Physics notes, chapter 1
Force and Motion II
Textbook Physics Form 5 (DLP), pages 1–37
What this chapter covers
- Determine the resultant of forces acting on a point, in a line, at right angles or at an angle (1.1)
- Draw free body diagrams and use F = ma for objects that are stationary, at constant velocity or accelerating (1.1)
- Resolve a force into horizontal and vertical components, and on an inclined plane (1.2)
- Explain forces in equilibrium and use the triangle of forces to find unknown forces (1.3)
- State Hooke's law, find the spring constant and elastic potential energy from a graph (1.4)
- Solve problems on springs in series and in parallel (1.4)
1.1Resultant force
The resultant force is the single force that represents the vector sum of two or more forces acting on an object. A force has both magnitude and direction, so it must be added as a vector.
| Situation | How to find the resultant, F | Example |
|---|---|---|
| Same direction | Add the magnitudes. Direction is the same as the forces. | 6 N and 8 N to the right: F = 14 N to the right |
| Opposite directions | Subtract. Direction is that of the larger force. | 8 N left and 6 N right: F = 2 N to the left |
| Perpendicular | Pythagoras: F = √(F₁² + F₂²). Angle: tan θ = F₂ ÷ F₁. | 8 N and 6 N: F = 10 N, θ = 36.9° |
| At any other angle | Scaled drawing: triangle of forces or parallelogram of forces. Measure F and θ. | Choose a scale, e.g. 1 cm = 10 N |
The resultant of two forces is largest when they act in the same direction (F₁ + F₂) and smallest when they act in opposite directions (F₁ − F₂). At any other angle it lies between the two.
Methods: textbook Activity 1.2 and Table 1.2, pp. 4–6.
Resultant force and motion
A free body diagram shows only the forces acting on the object. Newton's second law, F = ma, links the resultant force to acceleration.
| State of motion | Acceleration | Resultant force | Forces |
|---|---|---|---|
| Stationary | a = 0 | F = 0 | Balanced (weight = normal reaction) |
| Constant velocity | a = 0 | F = 0 | Engine thrust = friction |
| Accelerating | a ≠ 0 | F ≠ 0, in the direction of acceleration | Thrust > friction |
"The object is moving, so there must be a resultant force." No. At constant velocity the resultant force is zero. A resultant force is needed only to change the velocity.
Worked example: a lift
A 60 kg passenger stands in a lift. [g = 9.81 m s⁻²] Weight W = 60 × 9.81 = 588.6 N.Stationary or constant velocity: R = W = 588.6 N.
Accelerating upwards at 1.2 m s⁻²: R − W = ma, so R = 588.6 + 60 × 1.2 = 660.6 N.
Accelerating downwards at 1.2 m s⁻²: W − R = ma, so R = 588.6 − 72 = 516.6 N.
Worked example: falling coconut
A 2.0 kg coconut falls with a = 9.0 m s⁻². Resultant force F = ma = 2.0 × 9.0 = 18.0 N downwards. Weight = 2.0 × 9.81 = 19.62 N, so air resistance = 19.62 − 18.0 = 1.62 N.A trolley of mass 2.0 kg is pulled by a force of 12 N against a friction of 4.0 N. Find its acceleration.
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F = 12 − 4.0 = 8.0 N. a = F ÷ m = 8.0 ÷ 2.0 = 4.0 m s⁻².
1.2Resolution of forces
Resolution of forces is splitting one force into two perpendicular components, usually horizontal (Fx) and vertical (Fy). It is the reverse of finding a resultant.
cos goes with the angle next to the component (adjacent side) and sin with the component opposite the angle. If the angle is measured from the vertical, the two swap. Always sketch the triangle first.
Worked example: a block pulled at 30°
A 36 N pull acts 30° above the horizontal on a block of mass 2.4 kg (weight 23.5 N), friction 20 N and normal reaction 5.5 N.Tx = 36 cos 30° = 31.18 N (right). Ty = 36 sin 30° = 18.00 N (up).
Vertical: 18 + 5.5 − 23.5 = 0, so no vertical motion.
Horizontal resultant: 31.18 − 20 = 11.18 N. a = 11.18 ÷ 2.4 = 4.66 m s⁻².
Inclined plane
On a slope of angle θ, resolve the weight W into a component parallel to the plane, W sin θ, and a component perpendicular to the plane, W cos θ. The angle between the weight and the perpendicular to the plane equals the angle of the slope.
| Component of the weight | Formula | Role |
|---|---|---|
| Parallel to the plane | W sin θ | Pulls the object down the slope |
| Perpendicular to the plane | W cos θ | Balanced by the normal reaction, R = W cos θ |
Worked example: smooth slope
A block of mass 2.4 kg (weight 2.4 × 9.81 = 23.54 N) slides down a smooth slope of 60°. Parallel component = 23.54 sin 60° = 20.39 N; perpendicular component = 23.54 cos 60° = 11.77 N (equals R). a = 20.39 ÷ 2.4 = 8.50 m s⁻².A pushing force of 90 N acts at 60° below the horizontal on a lawn mower handle. Find the horizontal component and say what it does.
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Horizontal = 90 cos 60° = 45 N. It moves the mower forward against friction. The vertical component, 90 sin 60° = 77.9 N, presses the mower on the ground.
Worked examples follow textbook pp. 15–17.
1.3Forces in equilibrium
An object is in equilibrium of forces when the forces acting on it produce a zero resultant force. It is then either at rest or moving with constant velocity (a = 0).
When three forces are in equilibrium and are drawn head to tail, they form a closed triangle of forces. A lamp hung from two strings has weight W = 4 N and tensions P = 3 N (horizontal) and T = 5 N.
Two ways to solve problems
- Resolve and balance: horizontal forces sum to zero and vertical forces sum to zero. This is more accurate.
- Scaled triangle of forces: choose a scale, draw the known forces head to tail, close the triangle and measure. Alternatively use the sine rule or cosine rule.
Worked example: box on a slope
A box of weight 50 N is stationary on a 40° slope. Parallel forces balance: friction FR = 50 sin 40° = 32.14 N. Perpendicular forces balance: R = 50 cos 40° = 38.30 N. (A scaled drawing at 1 cm = 10 N gives about 32 N and 38 N.)Worked example: poster on a string
A poster of weight 12.0 N hangs from two equal string tensions, each at 35° above the horizontal. The triangle has angles 55°, 55° and 70°. By the sine rule, T ÷ sin 55° = 12 ÷ sin 70°, so T = 10.46 N.Equal tensions on a symmetrical pair of strings: resolve vertically, 2T sin θ = W (θ above the horizontal). A flatter string (smaller θ) needs a much larger tension for the same weight.
A lamp hangs from a string T = 8 N at 60° above the horizontal and a horizontal string P. Find P and the weight W.
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P = 8 cos 60° = 4 N. W = 8 sin 60° = 6.93 N.
1.4Elasticity
Elasticity is the property of a material that lets an object return to its original shape and size after the force applied to it is removed.
The extension of a spring is x = l − l₀, the increase in length. A spring is permanently stretched if it passes its elastic limit.
Hooke's law: the extension of a spring is directly proportional to the applied force, provided the elastic limit is not exceeded. F = kx, where k is the spring constant (N m⁻¹). The graph of F against x is a straight line through the origin and k = gradient.
Elastic potential energy
The work done to stretch a spring is stored as elastic potential energy. It equals the area under the F–x graph:
Ep = ½Fx = ½kx²
The F–x graph gives Ep as the area of a triangle (½ base × height), not F × x. The ½ is there because the force grows from 0 to F, so the average force is F ÷ 2.
What changes the spring constant
| Factor | Change | Effect on k |
|---|---|---|
| Length of spring | Shorter | Higher (stiffer) |
| Diameter of spring | Smaller | Higher |
| Thickness of wire | Thicker | Higher |
| Material | Different material | Depends on the material |
Springs in series and in parallel
For identical springs, each with a force F on the system and extension x for a single spring:
| Arrangement | Tension in each spring | Total extension |
|---|---|---|
| Series (end to end) | F (the full force on each) | 2x for two springs: it stretches more, so it is softer |
| Parallel (side by side) | F ÷ 2 (shared equally) | x ÷ 2 for two springs: it stretches less, so it is stiffer |
Worked example
A spring of original length 50 mm extends 6 mm under 12 N: k = 12 ÷ 6 = 2 N mm⁻¹. Two such springs in series under 12 N: each extends 6 mm, total extension 12 mm, total length 50 + 50 + 12 = 112 mm. In parallel: each carries 6 N and extends 3 mm, total length 50 + 3 = 53 mm.A spring (k = 200 N m⁻¹) is stretched by 0.10 m. Find the force and the elastic potential energy.
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F = kx = 200 × 0.10 = 20 N. Ep = ½kx² = ½ × 200 × 0.10² = 1.0 J.