Laws of Motion · Lesson 7 of 14
Vertical Motion, Apparent Weight and Release
“Apply force diagrams in lifts and accelerating vehicles and predict what changes when support is removed.”
• Distinguish gravitational weight from a scale’s support-force reading. • Calculate apparent weight for upward, downward and free-fall acceleration. • Use present interactions after release from an accelerating vehicle. • Explain the initial trajectory when a pendulum string is cut.
What a weighing scale measures
A scale responds to the contact force compressing it. For a person standing on it, the upward normal force N on the person equals the downward force on the scale by the person in magnitude, by the third law. This is the apparent-weight force reading. Gravitational weight remains mg. A scale calibrated in kilograms divides the measured force by its calibration value of g; its displayed mass can therefore differ from actual mass when the scale accelerates.
Constant upward or downward velocity gives a_y = 0, so N = mg. Upward acceleration gives N > mg; downward acceleration of magnitude a gives N = m(g − a). The direction of velocity alone does not decide the reading: an upward-moving lift that is slowing down accelerates downward and gives a reduced reading. A downward-moving lift that is slowing down accelerates upward and gives a larger reading.
Problem
Use g = 10 m/s². Find force reading and kilogram-calibrated display for constant upward speed, downward acceleration 5 m/s², upward acceleration 5 m/s², and free fall.
- 1.At constant speed, N = 70 × 10 = 700 N. A scale calibrated using 10 m/s² displays 70 kg.
- 2.For downward acceleration 5 m/s², N = 70(10 − 5) = 350 N, corresponding to a 35 kg display.
- 3.For upward acceleration 5 m/s², N = 70(10 + 5) = 1050 N, corresponding to a 105 kg display.
- 4.In free fall, a_y = −g, giving N = 0. The display is zero even though gravitational weight is still 700 N. This is support-force weightlessness, not absence of gravity.
A yielding floor beneath a combined system
Problem
A 2 kg block rests on a soft floor. A 25 kg cylinder is added and the pair descends with acceleration 0.1 m/s². Use g = 10 m/s². Find the force on the floor before loading and during descent.
- 1.Before adding the cylinder, the stationary 2 kg block has N = 20 N, so it pushes down on the floor with 20 N.
- 2.During descent, choose block plus cylinder as the system. Total weight is 270 N downward, and the floor’s normal force is N upward.
- 3.Using downward positive, 270 − N = 27 × 0.1. Thus N = 267.3 N. The pair pushes downward on the floor with the same 267.3 N magnitude.
- 4.The internal contact forces between cylinder and block cancel in the combined diagram. Separately, the cylinder pushes on the block with 25(10 − 0.1) = 247.5 N.
- 5.Weight and support are unequal during acceleration. The partner to the floor’s force is the force on the floor, not the gravitational weight.
Release preserves velocity, not the earlier acceleration
Before release, a body in an accelerating vehicle can be accelerated by contact forces from the vehicle. Immediately after separation, those contact forces vanish. Its initial velocity equals its velocity at release, while the new acceleration follows from the forces that remain. With air resistance neglected, a released stone experiences gravity only, whether the vehicle was stationary, moving steadily or accelerating.
Problem
A truck starts from rest with horizontal acceleration 2 m/s². At truck time 10 s, a stone is released from 6 m above ground. Use g = 10 m/s². Find the stone’s velocity and acceleration at truck time 11 s.
- 1.Truck velocity at release is 2 × 10 = 20 m/s horizontally. That is the stone’s initial horizontal velocity; initial vertical velocity is zero.
- 2.At truck time 11 s, only 1 s of free flight has elapsed. The velocity is 20 i-hat − 10 j-hat m/s, taking upward as +y.
- 3.Speed is √500 = 22.36 m/s at arctan(10/20) = 26.57° below horizontal. Acceleration is −10 j-hat m/s² with no horizontal component.
- 4.The fall in 1 s is 5 m, so the stone remains 1 m above the ground. The requested time lies before impact, validating the free-flight calculation.
On the floor and outside the window are different force situations
A 0.1 kg stone lying at rest relative to a horizontally accelerating train with a = 1 m/s² has horizontal net force 0.1 N in the train’s acceleration direction. Weight and normal force balance vertically. The instant the same stone is released out of the window, horizontal contact is removed; its net force becomes 1 N vertically downward for g = 10 m/s². Initial horizontal velocity does not act as a forward force.
Cutting a pendulum string
At an extreme point of a simple back-and-forth pendulum oscillation, the bob’s speed is zero. If the string is cut there, it begins to fall vertically under gravity. At the mean position, its velocity is horizontal and tangent to the arc; cutting the string there gives a horizontal-launch projectile trajectory. For a bob with 1 m/s speed at the mean point, its initial projectile horizontal velocity is 1 m/s. In either case, tension disappears and gravity remains.
The same rule explains an upward-thrown pebble. A 0.05 kg pebble has weight 0.5 N downward throughout free flight with g = 10 m/s²: on the way up, on the way down and at the top. An oblique launch changes the velocity, but not the weight vector. A zero velocity component at the top is not a reason to remove gravity from the force diagram.
Quiz
A lift moves downward while slowing. How does a standing person’s apparent weight compare with mg?
Key Takeaways
• A scale measures support force; gravitational weight remains mg near Earth. • Acceleration direction, rather than velocity direction, sets the apparent-weight change. • Free fall gives zero support force without removing gravity. • After release, retain the initial velocity and redraw the forces. • Cutting a string removes tension; the new path depends on the velocity at that instant.