A Level Mechanics: Forces, Moments and Kinematics

Mechanics is where mathematics meets the physical world. In A Level Mathematics, and in the mechanics content of FSc Physics, you model moving cars, sliding blocks and balanced beams using a small set of powerful equations. Once you learn to draw a clear diagram and choose the right equation, most questions follow a predictable path.

The Modelling Mindset

Every mechanics question starts with a model, which is a simplified picture of reality. Common modelling assumptions include:

• Particle: the object has mass but no size, so it does not rotate.
• Smooth surface: there is no friction.
• Light string or rod: its mass is ignored.
• Inextensible string: it does not stretch, so connected objects share the same acceleration.
• Uniform rod: its weight acts at its midpoint.

You should be able to state these assumptions and explain what effect they have. For example, ignoring air resistance usually means a model predicts a slightly greater speed than real life.

Throughout this article we take \( g = 9.8 \text{ m s}^{-2} \).

Kinematics: Motion with Constant Acceleration

Kinematics describes motion without asking what causes it. When acceleration is constant, five quantities are linked: displacement \( s \), initial velocity \( u \), final velocity \( v \), acceleration \( a \) and time \( t \). These are the SUVAT equations:

$$ v = u + at, \qquad s = ut + \tfrac{1}{2}at^2, \qquad v^2 = u^2 + 2as $$

$$ s = \tfrac{1}{2}(u + v)t, \qquad s = vt - \tfrac{1}{2}at^2 $$

Each equation leaves out one of the five quantities. The trick is to list what you know, identify what you need, and pick the equation that skips the quantity you do not care about.

Worked Example 1: An Accelerating Car

A car travelling at \( 4 \text{ m s}^{-1} \) accelerates uniformly at \( 2.5 \text{ m s}^{-2} \) for 6 seconds. Find its final speed and the distance it covers.

Step 1. List the values: \( u = 4 \text{ m s}^{-1} \), \( a = 2.5 \text{ m s}^{-2} \), \( t = 6 \text{ s} \). We want \( v \) and \( s \).

Step 2. Final speed:

$$ v = u + at = 4 + 2.5 \times 6 = 4 + 15 = 19 \text{ m s}^{-1} $$

Step 3. Distance:

$$ s = ut + \tfrac{1}{2}at^2 = 4 \times 6 + 0.5 \times 2.5 \times 36 = 24 + 45 = 69 \text{ m} $$

Step 4. Check with a different equation. Using \( v^2 = u^2 + 2as \): \( 16 + 2 \times 2.5 \times 69 = 16 + 345 = 361 \), and \( 19^2 = 361 \). The answers agree.

Graphs of Motion

On a velocity-time graph, the gradient gives acceleration and the area under the graph gives displacement. On a displacement-time graph, the gradient gives velocity. When acceleration is not constant, calculus takes over: \( v = \frac{ds}{dt} \) and \( a = \frac{dv}{dt} \), so you integrate to go back from acceleration to velocity to displacement.

Forces and Newton's Laws

Newton's three laws are the foundation of dynamics:

1. An object stays at rest or moves at constant velocity unless a resultant force acts on it.
2. The resultant force equals mass times acceleration, \( F = ma \).
3. When two objects interact, they exert equal and opposite forces on each other.

The most common forces are weight \( mg \), the normal reaction \( R \), tension \( T \), and friction \( F \). For friction, the limiting value is

$$ F_{\max} = \mu R $$

where \( \mu \) is the coefficient of friction. Friction only reaches \( \mu R \) when the object is moving or about to move.

Worked Example 2: A Block Sliding Down a Rough Slope

A block of mass 5 kg is released from rest on a rough slope inclined at \( 30^\circ \) to the horizontal. The coefficient of friction is 0.2. Find the acceleration of the block.

Step 1. Resolve perpendicular to the slope. There is no acceleration in this direction, so

$$ R = mg\cos 30^\circ = 5 \times 9.8 \times 0.8660 = 42.44 \text{ N} $$

Step 2. Find the friction force, which acts up the slope because the block slides down:

$$ F = \mu R = 0.2 \times 42.44 = 8.49 \text{ N} $$

Step 3. Find the component of weight down the slope:

$$ mg\sin 30^\circ = 5 \times 9.8 \times 0.5 = 24.5 \text{ N} $$

Step 4. Apply Newton's second law along the slope:

$$ 24.5 - 8.49 = 5a \quad\Rightarrow\quad a = \frac{16.01}{5} = 3.20 \text{ m s}^{-2} $$

Sanity check: on a smooth slope the acceleration would be \( g\sin 30^\circ = 4.9 \text{ m s}^{-2} \). Friction should reduce this, and 3.20 is indeed smaller. The block slides because 24.5 N is greater than the maximum friction of 8.49 N.

Moments: Turning Effects

A moment measures how much a force tends to turn an object about a point:

$$ \text{Moment} = \text{force} \times \text{perpendicular distance from the pivot} $$

For a rigid body in equilibrium, two conditions hold:

• The resultant force is zero (forces up equal forces down).
• The total clockwise moment about any point equals the total anticlockwise moment about that point.

A smart choice of pivot removes unknown forces from the equation, because a force acting through the pivot has zero moment.

Worked Example 3: A Beam on Two Supports

A uniform beam AB is 4 m long and has mass 20 kg. It rests horizontally on a support at A and a support at C, where AC = 3 m. A 10 kg load is placed at B. Find the reaction forces at A and C.

Step 1. Identify the forces. The beam's weight is \( 20g = 196 \text{ N} \) at its midpoint, 2 m from A. The load is \( 10g = 98 \text{ N} \) at B, 4 m from A. The reactions are \( R_A \) and \( R_C \).

Step 2. Take moments about A, which removes \( R_A \):

$$ R_C \times 3 = 196 \times 2 + 98 \times 4 = 392 + 392 = 784 $$

$$ R_C = \frac{784}{3} = 261.3 \text{ N} $$

Step 3. Resolve vertically:

$$ R_A + R_C = 196 + 98 = 294 \quad\Rightarrow\quad R_A = 294 - 261.3 = 32.7 \text{ N} $$

Check: take moments about C instead. With A on the left, the upward force at A (3 m left of C) turns the beam clockwise, the beam's weight (1 m left of C) turns it anticlockwise, and the load at B (1 m right of C) turns it clockwise. Clockwise: \( 32.7 \times 3 + 98 \times 1 = 98 + 98 = 196 \text{ N m} \). Anticlockwise: \( 196 \times 1 = 196 \text{ N m} \). Balanced. Notice how carefully each turning direction had to be decided; this is where most sign errors creep in.

The small value of \( R_A \) also makes sense: the heavy load at B pulls the beam's weight towards C, so the support at C carries most of it. If the load moved further out, \( R_A \) would fall to zero and the beam would start to tip about C.

Common Mistakes to Avoid

• Mixing up mass (kg) and weight (N). Weight is \( mg \).
• Forgetting that friction always opposes motion or the tendency to move.
• Using SUVAT when acceleration is not constant.
• Taking moments with distances that are not perpendicular to the force.
• Skipping the diagram. A clear force diagram prevents most sign errors.

Did You Know?

• Galileo is traditionally said to have studied falling bodies by rolling balls down inclined planes, which slowed the motion enough to time it.
• Archimedes stated the law of the lever over 2,200 years ago, and the principle of moments is essentially his result.
• The value of \( g \) varies slightly over the Earth's surface, being a little larger at the poles than at the equator.

Key Takeaways

• Mechanics questions begin with a model and a clear diagram showing every force.
• The SUVAT equations apply only when acceleration is constant; choose the one that omits the quantity you do not need.
• Newton's second law, \( F = ma \), is applied separately along each direction you resolve.
• Limiting friction is \( \mu R \), and friction always opposes the direction of motion.
• For equilibrium, resultant force is zero and moments about any point balance.

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