{o}}g$

• When an object is fully submerged, Vf = Vo. This also represents the maximum upthrust the object can generate.

• If ρo > ρf , then W > Umax.
• The object will sink all the way down.
• If ρo = ρf, then W = Umax.
• The object will neither sink nor float.
• It can hover at equilibrium at any depth in the fluid as long as it is fully submerged.
• If ρo < ρf, then W < Umax.
• The object will float partially submerged.
• It will displace just enough fluid to generate the amount of upthrust exactly equal to its weight.
• The lower the object’s density, the higher it will float.

# 4.3 Upthrust

What does this video tell us about pressure vs pressure forces?

Just a video explanation.

# 4.2.1 Atmospheric Pressure

At 101 kPa, there is (0.30 x 0.30 x 101,000) = 9090 N, or about 900 kg worth of weight pressing down on the 30 cm by 30 cm pad. Because there is about 900 kg of air sitting on top of the pad!

But of course, we can’t expect the pad to lift 900 kg. Two reasons: (1) It is NOT a total vacuum under the pad. So the net downward force would be much less. (2) The pad itself (the rubber, handle knob, etc) would have been pulled apart long before that.

# 4.2 Hydrostatic Pressure

Fluids have amazing properties. Nobody says it better than Bruce Lee.

Can you explain why the water stops leaking once the bottle is dropped?

In the following video, the pig is touring the ocean in an air shaft. When is the pig in danger, when it is near the bottom or near the surface?

# 4.1.3.1 Centre of Mass and Centre of Gravity

In this video, you can see a rigid body, formed by joining two balls, being thrown across the screen. Tracing the trajectory of either ball shows a complicated path. On the other hand, the trajectory of the CM is the familiar parabolic arc of a projectile motion. What does this tell us? When analysing the translational motion of a rigid body, we should can the entire mass of the body to be concentrated at its CM.

We can also assume that the entire weight of a rigid body acts at its CG. For this toy bird, weights cleverly positioned in the tips of the wings positions the CG of the bird just BELOW its beak. With the CG below the pivot, this bird is basically a pendulum.