IP3 03 Dynamics

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, [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.

DYNAMICS2
Considering 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.

DYNAMICS3
Considering 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:

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.

Newton’s 3rd Law

Activity 1: Propeller Cart (Predict, Observe and Explain)

Activity 2: Untied Balloon

A balloon is filled with air and released with its mouth downwards. Explain

  1. why it moves upwards.                      
  2. why it stops rising after some time.   

Activity 3: Water Rocket

Elevator Physics

In a recent IP3 class on Newton’s 2nd Law, the students were presented the “Elevator Problem” based on the THINK Cycle approach – a version of inquiry-based learning that was started in Temasek Junior College, Singapore.

The “Elevator Problem” is a physics phenomenon observed in an everyday experience that students can relate to quite easily. It is presented to our IP3 (K9 students) right after the introduction of Newton’s 2nd Law, with the students having a good understanding of the forces of weight and normal contact as well as what makes a resultant force.

TRIGGER

The THINK Cycle kicks off with a Trigger: a problem or phenomenon for which students have to solve or explain. In the “Elevator Problem”, the Trigger is the observation that as I stand on a bathroom scale in a lift going from one floor to another, the reading on the scale changes in such a way:

  1. When the lift starts moving, the reading on the scale increases momentarily.
  2. For most of the journey, the reading is constant.
  3. When the lift is stopping, the reading on the scale decreases momentarily

The video below (taken by myself) shows what happens:

The students are supposed to work in groups to explain this observation and hence, to deduce whether the elevator is on its way up or down.

HARNESS

In the Harness stage of the THINK Cycle, students would work in groups to answer some guiding questions to help them arrive at a conclusion:

  1. What are the forces acting on the boy?
  2. Which of these forces are constant and which can change?
  3. How does the motion of the lift affect the changing force?
  4. What force is the weighing scale showing?

I find that providing students with a small portable whiteboard or a few pieces of rough paper is necessary for them to represent their ideas in diagram form, especially when the objectives of this activity is best achieved with the help of free-body diagrams.

INVESTIGATE

After coming up with a hypothesis based on their discussions, they will then seek to verify their hypothesis. Task number 2, which is for students to determine whether the elevator is going up or down, can be tested by hanging a 500 g mass on a force meter attached to a datalogger. We use the Addestation aMixer in our school, which is a handy portable datalogger with a plug-and-play range of user-friendly sensors. It gives us a graph that looks like that shown below when the mass is being pulled upwards, thus confirming that the movement of the elevator is also upward.

Variation of tension with time as the mass is pulled upwards.

The initial increase in tension acting on the mass is similar to that of the normal contact force on the man standing on the bathroom scale on the elevator. This is because both systems are accelerating upward.

The graph looks rather haphazard as the pulling is done manually and over a small height. By the time one pulls the mass up, he will have to decelerate already, which explains the dip in tension that follows right after the peak. Hence, we are unable to observe a stage where the tension is equal to weight, as we did for the scale in the elevator.

Nevertheless, students should be able to appreciate that a rise followed by a drop is observed for a mass being pulled upward.

NETWORK

For the sake of checking what the students have learnt collaboratively, each group is tasked to explain their observation and results on a A2-sized poster, with half the group staying at their own posters to answer questions while the other half going around to study the results from other groups. Their roles can be reversed after some time.

KNOW

In the final stage of our activity, the teacher will address the class and point out the common misconceptions that arose during the class discussions. For instance, many students are unaware that the upward force acting on the person standing on a weighing scale is the normal contact force and not the gravitational pull. This requires the teacher to introduce the terms “apparent weight” and “true weight” and making a distinction between the two.