IP Combined Science Notes (Lower Sec, Year 1-2): 07) Forces, Motion & Simple Machines

Physics of everyday motion hinges on clear diagrams and consistent units. Develop habits for vector addition, graph interpretation, and machine efficiency.

Learning targets

• Differentiate scalar and vector quantities with examples.
• Use \( \pu{m.s-1} \) and \( \pu{m.s-2} \) calculations for speed and acceleration.
• Apply \( F = ma \), resultant force reasoning, and Newton's laws to motion scenarios.
• Analyse moments, levers, and pulleys to compute mechanical advantage and velocity ratios.

1. Motion basics

Distance-time & velocity-time graphs

• Gradient of distance-time graph gives speed.
• Area under velocity-time graph gives displacement.
• Gradient of velocity-time graph gives acceleration.

Worked example — Velocity-time area

A cyclist accelerates uniformly from rest to \( \pu{12 m.s-1} \) in \( \pu{6 s} \), travels at constant speed for \( \pu{10 s} \), then decelerates uniformly to rest in \( \pu{4 s} \).

1. Displacement during acceleration: area of triangle \( = \frac{1}{2} \times 6 \times 12 = \pu{36 m} \).
2. Constant speed: rectangle area \( = 10 \times 12 = \pu{120 m} \).
3. Deceleration: triangle \( = \frac{1}{2} \times 4 \times 12 = \pu{24 m} \).

Total displacement \( = \pu{180 m} \).

2. Forces & Newton's laws

• Resultant force determines acceleration: \( F = ma \).
• Balanced forces (resultant zero) mean constant velocity or rest.
• Action-reaction pairs act on different bodies, equal in magnitude but opposite in direction.

Worked example — Elevator problem

A \( \pu{650 kg} \) lift accelerates upward at \( \pu{1.5 m.s-2} \).

1. Weight: \( W = mg = 650 \times 9.81 = \pu{6.38 \times 10^{3} N} \).
2. Resultant force: \( F = ma = 650 \times 1.5 = \pu{975 N} \) upward.
3. Tension in cable: \( T = W + F = 6.38 \times 10^{3} + 9.75 \times 10^{2} = \pu{7.36 \times 10^{3} N} \).

3. Moments & equilibrium

Moment about a pivot:

\[ \text{Moment} = \text{Force} \times \text{Perpendicular distance}. \]

For equilibrium: sum of clockwise moments = sum of anticlockwise moments; resultant force = 0.

Worked example — Seesaw balance

A \( \pu{350 N} \) student sits \( \pu{1.8 m} \) from the pivot. What force must act \( \pu{1.2 m} \) on the other side to balance?

\[ 350 \times 1.8 = F \times 1.2 \Rightarrow F = \frac{350 \times 1.8}{1.2} = \pu{525 N}. \]

4. Simple machines

• Mechanical advantage (MA): \( \frac{\text{Load}}{\text{Effort}} \).
• Velocity ratio (VR): \( \frac{\text{Distance moved by effort}}{\text{Distance moved by load}} \).
• Efficiency: \( \frac{\text{Useful output work}}{\text{Input work}} \times 100% \).

Pulley example

A block-and-tackle with VR = 4 lifts a \( \pu{800 N} \) load using \( \pu{260 N} \) effort.

• \( \text{MA} = \frac{800}{260} = 3.08 \).
• Efficiency \( = \frac{3.08}{4} \times 100 = 77.0% \).

Try it yourself

1. Sketch a free-body diagram for a box dragged across a horizontal surface at constant speed. Label forces and explain why acceleration is zero.
2. A car decelerates from \( \pu{22 m.s-1} \) to rest in \( \pu{3.5 s} \). Calculate deceleration and stopping distance assuming constant deceleration.
3. A lever has effort arm \( \pu{0.80 m} \) and load arm \( \pu{0.25 m} \). Determine the effort needed to lift \( \pu{520 N} \) ignoring losses. Then comment on real-world efficiency losses.

Move to energy and pressure at https://eclatinstitute.sg/blog/ip-combined-sciences-lower-sec-notes/IP-Combined-Science-Lower-Sec-08-Work-Energy-Power-and-Pressure.