03 Dynamics

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.

Collision Simulation

I created this post here to bookmark some useful tools for use during my upcoming JC1 lectures on Dynamics.

This is a simulation for collisions that show the momenta before and after collisions. It requires registration after one visit.

A better choice for now could be the EJSS version (created by my ex-colleague Lawrence) which is far more detailed.

I had wanted to build one using GeoGebra and in fact, was halfway through it, but the Covid-19 pandemic has created other areas of work that now take priority.

Two Body Problems in Dynamics

Problems involving two bodies moving together usually involve asking for the magnitude of the force between the two.

For example:

A 1.0 kg and a 2.0 kg box are touching each other. A 12 N horizontal force is applied to the 2.0 kg box in order to accelerate both boxes across the floor. Ignoring friction, determine:

(a) the acceleration of the boxes, and

(b) the force acting between the boxes.

To solve for (b) requires an understanding that the free-body diagram of the 1.0 kg box can be considered independently as only the force acting between the two boxes contributes to its acceleration since it is the only force acting on it in the horizontal direction.

This interactive app allows for students to visualise the forces acting on the boxes separately as well as a single system.

The codes for embedding into SLS:

<iframe scrolling="no" title="Two Mass Problem" src="https://www.geogebra.org/material/iframe/id/fh5pwc37/width/638/height/478/border/888888/sfsb/true/smb/false/stb/false/stbh/false/ai/false/asb/false/sri/false/rc/false/ld/false/sdz/false/ctl/false" width="638px" height="478px" style="border:0px;"> </iframe>

Man in Elevator Simulation

man_in_elevator_simulation

In this simulation, students can observe the variation of the normal contact force (N) and its effect on acceleration and velocity as an elevator moves upward.

Questions for students to work on can include:

  1. Express the acceleration as a function of Normal Contact Force (N), Weight (W) and mass of the man.
  2. Determine the distance travelled by the elevator.
  3. Predict how the forces, acceleration and velocity will differ if the elevator was moving down instead.

Newton’s 2nd Law Experiment using Motion Sensor

For my students: To download the file and video for analysis using Tracker, right-click the file here…


To verify the equation F = ma, where F is the resultant force on an object, m is the mass of the object and a is the acceleration, this is one of the ways to do so:

Equipment:
1. Motion Sensor
2. Datalogger
3. Cart with variable mass
4. End Stop
5. Pulley with clamp
6. Hanger Mass Set
7. String (about 1.2 m)

For a system of a cart of mass M on a horizontal track that is connected to a hanging mass m with a string over a pulley, the net force F on the entire system (cart and hanging mass) is the weight of hanging mass. F = mg (no friction assumed).

newton 2nd law experiment

According to Newton’s Second Law, mg = (M+ m)a. We will try to prove experimentally that this is true in the video below.