Ch 1 · 4 Forces, Density and Pressure · A Level Physics 9702
Topic 1 of 3 in Forces, Energy and Materials — A Level Physics 9702 — 67 lessons.
Introduction
Why does a ladder resting against a wall not slip? How does a ship made of steel float? This topic covers turning effects, the conditions for equilibrium, and the pressure in fluids that produces upthrust. (Syllabus topic 4: sections 4.1–4.3.)
Key ideas — 4.1 (a) Centre of gravity and moments
The centre of gravity of an object is the single point at which its whole weight may be taken to act. For a uniform rod it is at the midpoint.
🧑🏫 Teacher's explanation
Imagine you are trying to open a heavy door. If you push near the hinges, it is incredibly hard; if you push at the handle, it is easy. This is because the 'turning effect' depends not just on how hard you push, but how far from the hinge you push. This 'turning effect' is what we call a moment.
📖 Definitions
🔍 Worked example — Easy
A uniform rod of length 2.0 m and weight 15 N is pivoted at its centre. A force of 5.0 N is applied perpendicularly at one end. Calculate the moment of this force about the pivot.
🔍 Worked example — Standard
A uniform beam of length 4.0 m and weight 20 N is supported by a pivot at one end (Point A) and a vertical reaction force R at a point 3.0 m from A. Calculate the magnitude of R when the beam is in equilibrium.
⚠️ Common errors
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🎯 Exam question
Key ideas — 4.1 (b) Couples and torque
A couple is a pair of forces that are equal in magnitude, opposite in direction and not acting along the same line. A couple produces rotation only: its resultant force is zero, so it causes no translation (e.g. turning a steering wheel or a tap).
🧑🏫 Teacher's explanation
Imagine you are trying to open a heavy door. If you push directly towards the hinges, the door won't move. If you push at the handle, it swings easily. This 'turning effect' is what we call a moment. Now, imagine you are using both hands to turn a screwdriver. Your left hand pushes one way, and your right hand pushes the opposite way. Because the forces are equal and opposite, the screwdriver doesn't fly across the room (no net force), but it does spin (net torque). This specific setup—two parallel, equal, and opposite forces that are not on the same line—is called a 'couple'.
📖 Definitions
🔍 Worked example — Easy
A technician applies two equal forces of 15.0 N to a wrench to turn a bolt. The forces are acting perpendicular to the handle, and the distance between the two lines of action of the forces is 0.250 m. Calculate the torque of the couple.
🔍 Worked example — Standard
Two forces of 40.0 N each act on a steering wheel. The forces are equal and opposite. The perpendicular distance between the lines of action of these forces is 0.30 m. Calculate the torque.
🔍 Worked example — Exam-style
A student applies a couple to a wheel. One force is 12.5 N and the other is -12.5 N. The perpendicular distance between the lines of action is 0.45 m. Calculate the magnitude of the torque produced.
⚠️ Common errors
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🎯 Exam question
Key ideas — 4.2 Equilibrium
Principle of moments: for an object in equilibrium, the sum of the clockwise moments about any point equals the sum of the anticlockwise moments about that same point.
🧑🏫 Teacher's explanation
Imagine a seesaw. If you sit on one end and your friend sits on the other, the seesaw only stays level if the 'turning effect' on both sides is balanced. This 'turning effect' is what we call a moment. For an object to be perfectly still and not spinning, two things must happen: all the pushes and pulls must cancel out so it doesn't move left/right or up/down (no resultant force), and all the twists must cancel out so it doesn't spin (no resultant torque).
📖 Definitions
🔍 Worked example — Easy
A uniform plank of length 2.0 m and mass 10 kg is supported by a pivot at its center. A 5.0 kg mass is placed 0.50 m from the pivot on the left side. Calculate the force required at the right end, 0.50 m from the pivot, to keep the plank in equilibrium. Use g = 9.81 m/s^2.
🔍 Worked example — Standard
Two forces, F_1 = 10 N and F_2 = 15 N, act on a particle. The angle between them is 60^°. If the particle is in equilibrium, find the magnitude of the third force F_3.
🔍 Worked example — Exam-style
A uniform beam of length L is balanced on a pivot at one end. A weight W_1 is placed at distance d_1 from the pivot. A second weight W_2 is placed at distance d_2 from the pivot on the opposite side. If the beam is in equilibrium, derive an expression for W_2 in terms of W_1, d_1, and d_2.
⚠️ Common errors
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🎯 Exam question
Key ideas — 4.3 (a) Density and pressure
Density is mass per unit volume: rho = m/V (kg m⁻³).
🧑🏫 Teacher's explanation
Imagine you are at the bottom of a deep swimming pool. You feel a heavy sensation in your ears. This isn't just 'weight'; it is the result of the water above you pushing down on you. Because water has mass, it exerts weight, and that weight creates pressure. The deeper you go, the more water is sitting above you, meaning the pressure increases.
📖 Definitions
🔍 Worked example — Easy
A solid metal block has a mass of 2.5 kg and a volume of 0.0005 m^3. Calculate its density rho.
🔍 Worked example — Exam-style
An object with a volume of 0.020 m^3 is fully submerged in seawater. Calculate the upthrust acting on the object. Assume the density of seawater is 1025 kg m^-3 and g = 9.81 ms^-2.
⚠️ Common errors
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🎯 Exam question
Key ideas — 4.3 (b) Hydrostatic pressure
Derivation. Consider a column of fluid of density rho, cross-sectional area A and height Δ h:
🧑🏫 Teacher's explanation
Imagine you are scuba diving. As you descend deeper, you feel a 'squeeze' on your ears. This is because the weight of the water above you increases as you go deeper. Because pressure is simply force divided by area, and that force is caused by the weight of the liquid column above you, the deeper you go, the higher the pressure becomes.
📖 Definitions
🔍 Worked example — Easy
A diver is at a depth of 2.5 m in seawater. Calculate the hydrostatic pressure at this depth. Assume the density of seawater is 1025 kg m^-3 and g = 9.81 m s^-2.
🔍 Worked example — Standard
A solid wooden sphere with a volume of 0.045 m^3 is fully submerged in a lake. Calculate the upthrust acting on the sphere. Assume the density of water is 1000 kg m^-3 and g = 9.81 m s^-2.
🔍 Worked example — Exam-style
A rectangular tank is filled with oil (density rho_oil = 850 kg m^-3) to a height of 2.0 m. The bottom of the tank is open to the atmosphere (p_atm = 1.01 × 10^5 Pa). Calculate the total pressure at the bottom of the tank. Use g = 9.81 m s^-2.
⚠️ Common errors
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🎯 Exam question
Key ideas — 4.3 (c) Upthrust and Archimedes' principle
An object in a fluid feels a greater pressure on its lower surface (deeper) than on its upper surface. This difference in hydrostatic pressure gives a net upward force: the upthrust.
🧑🏫 Teacher's explanation
Imagine you are diving to the bottom of a swimming pool. You notice that your ears feel a lot more pressure at the bottom than they did at the surface. This is because the weight of the water above you is pushing down on you. Because water is a fluid, it pushes in all directions, but the pressure at the bottom of your body is higher than the pressure at the top. This difference in pressure creates a net force pointing upwards.
📖 Definitions
🔍 Worked example — Easy
A solid wooden cube with a volume of 0.002 m^3 is fully submerged in water. Calculate the upthrust acting on the cube. (Density of water rho = 1010\ ext kg m^-3, g = 9.81 m/s^2)
🔍 Worked example — Standard
A metal sphere of volume 4.5 × 10^-4 m^3 is lowered into a tank of oil. The density of the oil is 850 kg m^-3. Calculate the upthrust. (Take g = 9.81 m/s^2)
🔍 Worked example — Exam-style
An object of mass 2.5 kg and volume 0.003 m^3 is placed in a liquid of density 1100 kg m^-3. Determine whether the object sinks or floats, and calculate the apparent weight of the object when fully submerged. (Take g = 9.81 m/s^2)
⚠️ Common errors
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🎯 Exam question
(a) Calculate the weight of the sphere. (b) Calculate the upthrust acting on the sphere. (c) Calculate the apparent weight of the sphere while submerged. (Take density of water rho = 1000 kg m^-3 and g = 9.81 m/s^2). [5 marks]
Worked example
Example 1: equilibrium of a beam. A uniform beam of weight 200 N and length 4.0 m rests on supports at its ends. A 500 N load sits 1.0 m from the left end. Find the support forces.
🔍 Worked example — Easy
A uniform beam of mass 50.0 kg and length 3.00 m is supported at both ends. A mass of 80.0 kg is placed 1.00 m from the left support. Calculate the upward reaction force at the right support. (Use g = 9.81 m/s^2)
🔍 Worked example — Standard
A diver is at a depth of 30.0 m in seawater with a density of 1030 kg m^-3. Calculate the total pressure acting on the diver. (Atmospheric pressure P_atm = 1.01 × 10^5 Pa, g = 9.81 m/s^2)
🔍 Worked example — Exam-style
A wooden block of volume 4.50 × 10^-4 m^3 and density 600 kg m^-3 is fully submerged in water (density 1000 kg m^-3). Calculate the apparent weight of the block. (Use g = 9.81 m/s^2)
⚠️ Common errors
Common mistakes
Practice
Answers: (1) One of the forces × the perpendicular distance between the lines of action of the two forces. (2) No resultant force and no resultant torque. (3) See section 4.3 (b): weight of column rho A Δ h g divided by area A. (4) The plank's weight acts 0.50 m beyond the pivot: W × 0.50 = 150 × 0.50 ⇒ W = 150 N. (5) Upthrust = 1000 × 9.81 × 3.0 × 10^-3 = 29.4 N = weight, so m = 3.0 kg.