One common misconception among new learners of kinematics is that acceleration of an object being thrown upward is zero at the top of the path when it is momentarily at rest. I created this interactive, along with the 3 graphs in order to help students relate the vectors to the graphical representation of motion.

I am using this post as a way to document my brief plans for tomorrow’s Google Meet lecture with the LOA students as well as to park the links to the resources and tools that I intend to use for easy retrieval.

Instruction Objectives:

apply the principle of moments to new situations or to solve related problems.

show an understanding that, when there is no resultant force and no resultant torque, a system is in equilibrium.

use a vector triangle to represent forces in equilibrium.

*derive, from the definitions of pressure and density, the equation ?=??ℎ.

*solve problems using the equation ?=??ℎ.

*show an understanding of the origin of the force of upthrust acting on a body in a fluid.

Activity 1: Find CG of ruler demonstration

Having shown them the demonstration last week, I will explain the reason why one can find the CG this way:

As I move the fingers inward, there is friction between the ruler and my finger. This friction depends on the normal contact force as $f=\mu N$.

Drawing the free-body diagram of the ruler, there are two normal contact forces acting on the ruler by my fingers. The sum of these two upward forces must be equal to the weight of the ruler. These forces vary depending on their distance from the CG. Taking moments about the centre of gravity, $$N_1\times d_1=N_2 \times d_2$$

The finger that is nearer to the CG will always have a larger normal contact force and hence, more friction. Hence, the ruler will tend to stop sliding along that finger and allow the other finger to slide nearer. When that other finger becomes closer to the CG, the ruler also stops sliding along it and tends to then slide along the first finger.

This keeps repeating until both fingers reach somewhere near the CG.

Activity 2: Moments of a Force at an Angle to the line between Pivot and Point of Action.

Recollection of the slides on moment of a force and torque of a couple.

Ask students to sketching on Nearpod’s “Draw It” slides the “perpendicular distance between axis of rotation and line of action of force” and “perpendicular distance between the lines of action of the couple” for Example 5 and 6 of the lecture notes respectively.

Mention that

axis of rotation is commonly known as where the pivot is

perpendicular distance is also the “shortest distance”

Activity 3: Conditions for Equilibrium

State the conditions for translational and rotational equilibrium

Show how translation equilibrium is due to resultant force being zero using vector addition

Show how rotational equilibrium is due to resultant moment about any axis being zero by equating sum of clockwise moments to sum of anticlockwise moments.

Go through example 7 (2 methods: resolution of vectors and closed vector triangle)

Useful tip: 3 non-parallel coplanar forces acting on a rigid body that is in equilibrium must act through the same point. Use 2006P1Q6 as example.

Go through example 8. For 8(b), there are two methods: using concept that the 3 forces pass through the same point or closed triangle.

For next lecture (pressure and upthrust):

Activity 4: Hydrostatic Pressure

Derive from definitions of pressure and density that $p = h\rho g$

Note that this is an O-level concept.

Activity 5: Something to sink about

Students are likely to come up with answers related to relative density. As them to draw a free body diagram of the ketchup packet. However, we will use the concept of the forces acting on the ketchup packet such as weight and upthrust to explain later.

Activity 6: Origin of Upthrust

I designed this GeoGebra app to demonstrate that forces due to pressure at different depths are different. For a infinitesimal (extremely small) object, the forces are equal in magnitude even though they are of different directions, which is why we say pressure acts equally in all directions at a point. However, when the volume of the object increases, you can clearly see the different in magnitudes above and below the object. This gives rise to a net force that acts upwards – known as upthrust.

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.

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, [latex]F=(m_A+m_B) \times a[/latex], where [latex]m_A[/latex] is the mass of box A, [latex]m_B[/latex] 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.

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

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

The equation is [latex]F_{BA}=m_B \times a[/latex], where [latex]F_{BA}[/latex] is the force exerted on box B by box A. Applying Newton’s 3rd law, [latex]F_{BA}=F_{AB}[/latex] 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:

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.

To catch the annular solar eclipse in MOE HQ on 26 Dec 2019, we ordered some solar glasses two weeks beforehand. Good thing they came in time despite the expected delay due to the Christmas season.
I was not equipped to take photos of the eclipse but that does not matter as there are plenty of quality photos all over social media.

The photos taken were done using my camera phone with the solar filter in front of it.

The thin “ring of fire” looks rather thick when captured by my camera as the light tends to be slightly diffused.

After going through the topic of Dynamics in the A-level syllabus recently, it strikes me that there are several common mistakes or misconceptions that students whom I have taught over the years have always struggled with. I shall lay them out here in the hope that students who read this do not repeat the same mistakes.

1. Mixing up free-body diagrams involving multiple bodies in 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. A common mistake that students make is to draw all the forces, including the internal forces within a system, on a single diagram.

To solve a two-body dynamics problems, students need to learn how to draw the 3 possible free-body diagrams. A detailed set of instructions is found in an earlier post.

2. Adding instead of subtracting momenta to find impulse

Students who do not have a strong foundation in Math may not appreciate the fact that the addition and subtracting of vectors is not done by simply adding or subtracting the magnitudes. Sometimes, students are asking to find the force acting on an object when given the initial and final momenta. The correct way to find the vector of the change in momentum is to place the tails of the initial and final momenta together, and to join the arrowhead of the initial momentum with that of the final momentum with a third vector in that direction.

They are often confused between finding the change in momentum and the act of adding two vectors together. When that happens, what teachers see is the following triangle instead.

While the magnitude of the vector found in both diagrams are the same, students will get the wrong answer if asked for the direction of the change in momentum. This is actually an important notion because the direction of change in momentum is really the direction of the net force applied on the body.

3. Using wrong signs for momentum for head-on collisions

In A-level physics, in the study of collision problems, we will deal only with head-on collisions where two objects collide along the same line. Even then, students can get confused with the sign convention associated with the direction of motion of the objects.

Applying Conservation of Momentum

For a two body head-on collision problem, always start by writing down the following equation for the conservation of momentum:

[latex]m_Au_A +m_Bu_B = m_Av_A + m_Bv_B[/latex]

where [latex]m_X[/latex] is the mass of X, [latex]u_X[/latex] is the initial velocity of X, [latex]v_X[/latex] is the final velocity of X, where X is either A or B.

Follow this up by deciding on which direction you want to define as positive. For example, if you decide that the rightward direction is positive, the value of initial velocity for mass A is positive and that for mass B is negative.

What students sometimes do is to include the negative signs into the equation before substitution, e.g. [latex]m_Au_A +m_B(-u_B) = m_Av_A + m_Bv_B[/latex]. This causes confusion as they do not know where the directions of the final velocities are and hence, do not know which vectors to assign as negative.

What we should do instead is to determine the final direction of the velocities after we have found the values. Before we are able to do so, we shall assume that the final velocities are to the right. If the final value of velocity of a mass is negative, it means that its final velocity is to the left.

Substitute the values into equation.

Assuming that the initial speeds are both 4.0 m s^{-1};, the mass of A is 3.0 kg, mass of B is 5.0 kg, and that the collision is perfectly inelastic, we have:

Concept cartoons useful for eliciting rich questioning and classroom dialogue in a non-threatening manner. Speech bubbles can either represent possible problems and questions arising from daily interactions with science or surface alternative concepts of scientific ideas.

Here’s my attempt at doing one! It’s on the Physics topic of forces. There is both truth and error in what each student says. Can you identify them?

If you agree with Michael, how do you reconcile the reason given with the observation that in vacuum, both objects fall to the bottom in the same time despite having different weights?

Stuart Naylor from Milgate House Publishing and Consultancy Ltd will be coming to Singapore from 11 to 12 June 2009 to conduct a Concept Cartoon Workshop for Sec Science Teachers.

Suggested solution

While it is true that in vacuum, both balls will reach the bottom at the same time, in a fluid, there are other forces such as viscous forces (drag) and upthrust. Viscous forces would increase with velocity until the object reaches terminal velocity, when the sum of the upward forces of drag and upthrust and the downward force of weight are equal in magnitude. With the weight of the lead ball being bigger, it will take a longer while for the sum of the upward forces to equal its weight, assuming that both balls have the same acceleration in the beginning. Hence, the terminal velocity of the lead ball is larger than that of the aluminum ball and it will fall faster.