Free Body Diagrams: Complete Tutorial with Examples

What is a Free Body Diagram?

A free body diagram (FBD) is a graphical representation used in physics and engineering to visualize the forces acting on a single object. By isolating the object from its surroundings and representing all external forces as vectors, FBDs provide a powerful tool for applying Newton's laws of motion:

• Newton's First Law (\( \sum \vec{F} = 0 \) for objects at rest or moving with constant velocity)
• Newton's Second Law (\( \sum \vec{F} = m\vec{a} \) for accelerating objects)

Each force vector is drawn with:

1. Appropriate direction (indicating the force's line of action)
2. Reasonable relative magnitude (longer arrows for larger forces)
3. Clear labeling identifying the force type and acting object

The key principle is to include only forces acting ON the object, never forces exerted BY the object on other entities.

Examples of Free Body Diagrams with Detailed Explanations

Example 1: Book resting on a horizontal table

Forces acting on the stationary book:

1. The gravitational force (weight) \( \vec{W} = m\vec{g} \) exerted by Earth
2. The normal force \( \vec{N} \) exerted by the table surface, perpendicular to the contact interface

Since the book is in equilibrium (\( \vec{a} = 0 \)), Newton's First Law gives: \[ \sum F_y = N - W = 0 \quad \Rightarrow \quad N = W = mg \] The horizontal forces sum to zero trivially.

Example 2: Block suspended by a rope

Forces on the stationary suspended block:

1. Weight \( \vec{W} = m\vec{g} \) (downward, gravitational)
2. Tension \( \vec{T} \) (upward, along the rope)

Equilibrium condition yields: \[ \sum F_y = T - W = 0 \quad \Rightarrow \quad T = W = mg \] The tension force arises from the rope's molecular bonds resisting stretching.

Example 3: Block pulled horizontally with friction

Forces acting on the block (assuming motion or impending motion):

1. Weight \( \vec{W} \) (downward)
2. Normal force \( \vec{N} \) (upward, perpendicular to surface)
3. Applied force \( \vec{F}_a \) (horizontal, direction of pulling)
4. Kinetic friction \( \vec{f}_f \) (opposing motion, parallel to surface)

The friction magnitude is given by \( f_f = \mu_k N \), where \( \mu_k \) is the kinetic friction coefficient. Vertical equilibrium gives \( N = W \). Horizontally: \[ \sum F_x = F_a - f_f = ma_x \] where \( a_x \) is the block's acceleration.

Example 4: Object in free fall (neglecting air resistance)

Forces on the falling object:

1. Weight \( \vec{W} = m\vec{g} \) (downward, only significant force)

Applying Newton's Second Law: \[ \sum F_y = W = mg = ma \quad \Rightarrow \quad a = g \] The object accelerates downward with gravitational acceleration \( g \approx 9.8 \, \text{m/s}^2 \).

Example 5: Box on frictionless inclined plane

Forces acting on the box:

1. Weight \( \vec{W} = m\vec{g} \) (vertically downward)
2. Normal force \( \vec{N} \) (perpendicular to the incline)

Resolving weight into components: \[ W_{\parallel} = mg\sin\theta \quad \text{(down the incline)} \] \[ W_{\perp} = mg\cos\theta \quad \text{(into the incline)} \] Since there's no friction, the net force along the incline is \( mg\sin\theta \), causing acceleration: \[ a = g\sin\theta \] Perpendicular to the incline: \( N = mg\cos\theta \).

Example 6: Box on inclined plane with applied force and friction

Forces on the box:

1. Weight \( \vec{W} = m\vec{g} \)
2. Normal force \( \vec{N} \)
3. Applied force \( \vec{F}_a \) (up the incline)
4. Static friction \( \vec{f}_s \) (direction depends on motion tendency)

If the box is stationary, friction adjusts to satisfy equilibrium: \[ \sum F_{\parallel} = F_a + f_s - mg\sin\theta = 0 \] \[ \sum F_{\perp} = N - mg\cos\theta = 0 \] The maximum static friction is \( f_{s,\text{max}} = \mu_s N \), where \( \mu_s \) is the static friction coefficient.

Example 7: Block suspended by three strings

A) Free body diagram for the block (two forces):

1. Weight \( \vec{W} \)
2. Tension \( \vec{T}_3' \) from the vertical string

Vertical equilibrium: \( T_3' = W \).

B) Free body diagram for junction point P (three forces):

1. Tension \( \vec{T}_1 \) (left string)
2. Tension \( \vec{T}_2 \) (right string)
3. Tension \( \vec{T}_3 \) (vertical string, equal and opposite to \( T_3' \))

The junction is in equilibrium: \[ \sum F_x = T_{2,x} - T_{1,x} = 0 \] \[ \sum F_y = T_{1,y} + T_{2,y} - T_3 = 0 \] where \( T_3 = W \). This system solves for \( T_1 \) and \( T_2 \) given the angles.

Example 8: Two-block system with inclined plane and pulley

A) Free body diagram for block \( m_1 \) on incline:

1. Weight \( \vec{W}_1 = m_1\vec{g} \)
2. Normal force \( \vec{N} \) (perpendicular to incline)
3. Kinetic friction \( \vec{f}_f \) (opposing motion)
4. Tension \( \vec{T} \) (up the incline, from the string)

Along the incline (x-direction): \[ \sum F_x = T - f_f - m_1g\sin\theta = m_1a \] Perpendicular to incline: \( N = m_1g\cos\theta \), with \( f_f = \mu_k N \).

B) Free body diagram for hanging block \( m_2 \):

1. Weight \( \vec{W}_2 = m_2\vec{g} \) (downward)
2. Tension \( \vec{T}' \) (upward, magnitude \( T' = T \) for massless string)

Vertical equation: \[ \sum F_y = m_2g - T' = m_2a \] assuming downward is positive for \( m_2 \). The two equations combine to solve for acceleration \( a \) and tension \( T \).