Additional: Free Body Diagrams and Problem Solving

Drawing Free Body Diagrams

Solving physics problems involving forces often begins with clearly identifying all the forces acting on the object(s) of interest. A Free Body Diagram (FBD) is an indispensable tool for this purpose.

What is a Free Body Diagram?

A Free Body Diagram is a diagram that represents a single object (or a system of objects treated as a single unit) isolated from its surroundings, with all the external forces acting ON that object shown as vectors. It is a visual representation used to analyse the forces and predict the motion of the object according to Newton's laws.

Purpose of Drawing an FBD

The main purposes of drawing an FBD are:

To clearly identify all the forces acting on a specific object.
To visualise the directions of these forces.
To help apply Newton's Second Law ($ \sum \vec{F} = m\vec{a} $) correctly by summing up all the vectors representing the forces shown in the diagram.

Steps to Draw a Free Body Diagram

Follow these steps to construct an accurate FBD:

1. Identify the Object: Choose the specific object or system you want to analyse. Draw a simple sketch of this object (a box, a dot representing a particle, etc.). Isolate this object mentally from everything else.

2. Identify All External Forces: Think about everything outside the object that is interacting with it and exerting a force on it. These are the external forces. Common external forces include:

Weight ($ \vec{W} $ or $ m\vec{g} $): Force due to gravity, always directed vertically downwards towards the centre of the Earth.
Normal Force ($ \vec{N} $): Force exerted by a surface, perpendicular to the surface, pushing *on* the object.
Tension ($ \vec{T} $): Force exerted by a stretched string or rope, pulling *on* the object, along the string.
Friction ($ \vec{f} $): Force exerted by a surface, parallel to the surface, opposing relative motion or impending motion.
Applied Force ($ \vec{F}_{app} $): Any other push or pull exerted by an external agent (hand, engine, etc.).
Air Resistance/Drag ($ \vec{F}_D $): Force exerted by the fluid (air/water), opposing the motion through the fluid.

3. Draw Force Vectors: Draw an arrow from the centre of the object (for simplicity, if treating as a particle) representing each identified external force. The length of the arrow can roughly indicate the magnitude of the force, but the direction is critical. Label each vector with the appropriate symbol ($ \vec{W}, \vec{N}, \vec{T}, \vec{f} $, etc.).

4. Do NOT Include:

Forces exerted *by* the object on its surroundings (these would be shown on the FBDs of other objects, according to Newton's Third Law).
Internal forces within the object if treating it as a single rigid body.
Velocity or acceleration vectors (these describe the motion, not the forces causing the motion. While you might indicate the direction of acceleration *next* to the FBD, the FBD itself shows only forces).

5. Indicate Coordinate System (Optional but Recommended): For applying Newton's laws, it is helpful to draw a suitable coordinate system near the FBD.

Examples of Free Body Diagrams

Example 1: Book on a Horizontal Table

Forces on the book:

Weight ($ \vec{W} $) downwards.
Normal force ($ \vec{N} $) upwards, perpendicular to the table surface.

Example 2: Mass Hanging by a String

Forces on the mass:

Weight ($ \vec{W} $) downwards.
Tension ($ \vec{T} $) upwards, along the string.

Example 3: Block on an Inclined Plane (No Friction, No Applied Force)

FBD of a block on an inclined plane without friction

Forces on the block:

Weight ($ \vec{W} $) vertically downwards.
Normal force ($ \vec{N} $) perpendicular to the inclined surface, upwards.

Example 4: Block being Pulled on a Horizontal Surface with Friction

FBD of a block being pulled on a horizontal surface with friction

Forces on the block:

Weight ($ \vec{W} $) downwards.
Normal force ($ \vec{N} $) upwards, perpendicular to the surface.
Applied force ($ \vec{F}_{app} $) in the direction of pull.
Friction force ($ \vec{f}_k $ or $ \vec{f}_s $) opposing motion, parallel to the surface. (Shown here opposing rightward motion).

Drawing accurate FBDs is the critical first step in solving almost any force-related problem in mechanics. It helps prevent missing forces or including spurious ones.

Applying Newton's Laws in Different Coordinates

Newton's Second Law, $ \vec{F}_{net} = m\vec{a} $, is a vector equation. This means that the net force and the acceleration are vectors and must be treated accordingly. To use this equation in calculations, it is usually most convenient to work with the components of the vectors along chosen coordinate axes.

Vector Equation to Scalar Equations

The vector equation $ \vec{F}_{net} = m\vec{a} $ can be broken down into a set of independent scalar equations, one for each dimension, in a chosen coordinate system. For a 3D problem with x, y, and z axes:

$ \vec{F}_{net} = \sum \vec{F} = (\sum F_x) \hat{i} + (\sum F_y) \hat{j} + (\sum F_z) \hat{k} $

$ \vec{a} = a_x \hat{i} + a_y \hat{j} + a_z \hat{k} $

So, $ (\sum F_x) \hat{i} + (\sum F_y) \hat{j} + (\sum F_z) \hat{k} = m (a_x \hat{i} + a_y \hat{j} + a_z \hat{k}) $.

This equality holds if and only if the components along each direction are equal:

$ \sum F_x = ma_x $ $ \sum F_y = ma_y $ $ \sum F_z = ma_z $

These are the scalar component equations of Newton's Second Law. Solving a mechanics problem often involves applying these equations after resolving all forces into components.

Choosing the Right Coordinate System

The choice of the coordinate system does not affect the physics of the problem, but a smart choice can significantly simplify the mathematical calculations. Here are some common strategies:

Rectangular (Cartesian) Coordinates (x-y): Best for problems where motion or forces are primarily horizontal and vertical (e.g., objects on a flat surface, projectiles, hanging masses). You might align x-axis horizontally and y-axis vertically.
Tilted Coordinates: For problems involving inclined planes, it is often easiest to align one axis (say, the x-axis) parallel to the incline and the other axis (y-axis) perpendicular to the incline. This way, two of the three forces (normal force and friction) acting on the object on the incline will lie along the axes, simplifying their resolution. Only the weight vector needs to be resolved into components along the tilted axes.
Polar or Cylindrical Coordinates: Useful for problems involving circular motion or radial symmetry. In uniform circular motion, choosing one axis along the instantaneous radius and the other tangential can be helpful.

Resolving Forces

Once the coordinate system is chosen and the FBD is drawn, resolve each force vector into components along the chosen axes. Trigonometry (sine, cosine, tangent) is used for this. Remember that vector components can be positive or negative depending on the direction relative to the positive axis.

Example: Block on an Inclined Plane (Revisited)

Let's revisit the FBD for a block on an inclined plane at angle $ \theta $ with the horizontal, showing how to choose coordinates and resolve forces for applying Newton's laws.

FBD of a block on an inclined plane with tilted coordinates and resolved weight

Forces on the block:

Weight ($ \vec{W} $): Magnitude $ W = mg $, vertically downwards.
Normal force ($ \vec{N} $): Magnitude $ N $, perpendicular to the incline, upwards.
Friction force ($ \vec{f} $): Magnitude $ f $, parallel to the incline, opposing motion (let's assume it's static friction opposing potential downward motion, so it acts up the incline).

Choose a coordinate system with the x-axis parallel to the incline (positive direction up the incline) and the y-axis perpendicular to the incline (positive direction upwards, away from the plane).

Resolve forces:

Normal Force ($ \vec{N} $): Already along the y-axis. $ N_x = 0 $, $ N_y = +N $.
Friction Force ($ \vec{f} $): Already along the x-axis. $ f_x = +f_s $, $ f_y = 0 $ (assuming static friction pointing up-incline).
Weight ($ \vec{W} $): This needs to be resolved. The angle between the vertical weight vector and the negative y-axis (which is perpendicular to the incline) is equal to the angle of the incline, $ \theta $.

Component of Weight parallel to incline (x-component): $ -W \sin\theta = -mg \sin\theta $ (down the incline, hence negative)

Component of Weight perpendicular to incline (y-component): $ -W \cos\theta = -mg \cos\theta $ (into the incline, hence negative)

Applying Newton's Second Law components ($ \sum F_x = ma_x, \sum F_y = ma_y $):

Sum of x-components (parallel to incline):

$ \sum F_x = T_x + N_x + f_x + W_x = ma_x $

If there is no tension, $ T_x=0 $.

$ 0 + 0 + f_s - mg \sin\theta = ma_x $

$ f_s - mg \sin\theta = ma_x $

Sum of y-components (perpendicular to incline):

$ \sum F_y = T_y + N_y + f_y + W_y = ma_y $

If the block stays on the incline, $ a_y = 0 $.

$ 0 + N + 0 - mg \cos\theta = m(0) $

$ N - mg \cos\theta = 0 \implies N = mg \cos\theta $

From these equations, we can analyse different scenarios:

If the block is at rest ($ a_x=0 $), then $ f_s = mg \sin\theta $ (as long as $ mg \sin\theta \le \mu_s N = \mu_s mg \cos\theta $).
If the block slides down ($ a_x $ is negative, or we choose down as positive x), the friction is kinetic ($ f_k = \mu_k N = \mu_k mg \cos\theta $), and $ mg \sin\theta - f_k = ma_x $.

Choosing an appropriate coordinate system and carefully resolving forces are essential steps for transforming the vector law into solvable scalar equations.

Solving Problems Involving Connected Bodies

Many mechanics problems involve two or more objects that are physically connected, such as being tied together by strings, in direct contact, or linked by rods and pulleys. These are called connected bodies. The motion of one body in the system is often related to the motion of the others due to the connections.

Strategy for Connected Bodies

Problems with connected bodies are solved by applying Newton's Second Law to each individual body separately or sometimes to the entire system as a whole. The forces acting between the connected bodies (like tension in a string or contact force between blocks) are internal forces for the entire system but external forces for the individual bodies.

The general strategy involves:

1. Identify Each Body: Treat each object or distinct part of the system as a separate body for analysis.

2. Draw FBD for Each Body: Draw a complete Free Body Diagram for *each* chosen body, showing all external forces acting on that specific body.

3. Choose Coordinate Systems: Select appropriate coordinate systems for each body. Be consistent with directions, especially if their motions are linked (e.g., if one mass moves up, the connected mass moves down). Often, it's helpful to align an axis with the expected direction of motion for each body.

4. Apply Newton's Second Law to Each Body: Write down the vector equation $ \sum \vec{F} = m\vec{a} $ for each body, and then express it as scalar component equations based on the chosen coordinate system for that body.

5. Identify Constraint Equations: Write down any equations that relate the motion of the bodies. These are usually kinematic constraints based on the connections:

If connected by an inextensible string passing over a fixed pulley, the magnitudes of their velocities and accelerations along the string are the same ($ |\vec{v}_1| = |\vec{v}_2|, |\vec{a}_1| = |\vec{a}_2| $). The directions are linked by how they are connected (e.g., one moves up, the other moves down).
If blocks are in contact and move together, they have the same velocity and acceleration.

6. Use Newton's Third Law for Interaction Forces: Forces acting between connected bodies (tension, contact forces) are action-reaction pairs by Newton's Third Law. Ensure these forces are shown in the FBDs of the respective bodies with equal magnitude and opposite direction.

7. Solve the System of Equations: You will have a set of linear equations from applying Newton's Second Law to each body, plus any constraint equations. Solve this system of equations to find the unknown forces (like tension, normal contact forces) and the acceleration(s).

Example: Two Blocks in Contact on a Smooth Horizontal Surface

Example 1. Two blocks of masses $ m_1 = 5 $ kg and $ m_2 = 3 $ kg are placed side-by-side in contact with each other on a smooth horizontal surface. A horizontal force of $ F = 40 $ N is applied to the 5 kg block, pushing it towards the 3 kg block. Find (a) the acceleration of the system, and (b) the magnitude of the contact force between the two blocks.

Answer:

1. Identify Bodies: The two bodies are the 5 kg block ($ m_1 $) and the 3 kg block ($ m_2 $).

2. Draw FBD for Each Body:

Assume positive direction is to the right.

FBD for $ m_1 $ (5 kg block):

Forces on $ m_1 $:

Applied force ($ \vec{F} $): 40 N to the right.
Weight ($ \vec{W}_1 $): $ W_1 = m_1 g $ downwards.
Normal force from surface ($ \vec{N}_1 $): upwards.
Contact force from $ m_2 $ on $ m_1 $ ($ \vec{F}_{2on1} $): to the left (pushes back on $m_1$).

FBD of block 1 pushed by external force and contacted by block 2

FBD for $ m_2 $ (3 kg block):

Forces on $ m_2 $:

Weight ($ \vec{W}_2 $): $ W_2 = m_2 g $ downwards.
Normal force from surface ($ \vec{N}_2 $): upwards.
Contact force from $ m_1 $ on $ m_2 $ ($ \vec{F}_{1on2} $): to the right (pushes $m_2$).

By Newton's Third Law, the contact forces form an action-reaction pair: $ \vec{F}_{1on2} = - \vec{F}_{2on1} $. Let $ P = |\vec{F}_{1on2}| = |\vec{F}_{2on1}| $ be the magnitude of the contact force.

3. Choose Coordinate Systems: Use standard x-y axes for both blocks, with positive x to the right and positive y upwards.

4. Apply Newton's Second Law to Each Body:

For $ m_1 $:

Vertical (y-axis): $ \sum F_y = N_1 - W_1 = m_1 a_y $. Since there is no vertical acceleration, $ a_y = 0 $. $ N_1 - m_1 g = 0 \implies N_1 = m_1 g $.

Horizontal (x-axis): $ \sum F_x = F - F_{2on1} = m_1 a_x $. Let the magnitude of contact force be $P$. $ F_{2on1} $ is to the left, so its component is $-P$.

$ F - P = m_1 a_x \quad \text{(Equation 1)} $

For $ m_2 $:

Vertical (y-axis): $ \sum F_y = N_2 - W_2 = m_2 a_y $. Since $ a_y = 0 $. $ N_2 - m_2 g = 0 \implies N_2 = m_2 g $.

Horizontal (x-axis): $ \sum F_x = F_{1on2} = m_2 a_x $. $ F_{1on2} $ is to the right, so its component is $+P$.

$ P = m_2 a_x \quad \text{(Equation 2)} $

5. Identify Constraint Equations: Since the blocks are in contact and pushed together, they must move with the same acceleration. Let $ a_x = a $.

$ a_1 = a_2 = a $

6. Use Newton's Third Law: Already incorporated by setting magnitude of $ \vec{F}_{1on2} $ and $ \vec{F}_{2on1} $ equal to $ P $.

7. Solve the System of Equations:

Substitute $ m_1 = 5 $ kg, $ m_2 = 3 $ kg, $ F = 40 $ N, and $ a_x = a $ into Equations 1 and 2:

Equation 1: $ 40 - P = 5a $

Equation 2: $ P = 3a $

(a) To find the acceleration $ a $, substitute $ P = 3a $ from Equation 2 into Equation 1:

$ 40 - (3a) = 5a $

$ 40 = 5a + 3a $

$ 40 = 8a $

$ a = \frac{40}{8} = 5 \, \text{m/s}^2 $

The acceleration of the system is $ 5 \, \text{m/s}^2 $ to the right.

(b) To find the contact force $ P $, substitute the value of $ a $ back into Equation 2:

$ P = 3a = 3 \times (5 \, \text{m/s}^2) = 15 \, \text{N} $

The magnitude of the contact force between the blocks is 15 N.

10. Check the Answer:

Let's check if Equation 1 is satisfied with $ a = 5 $ m/s$^2$ and $ P = 15 $ N: $ 40 - 15 = 25 $, and $ 5a = 5 \times 5 = 25 $. Yes, $ 25=25 $. The values are consistent. The contact force (15 N) is less than the applied force (40 N), which makes sense because the applied force is pushing both masses, while the contact force is only pushing the second mass.

This example illustrates the general approach: draw FBDs for each component, apply Newton's laws to each, identify kinematic links, and solve the resulting system of equations.