IP Topics

Forces in Equilibrium

While preparing for a bridging class for those JAE JC1s who did not do pure physics in O-levels, I prepared an app on using a vector triangle to “solve problems for a static point mass under the action of 3 forces for 2-dimensional cases”.

For A-level students, they can be encouraged to use either the sine rule or the cosine rule to solve for magnitudes of forces instead of scale drawing, which is often unreliable.

For students who are not familiar with these rules, here is a simple summary:

Sine Rule

If you are trying to find the length of a side while knowing only two angles and one side, use sine rule:

$$\dfrac{A}{\sin{a}}=\dfrac{B}{\sin{b}}$$

Cosine Rule

If you are trying to find the length of a side while knowing only one angle and two sides, use cosine rule:

$$A^2 = B^2 + C^2 – 2BC\cos{a}$$

Newton’s 2nd Law Questions

Here are the steps to solving a problem involving Newton’s 2nd law.

Step 1: Draw a free body diagram – where possible, sketch a free body diagram to represent:

  1. the body isolated from other objects
  2. the forces acting on the body (ignore all internal forces) along the line of motion.
  3. the direction of acceleration.

Step 2: Write down the equation: $$F_{net}=ma$$

Step 3: Add up the forces on the left-hand side of the equation, making sure that forces acting opposite to the direction of acceleration are subtracted.

Step 4: Solve the equation for the unknown.

For example,

A horizontal force of 12 N is applied to a wooden block of mass 0.60 kg on a rough horizontal surface, and the block accelerates at 4.0 m s-2. What is the magnitude of the frictional force acting on the block?

Step 1: Free-body diagram

Newton's 2nd law

Step 2: Write down the 2nd law equation

By Newton’s 2nd law, $$F_{net}=ma$$

Step 3: Add up the forces making up the net force

$$12 – f = 0.60 (4.0)$$

Step 4: Solve for unknown

$$ f = 12 – 2.4 = 9.6 N$$

Free-Body Diagrams in Two-Body Motion


Students are often confused about the forces in drawing free-body diagrams, especially so when they have to consider the different parts of multiple bodies in motion.

Two-Body Motion

Let’s consider the case of a two-body problem, where, a force F is applied to push two boxes horizontally. If we were to consider the free-body diagram of the two boxes as a single system, we only need to draw it like this.

two-body
Considering both boxes as a single system

For the sake of problem solving, there is no need to draw the normal forces or weights since they cancel each other out, so the diagram can look neater. Applying Newton’s 2nd law of motion, $F=(m_A+m_B) \times a$, where $m_A$ is the mass of box A, $m_B$ is the mass of box B, F is the force applied on the system and a is the acceleration of both boxes.

You may also consider box A on its own.

DYNAMICS2
Considering box A on its own

The equation is $F-F_{AB}=m_A \times a$, where $F_{AB}$ is the force exerted on box A by box B.

The third option is to consider box B on its own.

DYNAMICS3
Considering box B on its own

The equation is $F_{BA}=m_B \times a$, where $F_{BA}$ is the force exerted on box B by box A. Applying Newton’s 3rd law, $F_{BA}=F_{AB}$ in magnitude.

Never Draw Everything Together

NEVER draw the free-body diagram with all the forces and moving objects in the same diagram, like this:

wrong freebody diagram

You will not be able to decide which forces acting on which body and much less be able to form a sensible equation of motion.

Interactive

Use the following app to observe the changes in the forces considered in the 3 different scenarios. You can vary the masses of the bodies or the external force applied.

Multiple-Body Problems

For the two-body problem above, we can consider 3 different free-body diagrams.

For three bodies in motion together, we can consider up to 6 different free-body diagrams: the 3 objects independently, 2 objects at a go, and all 3 together. Find the force between any two bodies by simplifying a 3-body diagram into 2 bodies. This trick can be applied to problems with even more bodies.

Relationship between displacement-time and velocity-time graphs

Through this GeoGebra app, students can observe how the gradient of the displacement-time graph gives the instantaneous velocity and how the area under the velocity-time graph gives the change in displacement.

In the GeoGebra app below, you will see a displacement-time graph on the left and its corresponding velocity-time graph on the right. These graphs will be referring to the same motion occuring in a straight line. Instructions

  1. Click “Play” and observe the values of displacement and velocity change in each graph over time.
  2. Note the relationship between the gradient in the displacement-time graph and the value of velocity.
  3. Note the relationship between the area under the velocity-time graph and the value of displacement.