01 · Rising faster
A 50 kg passenger accelerates upward at 0.8 m/s². Find the floor reaction using g = 9.8.
Hint
Use upward positive.
Worked solution
R − 490 = 50 × 0.8, so R = 530 N.
Understand · explore · practise
Lifts, normal reactions and apparent weight
Use R − mg = ma for passengers and cable equations for whole lifts, keeping velocity, acceleration and apparent weight distinct.
Before you startNewton’s second law, weight and signed vertical motion.
01 / Choose the passenger or the whole lift
Different bodies require different force diagrams.
For a passenger: R − mg = ma, taking upward positive.
The passenger has weight downward and the floor’s normal reaction upward. The cable tension acts on the lift, not directly on the passenger. The floor reaction is often called apparent weight; true gravitational weight remains mg.
A person on a lift floorExplore
The person has mass 60 kg and g = 9.8 m/s². Take upward positive. The floor reaction is R = 60(9.8 + a) N while contact is maintained. Motion direction and acceleration direction are separate pieces of information.
02 / Use acceleration direction, not travel direction
Descending does not necessarily mean accelerating downward.
A 60 kg passenger is moving downward but slowing at 2 m/s².Worked example
a = +2 m/s² with upward positive
Slowing downward motion requires upward acceleration.
R − 588 = 60 × 2
Weight is 60 × 9.8 = 588 N.
R = 708 N
The reaction exceeds weight even though the passenger is descending.
Watch: reaction changes with acceleration while weight stays fixed
Pause, replay or seek freely. The notes explain the same idea and stay in view.
02 · Rising but slowing
The same passenger moves upward while slowing at 0.8 m/s². Find R.
Hint
Acceleration is downward.
Worked solution
a = −0.8, so R − 490 = −40 and R = 450 N.
03 / Translate speeding and slowing into a signed acceleration
Opposite velocities can share the same acceleration.
03 · Descending but slowing
A 70 kg passenger descends while slowing at 1.5 m/s². Find R.
Hint
Acceleration is upward.
Worked solution
R = 70(9.8 + 1.5) = 791 N.
04 · Ascending but slowing
A 70 kg passenger ascends while slowing at 2 m/s². Find R.
Hint
Acceleration is downward.
Worked solution
R = 70(9.8 − 2) = 546 N.
05 · Steady descent
A 70 kg passenger descends at constant velocity. Find R.
Hint
Velocity is nonzero but acceleration is zero.
Worked solution
R = mg = 70 × 9.8 = 686 N.
04 / For cable tension, include the total lifted mass
Internal forces cancel when lift and passenger form one system.
A lift car of mass 450 kg carries a 60 kg passenger and accelerates upward at 1.2 m/s². Neglect other external forces.Worked example
Total mass = 450 + 60 = 510 kg
Treat lift and passenger together.
T − 510 × 9.8 = 510 × 1.2
Cable tension and total weight are the external vertical forces.
T = 5610 N
This differs from the floor reaction on the passenger.
Passenger reaction = 60(9.8 + 1.2) = 660 N
For the car alone: 5610 − 4410 − 660 = 450 × 1.2.
06 · Whole lift
A 400 kg lift car carries a 60 kg passenger. The system accelerates downward at 1 m/s². Find cable tension with no other external forces.
Hint
Total mass is 460 kg; upward acceleration is −1.
Worked solution
T − 460 × 9.8 = −460, so T = 4048 N.
07 · Passenger reaction
Find the passenger’s floor reaction in question 6.
Hint
Use the passenger mass only.
Worked solution
R = 60(9.8 − 1) = 528 N. The car feels an equal downward contact force from the passenger.
05 / Infer acceleration from a measured reaction
A larger reading identifies upward acceleration, not upward velocity.
08 · Infer acceleration
A 60 kg passenger’s reaction is 720 N. Find acceleration.
Hint
Rearrange R − mg = ma.
Worked solution
a = (720 − 588)/60 = 2.2 m/s² upward. The passenger may be rising and speeding up or descending and slowing down.
09 · Infer mass
A passenger accelerates upward at 1 m/s² and has reaction 432 N. Find mass.
Hint
Write 432 = m(9.8 + 1).
Worked solution
m = 432/10.8 = 40 kg.
10 · Cable measurement
The total lift-plus-load mass is 250 kg. Cable tension is 3000 N. Find upward acceleration, neglecting other external forces.
Hint
Use total weight in the system equation.
Worked solution
a = (3000 − 250 × 9.8)/250 = 2.2 m/s².
06 / Zero apparent weight does not mean zero gravity
The floor need not push when passenger and lift fall together.
In ideal free fall, a = −g and R = 0.
With air resistance neglected, both lift and passenger accelerate downward at g. The passenger still has gravitational weight mg. The zero reading refers to the normal reaction, not to the disappearance of gravity.
11 · Free-fall reading
For a 60 kg passenger in ideal free fall, state weight, acceleration and reaction.
Hint
Keep the three quantities separate.
Worked solution
Weight is 588 N downward, acceleration is 9.8 m/s² downward, and floor reaction is 0 N.
12 · Impossible negative reaction
A platform is prescribed acceleration 12 m/s² downward. Can an unrestrained 60 kg person stay on its ordinary floor with only weight and floor contact acting?
Hint
Calculate the reaction that shared acceleration would require.
Worked solution
R = 60(9.8 − 12) = −132 N. An ordinary floor cannot pull the person down, so that contact model fails: the floor moves away and the person falls freely once contact is lost.
07 / Read what a scale measures
A scale’s displayed kilograms are not a changing physical mass.
13 · Displayed mass
A scale divides force by 9.8 to display “kg”. If its force reading is 720 N for a 60 kg passenger, what does it display?
Hint
Divide the force reading by the calibration g.
Worked solution
720/9.8 = 73.5 kg to 3 significant figures. This is an apparent mass display; the passenger’s actual mass is still 60 kg.
14 · Force on which body?
Should cable tension T be added as an upward force in the passenger-only equation?
Hint
The cable is attached to the lift car.
Worked solution
No. The passenger is supported by the floor reaction R. For the combined system, cable tension replaces the internal floor-contact pair in the external force equation.
08 / Separate the body, the acceleration and the reading
A correct sign matters more than the direction of travel.
For a passenger, resolve R − mg = ma with upward positive. For the entire lift, use total mass and external cable tension. Check that any ordinary contact reaction is nonnegative and that the assumed contact or taut cable still exists.
Section 1 of 8 · Choose the passenger or the whole lift