Forces: Free Body Diagrams and Equilibrium

Mastering forces is the cornerstone of Newtonian mechanics and a non-negotiable skill for success in IB Physics. Whether predicting the motion of a car or the stability of a bridge, the analysis always begins with a single, critical step: identifying all the interactions on an object.

The Foundation: Constructing an Accurate Free Body Diagram

A free body diagram (FBD) is a simplified sketch that isolates a single object and represents all the external forces acting on it as vectors originating from a central point. This act of isolation is your most powerful analytical tool. To draw one correctly, you must mentally "cut" the object free from its environment and replace every connection with a force vector. Common forces include:

• Weight ($\vec{W}$ or $m\vec{g}$): Always acts vertically downward from the object's center of mass.
• Normal Force ($\vec{N}$ or ${\vec{F}}_N$): A perpendicular contact force exerted by a surface. It is not always equal to weight.
• Tension ($\vec{T}$): A pulling force exerted by strings, ropes, or cables. Tension always acts away from the object along the line of the string.
• Friction ($\vec{f}$): A force opposing motion or the tendency of motion, acting parallel to the contact surface.
• Applied Force (${\vec{F}}_{app}$): Any direct push or pull.

For example, consider a book resting on a table. Its FBD would show only two forces: weight ($mg$) downward and the normal force ($N$) upward. If you push the book horizontally, you add an applied force ($F_{app}$) in the direction of the push. The diagram must only include forces acting on the chosen object, never forces that the object exerts on something else.

Resolving Forces and Newton's Second Law on an Incline

Forces rarely align neatly with your chosen coordinate system. Resolving forces into perpendicular components is the mathematical technique that unlocks analysis. You break a force vector into two mutually perpendicular vectors (typically horizontal and vertical, or parallel and perpendicular to a surface) whose combined effect is identical to the original force.

This is indispensable for inclined plane problems. Here, the most efficient coordinate system is tilted: one axis parallel to the slope (x'), and one perpendicular to it (y'). The weight vector ($mg$) is resolved into these two components using trigonometry. The component parallel to the slope is $mg\sin \theta$, which causes the object to accelerate down the plane. The component perpendicular to the slope is $mg\cos \theta$, which is balanced by the normal force ($N$). Applying Newton's second law ($\sum \vec{F}=m\vec{a}$) along each axis separately allows you to solve for acceleration, normal force, or required applied forces.

Worked Example: A 5.0 kg block slides down a frictionless plane inclined at $30^{\circ }$. Find its acceleration.

1. FBD: Forces on the block are weight ($mg$) straight down and normal force ($N$) perpendicular to the slope.
2. Resolve Weight: $F_{\parallel }=mg\sin \theta$; $F_{\perp }=mg\cos \theta$.
3. Apply Newton's Second Law:

• Perpendicular (y'): $N-mg\cos \theta =0$ (no acceleration off the plane).
• Parallel (x'): $mg\sin \theta =ma$.

1. Solve: $a=g\sin \theta =(9.8{\text{m s}}^{-2})(\sin 30^{\circ })=4.9{\text{m s}}^{-2}$.

The Nature of Friction: Static and Kinetic

Friction is a contact force that opposes relative motion. Its nature changes depending on whether surfaces are sliding past each other.

• Static Friction ($f_s$): Acts to prevent motion from starting. It is a responsive force that matches the applied force up to a maximum value. The maximum static friction is given by $f_{s,max}={\mu }_{s}N$, where ${\mu }_s$ is the coefficient of static friction.
• Kinetic Friction ($f_k$): Acts to oppose motion that is already occurring. It is generally constant and given by $f_k={\mu }_{k}N$, where ${\mu }_k$ is the coefficient of kinetic friction (${\mu }_k<{\mu }_s$).

A critical concept is that the force of friction ($f_s$ or $f_k$) is proportional to the normal force ($N$), not the weight. On an inclined plane, $N=mg\cos \theta$, so friction decreases as the angle increases.

Conditions for Translational Equilibrium

An object is in translational equilibrium when its linear acceleration is zero. This does not necessarily mean it is at rest; it could be moving with constant velocity. According to Newton's first law, the condition for translational equilibrium is that the vector sum of all forces acting on the object is zero: $$\sum \vec{F}=0$$ This single vector equation breaks down into two powerful scalar conditions: the sum of forces in the x-direction is zero, and the sum of forces in the y-direction is zero ($\sum F_x=0$ and $\sum F_y=0$). You use these two equations to solve for up to two unknown forces in a static equilibrium problem.

Analyzing Systems: Connected Bodies and Pulleys

Real-world problems often involve multiple connected objects, like boxes linked by strings or masses hanging over pulleys. The analysis strategy is systematic:

1. Isolate Each Object: Draw a separate FBD for each mass.
2. Identify Interactions: A string or rope exerts a tension force ($T$) on each object it connects. For a light, inextensible string and a frictionless pulley, the tension has the same magnitude throughout the string.
3. Apply Newton's Second Law: Write $\sum F=ma$ for each object. For connected objects, their accelerations will have the same magnitude (they are linked).
4. Solve the System of Equations: Combine the equations from each object to solve for unknowns like acceleration and tension.

In a pulley system where one mass ($m_1$) hangs vertically and is connected over a pulley to another mass ($m_2$) on a horizontal surface, you treat each mass separately. For $m_1$: $m_{1}g-T=m_{1}a$. For $m_2$: $T-f_k=m_{2}a$. The acceleration $a$ is the same for both, allowing you to solve for $a$ and $T$.

Common Pitfalls

1. Including Non-Forces or Wrong Forces in the FBD: Velocity and acceleration are not forces. Do not draw them. The force of gravity (weight) is not the normal force. Ensure every force is an interaction with a specific, identifiable external agent (e.g., Earth, rope, surface).
2. Misapplying Friction Formulae: Remember $f_{s,max}={\mu }_{s}N$ is the maximum possible static friction. The actual static friction force matches the net applied force up to that maximum. Use $f_k={\mu }_{k}N$ only when objects are sliding. Always calculate $N$ correctly from force analysis; it is not automatically $mg$.
3. Incorrectly Handling Tension in Connected Systems: The most common error is assuming tension in a string is equal to the weight of a hanging mass. This is only true if the mass is in equilibrium or the string is massless and the pulley is frictionless and the system is not accelerating. Always derive tension from applying Newton's second law to the system.
4. Sign Errors in Component Resolution: Be consistent with your coordinate system. When applying $\sum F=ma$, decide which direction is positive for each axis and stick to it for all forces. A force component in the negative direction must have a negative sign in the equation.

Summary

• A free body diagram (FBD) is the essential first step, isolating an object to show only the external forces acting upon it.
• Resolving forces into perpendicular components (especially parallel and perpendicular to an incline) is required to apply Newton's second law effectively.
• Friction is modeled as static ($f_s\le {\mu }_{s}N$) or kinetic ($f_k={\mu }_{k}N$), is proportional to the normal force, and always opposes motion or its tendency.
• The condition for translational equilibrium is $\sum \vec{F}=0$, which provides two independent scalar equations ($\sum F_x=0$, $\sum F_y=0$) to solve for unknown forces.
• For connected bodies and pulley systems, draw separate FBDs for each mass, recognize that tension in a single string has constant magnitude (under ideal conditions), and solve the simultaneous equations from applying Newton's second law to each object.