Year 11 PhysicsModule 2⏱ ~35 min5 MC · 3 Short AnswerLesson 3 of 15
Inclined Planes
In 2560 BCE, Egyptian builders dragged approximately 2.3 million limestone blocks averaging 2.5 t each up earthen ramps inclined at 8–10°. At 8°, the force required along the ramp was $mg\sin8° \approx 3,400\text{ N}$ per block, less than one-seventh of the 24,500 N needed to lift the block vertically. The inclined-plane geometry reduced the required force by 86%, enabling construction with human labour alone.
Today's hook: In 2560 BCE, Egyptian builders dragged 2.5 t blocks up 8° ramps, reducing the required force from 24,500 N to ~3,400 N. Why does a gentler slope require so much less force to move the same mass?
0/5TASKS
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Warm up first
Warm up first
Three quick questions from earlier lessons. Pulling old material back to mind before you learn something new makes the new material stick better, so this is not busywork.
A block sits on a ramp. As the angle increases, at some point the block starts sliding on its own. What force causes it to slide, and why does the angle matter?
For a block on a frictionless inclined plane at angle $\theta$, the component of weight parallel to the slope is:
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Learning Intentions
Learning Intentions
goals
Know
$W_\parallel = mg\sin\theta$ (weight component along the slope)
$W_\perp = mg\cos\theta$ (weight component perpendicular to slope)
$N = mg\cos\theta$ (normal force on inclined plane)
$f = \mu mg\cos\theta$ (friction force on inclined plane)
Understand
Why objects on a slope accelerate down the incline
Why steeper angles produce greater acceleration
Why mass does not affect the frictionless acceleration
How friction changes depending on the normal force
Can Do
Draw an FBD for a block on an inclined plane
Calculate all force components on a slope
Find the net force and acceleration on a slope
Determine whether a block will slide given μ and θ
The normal force on an inclined plane equals the weight: $N = mg$.On a slope, the normal force is perpendicular to the surface: $N = mg\cos\theta$, which is less than $mg$ whenever $\theta > 0°$.
Heavier objects accelerate faster on frictionless slopes.On a frictionless slope, $a = g\sin\theta$, mass cancels out. All objects accelerate at the same rate regardless of mass.
Cross-lesson links: This lesson extends L02's vector resolution to a rotated coordinate system. The normal force $N = mg\cos\theta$ you calculate here determines the friction force in L04 ($f = \mu N$). The work done against that friction ($W = fs$) feeds directly into L06's energy analysis. Inclined plane problems therefore bridge every phase of Module 2.
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Resolving Weight on a Slope
Core Content
01
Resolving Weight on a Slope
+5 XP
Place a book on a tilted clipboard. As you tilt it further, the book starts to slide, without you pushing it at all. The weight has not changed, and no new force appeared. What changed is the direction of weight relative to the slope: a steeper angle means more of the weight acts along the surface and less pushes into it.
When a block of mass $m$ is on a slope at angle $\theta$, we use the slope axis (parallel and perpendicular to surface) instead of horizontal/vertical:
$$W_\parallel = mg\sin\theta \quad \text{(down the slope)}$$$$W_\perp = mg\cos\theta \quad \text{(into the slope)}$$
Draw the actual forces first. Then replace weight with its slope-axis components when writing equations; $W_\parallel$ and $W_\perp$ are parts of $W$, not additional forces.
Trace: Start at $W = mg$ in panel 2. Follow the purple and green arrows head-to-tail and check that their resultant is still vertical.
The normal force balances $W_\perp$, so $N = mg\cos\theta$. The friction force is $f = \mu N = \mu mg\cos\theta$.
Note that as $\theta$ increases: $\sin\theta$ increases (more drive down slope), $\cos\theta$ decreases (less normal force, so less maximum friction). This is why steeper slopes have both more driving force AND less available friction, the combination dramatically increases acceleration.
On a slope at angle $\theta$: weight component along slope $W_\parallel = mg\sin\theta$ (drives the block down), weight component into slope $W_\perp = mg\cos\theta$, normal force $N = mg\cos\theta$, and friction $f = \mu mg\cos\theta$.
Pause, copy the highlighted formulae into your book before moving on.
True or false: On a frictionless inclined plane, a 10 kg block and a 2 kg block will reach the bottom at the same time if released from the same height.
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Net Force and Acceleration on an Incline
02
Net Force and Acceleration on an Incline
+5 XP
We just saw how to resolve weight into parallel and perpendicular components on a slope. That raises a question: how do friction and the driving component combine to give a net force, and what acceleration does that produce? This card answers it → $a = g(\sin\theta - \mu\cos\theta)$, and mass cancels completely.
The net force along the slope is the driving force minus friction. Divide by mass to find acceleration.
Mass cancels from the acceleration equation. This confirms that all objects on the same frictionless slope have the same acceleration regardless of mass.
Worked ExampleInclined Plane with Friction
A 5 kg block sits on a 30° slope with $\mu = 0.25$. Find the acceleration. ($g = 9.8$ m/s²)
Net acceleration on a rough incline: $a = g\sin\theta - \mu g\cos\theta$ (mass cancels); the block slides when $\tan\theta > \mu$ (driving component exceeds maximum static friction); on a frictionless slope $a = g\sin\theta$ for all masses.
Pause, copy the highlighted formula and sliding condition into your book before moving on.
A 4 kg block on a frictionless 45° slope. Its acceleration is approximately:
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Practical Investigation, Motion on Inclined Planes
Activities
Practical Investigation, Motion on Inclined Planes
ApplyBand 3
Design and conduct (or simulate) an experiment to investigate how the angle of an incline affects the acceleration of a trolley or cart. Before conducting: state a hypothesis using $a = g\sin\theta$.
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Inclined Plane Analysis
Inclined Plane Analysis
AnalyseBand 4
A 10 kg crate sits on a 25° slope. $\mu = 0.30$. Draw the FBD and calculate: (a) normal force, (b) friction force, (c) net force, (d) acceleration, (e) will the crate slide?
Why does mass cancel from the acceleration equation $a = g\sin\theta - \mu g\cos\theta$ for an inclined plane?
✓
Show what you have learned
Quick recall, Inclined Planes
+5 XP
Pick your answer, then rate your confidence.
Short Answer, 10 marks
+5 XP
UnderstandBand 3(3 marks) 1. Explain why the normal force on an object resting on a 30° incline is less than the weight of the object. Use a force diagram in your explanation.
ApplyBand 4(3 marks) 2. A 6 kg block rests on a 40° frictionless slope. Calculate the normal force and the net force acting on the block along the slope. Determine its acceleration. ($g = 9.8$ m/s²)
AnalyseBand 5(4 marks) 3. A student claims: "A 20 kg block will reach the bottom of a frictionless slope before a 5 kg block released from the same height." Evaluate this claim and explain whether the student's reasoning is correct.
Show all answers
Short Answer, Model Answers
Q1 (3 marks): On an inclined plane, weight ($mg$) acts vertically downward. The surface can only exert a force perpendicular to itself (normal force $N$). The normal force must balance only the component of weight perpendicular to the slope: $N = mg\cos\theta$. Since $\cos30° = 0.866 < 1$, we have $N = 0.866mg < mg$. The remaining component $mg\sin\theta$ is parallel to the slope and causes acceleration.
Q3 (4 marks): The student is incorrect. For a frictionless slope, $F_{net} = mg\sin\theta$, and $a = F_{net}/m = g\sin\theta$. Mass cancels from the acceleration equation. Both the 20 kg and 5 kg blocks experience the same acceleration $g\sin\theta$ on the same frictionless slope. They would reach the bottom at the same time. This is analogous to Galileo's famous observation that objects of different masses fall at the same rate in the absence of air resistance.
✓
Retrieve, reflect and finish
Check what actually stuck
Take the full module quiz
quiz
A full module quiz covering every lesson in this module, not just this one. Set aside a decent block of time and treat it like a real assessment.
In 2560 BCE, Egyptian builders' choice of an 8–10° ramp angle was optimal: $mg\sin8° \approx 3,400\text{ N}$ per block, far below the 24,500 N vertical lift force. The key insight is that the angle, not mass, determines how much force is needed. Mass cancels from the acceleration formula ($a = g\sin\theta - \mu g\cos\theta$), but it multiplies the ramp-parallel component, making every degree of steepening more costly.
