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 ladder rests on rough ground and leans against a rough wall. Its weight W acts through the centre of gravity G. Forces also act on the ladder at P and Q. These forces are P and Q respectively.
Which vector triangle represents the forces on the ladder?
It’s Day 1 of the full home-based learning month in Singapore! As teachers all over Singapore scramble to understand the use of the myriad EdTech tools, I have finally come to settle on a few:
Google Meet to do video conferencing
Google Classroom for assignment that requires marking
Student Learning Space for students’ self-directed learning, collaborative discussion and formative assessment.
Loom for lecture recording
GeoGebra for visualisation
The following is a video that was created using Loom to explain a question on why tension in a rope on which a weight is balanced increases when the rope straightens.
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:
the body isolated from other objects
the forces acting on the body (ignore all internal forces) along the line of motion.
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?
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
Get students to explain how the ketchup packet sinks and floats.
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.
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.
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.
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:
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.
