Forces & Moments: interactions and equilibrium
Key idea: Represent every interaction first, then connect force laws, turning effects and equilibrium with consistent diagrams.
Before you start: Vector Addition & ComponentsTorque & Couples
By the end, you can
- Describe forces on masses, charges and current-carrying conductors in fields, and use complete force diagrams in equilibrium.
- Describe normal, buoyant, frictional and viscous forces qualitatively without introducing excluded coefficients.
- Apply Hooke’s law to new elastic situations.
- Use force diagrams, vector triangles and both equilibrium conditions consistently.
Starting-point self-check
1. Check your starting point
Draw or state the interaction model before calculating. Record any force whose source or direction you cannot justify. Use the recorded topic diagnostic above when you want scoring and a personalised repair plan.
Question 1
State the force and direction on (i) a mass in a gravitational field, (ii) a positive charge in an electric field and (iii) a perpendicular current-carrying wire in a magnetic field.
Check the model response
The mass experiences weight along the gravitational field; the positive charge experiences electric force along the electric field; the wire experiences magnetic force perpendicular to both current and magnetic field, with direction from Fleming’s left-hand rule.
Question 2
A submerged block is stationary in water but is not touching the container. Name the forces on it. Which force arises from the surrounding fluid?
Check the model response
Weight acts downward and buoyant force (upthrust) acts upward. Upthrust arises from the surrounding fluid’s pressure forces.
Question 3
A spring extends 0.060 m under an 18 N force. Find its force constant, assuming Hooke’s law applies.
Check the model response
k = F/x = 18/0.060 = 300 N m⁻¹.
Question 4
Distinguish the moment of one force from the torque of a couple. Find the torque of two opposite 12 N forces whose parallel lines of action are 0.25 m apart.
Check the model response
A force moment is force times perpendicular distance from a chosen pivot. A couple is two equal, opposite, parallel forces with separated lines of action; its torque is one force times their separation: 12 × 0.25 = 3.0 N m.
Question 5
A uniform 200 N beam is pivoted at one end and supported upward 2.5 m from the pivot. Its centre is 1.0 m from the pivot. Find the support force and state the second equilibrium condition still required.
Check the model response
Taking moments about the pivot: support × 2.5 = 200 × 1.0, so support = 80 N. Translational equilibrium also requires zero resultant force, giving a 120 N upward pivot reaction.
repair
2. Repair the five common breaks
Use only the repair matching an error, then redraw or recalculate the diagnostic model.
Check this idea
Misconception: A field name is itself a complete force description.
Repair: Name the body and interaction: gravitational force on mass, electric force on charge, or magnetic force on a current. Direction follows the relevant field and sign or current rule.
Check this idea
Misconception: Motion always has a force in the direction of motion.
Repair: Forces come from interactions, not from velocity. Friction or viscous force opposes relative motion; a normal force is perpendicular to contact; upthrust results from fluid pressure.
Check this idea
Misconception: Hooke’s law means every force–extension graph stays linear.
Repair: F = kx applies only over the Hooke-law region. Confirm proportionality and use extension, not total spring length.
Check this idea
Misconception: Any distance from a pivot can be used in a moment.
Repair: Use the perpendicular distance from the pivot to the force’s line of action. Couple torque uses the perpendicular separation of the two lines of action.
Check this idea
Misconception: Zero resultant force alone proves equilibrium.
Repair: Complete equilibrium requires both zero resultant force and zero resultant torque. A free-body diagram and a consistent moment equation must satisfy both.
worked example
3. Follow four worked models
Follow the representation, equation and physical check in each model.
Model 1
A −3.0 μC charge is in a 4.0 × 10⁴ N C⁻¹ electric field directed right. Find its force.
Check the model response
F = qE = (−3.0 × 10⁻⁶)(4.0 × 10⁴) = −0.12 N. The negative sign means 0.12 N left, opposite the field direction defined by a positive charge.
Model 2
A spring of force constant 250 N m⁻¹ has original length 0.180 m and loaded length 0.212 m. Find the elastic force.
Check the model response
Extension x = 0.212 − 0.180 = 0.032 m. F = kx = 250 × 0.032 = 8.0 N.
Model 3
A 30 N force acts at 40° to a 0.50 m spanner measured from its pivot. Find the moment.
Check the model response
The perpendicular component is 30 sin 40°, so moment = rF sin 40° = 0.50 × 30 × sin 40° = 9.64 N m.
Model 4
A 100 N uniform horizontal board is supported at both ends, 4.0 m apart. A 300 N load is 1.0 m from the left end. Find both reactions.
Check the model response
Weight of the board acts at its centre. About the left end: Rright(4.0) = 300(1.0) + 100(2.0), so Rright = 125 N. Vertical equilibrium gives Rleft = 400 − 125 = 275 N.
guided practice
4. Guided practice
Use each hint only to set up the representation.
Question 1
A 0.30 m wire carries 2.0 A perpendicular to a 0.50 T magnetic field. Find the magnetic-force magnitude.
Hint: For perpendicular current and field, use F = BIL.
Check the model response
F = 0.50 × 2.0 × 0.30 = 0.30 N. Its direction is perpendicular to current and field.
Question 2
Explain why a falling object approaching terminal speed can still move downward while its resultant force tends to zero.
Hint: Separate velocity from acceleration and identify viscous drag.
Check the model response
Weight remains downward while air resistance increases upward with speed. As they become equal, resultant force and acceleration tend to zero, but the existing downward velocity remains constant.
Question 3
A spring extends from 4.0 cm to 7.5 cm under 14 N. Find k.
Hint: Convert the extension—not the final length—to metres.
Check the model response
x = 3.5 cm = 0.035 m, so k = 14/0.035 = 400 N m⁻¹.
Question 4
A 60 N load is 0.80 m left of a pivot. Where should a 40 N load act on the right for rotational equilibrium?
Hint: Equate clockwise and anticlockwise moments.
Check the model response
60 × 0.80 = 40d, so d = 1.20 m to the right. Translational equilibrium would still require the pivot reaction.
independent practice
5. Independent practice
Solve without repair notes and include a free-body diagram where equilibrium is involved.
Question 1
A 2.0 kg mass is in a 9.8 N kg⁻¹ gravitational field. A +5.0 μC charge is in a 3.0 × 10⁴ N C⁻¹ electric field. Find both forces and state their directions relative to the fields.
Check the model response
Weight = mg = 19.6 N along the gravitational field. Electric force = qE = 0.15 N along the electric field because the charge is positive.
Question 2
Describe normal force, friction, upthrust and viscous force by their source and typical direction. Do not use coefficients.
Check the model response
Normal force is a contact force perpendicular to a surface; friction opposes relative sliding at contact; upthrust is the resultant fluid-pressure force, typically upward; viscous force opposes motion through a fluid.
Question 3
Two opposite parallel 20 N forces are 0.18 m apart. Find the couple torque and explain why the couple has no resultant force.
Check the model response
Torque = 20 × 0.18 = 3.6 N m. The forces have equal magnitudes and opposite directions, so their vector sum is zero while their separated lines of action produce rotation.
Question 4
A 500 N sign hangs from two cables making 30° and 50° above the horizontal. State the vector-equilibrium equations needed to find both tensions.
Check the model response
With tensions T30 and T50: horizontal equilibrium gives T30 cos 30° = T50 cos 50°. Vertical equilibrium gives T30 sin 30° + T50 sin 50° = 500 N. These equations correspond to a closed vector triangle.
Practice exit check
6. Practice assessment
Use this as extra closed-book practice, then complete the separate recorded assessment in your plan.
Question 1
A 0.40 m conductor carrying 3.0 A is perpendicular to a 0.20 T field. Calculate the force and describe how its direction is found.
Check the model response
F = BIL = 0.20 × 3.0 × 0.40 = 0.24 N. Fleming’s left-hand rule gives the direction perpendicular to current and magnetic field.
Question 2
A spring extends 24 mm under 9.6 N. Find k and predict the extension at 15 N if Hooke’s law still applies.
Check the model response
k = 9.6/0.024 = 400 N m⁻¹. At 15 N, x = F/k = 0.0375 m = 37.5 mm.
Question 3
A couple consists of 16 N forces separated by 0.30 m. Compare its torque with the moment of one 16 N force acting 0.30 m from a pivot.
Check the model response
Both numerical values are 16 × 0.30 = 4.8 N m. The couple torque is independent of chosen origin and has zero resultant force; the single-force moment is about a specified pivot and the force is unbalanced unless other forces act.
Question 4
A 240 N uniform beam of length 3.0 m is hinged at its left end and held horizontal by an upward force at the right end. Find that force and the hinge’s vertical reaction.
Check the model response
Weight acts at the 1.5 m centre of gravity. Moments about hinge: F(3.0) = 240(1.5), so F = 120 N upward. Vertical equilibrium gives hinge reaction = 120 N upward.
Question 5
A block slides right through oil while touching a horizontal guide. State the directions of its normal, frictional, viscous and buoyant forces, without using force coefficients.
Check the model response
The guide's normal force is perpendicular to the contact; friction opposes relative sliding at the contact; viscous force opposes motion through the oil; buoyant force acts upward because of the fluid-pressure difference. Their magnitudes cannot be inferred without further data.
Re-test practice
7. Delayed re-test practice
Return after at least three days. Use fresh diagrams and numbers. The recorded plan enforces the delay and uses a separate re-test family for selected-response skill-group evidence.
Question 1
A 1.5 kg object falls through oil at constant speed. Name the forces and find the viscous force if upthrust is 2.0 N and g = 9.8 N kg⁻¹.
Check the model response
Weight = 14.7 N downward. Upthrust and viscous force act upward. Constant speed means zero resultant force, so viscous force = 14.7 − 2.0 = 12.7 N upward.
Question 2
A spring with k = 320 N m⁻¹ lengthens from 0.150 m to 0.195 m. Find its force.
Check the model response
Extension = 0.045 m, so F = 320 × 0.045 = 14.4 N.
Question 3
A 75 N force acts 0.60 m from a pivot at 25° to the lever. Find its moment.
Check the model response
Moment = rF sin 25° = 0.60 × 75 × sin 25° = 19.0 N m.
Question 4
A 180 N uniform plank is supported 0.50 m from each end; the supports are 2.0 m apart. A 120 N load is 0.40 m from the left support. Find both reactions.
Check the model response
The plank’s weight acts midway between supports, 1.0 m from the left. About the left support: Rright(2.0) = 120(0.40) + 180(1.0), so Rright = 114 N. Then Rleft = 300 − 114 = 186 N.
Question 5
A negative charge and a current-carrying wire are placed separately in uniform electric and magnetic fields. State how the force direction is determined in each case.
Check the model response
The electric force on a negative charge is opposite to the electric-field direction. The magnetic force on a perpendicular current is perpendicular to current and field, with its direction found using Fleming's left-hand rule.