Choose acceleration direction from the changing velocity, not from motion direction alone.
Syllabus 4.4
Notes and key methods
For a block remaining on a lift floor, take upward positive and write R−Mg=Ma. A lift moving upward but slowing down has downward acceleration, so R is below Mg. Its upward velocity does not determine the acceleration sign.
A force-calibrated scale measures the normal reaction. If it displays a mass using the model g, its indicated value is \(\displaystyle \frac{R}{g}\), while the actual mass is unchanged. Contact requires \(\displaystyle R\ge 0\); free fall gives \(\displaystyle R=0\) and faster downward floor acceleration would invalidate this simple contact assumption.
Check the conditions
Choose and retain a positive direction; distinguish signed quantities from magnitudes.
Use SI units and \(\displaystyle g= \frac{10m}{s^{2}}\) wherever gravity is relevant.
Apply only the model and time interval stated in the question.
Three worked examples
Example 1 · Contact forces in a moving lift
A 7 kg block stays on the horizontal floor of a lift that is moving upward and slowing at \(\displaystyle \frac{3m}{s^{2}}\). Use \(\displaystyle g= \frac{10m}{s^{2}}.A\) scale is calibrated to display reaction divided by g as a mass reading.
Schematic force diagram for the selected particle; arrow lengths are not a force scale.
Find the normal reaction on the block in N.
Find the scale’s displayed mass in kg.
Find the magnitude of the force exerted by the block on the floor, in N.
In a separate stage with upward acceleration \(\displaystyle \frac{3m}{s^{2}}\), find the normal reaction in N.
Show the reasoning
(a)
Use the downward acceleration sign in the vertical force equation.
Upward-positive acceleration is −3, so \(\displaystyle R=Mg\)−Md=49 N.
Answer: \(49\).
(b)
Convert the measured support force using the stated scale calibration.
The calibrated display is \(\displaystyle \frac{R}{g}=\)\(\displaystyle \frac{49}{10}\) kg. The actual mass remains 7 kg.
Answer: \(\displaystyle \frac{49}{10}\).
(c)
Use Newton’s third law for the block-floor pair, keeping the recipient explicit.
The floor interaction partner has the same magnitude 49 N, directed downward on the floor.
Answer: \(49\).
(d)
Change the acceleration sign for the explicitly different stage.
Now a is positive, giving \(\displaystyle R=M(g+a)=91 N\).
Answer: \(91\).
Example 2 · Contact forces in a moving lift
A 8 kg block stays on the horizontal floor of a lift that is moving upward and slowing at \(\displaystyle \frac{4m}{s^{2}}\). Use \(\displaystyle g= \frac{10m}{s^{2}}.A\) scale is calibrated to display reaction divided by g as a mass reading.
Schematic force diagram for the selected particle; arrow lengths are not a force scale.
Find the normal reaction on the block in N.
Find the scale’s displayed mass in kg.
Find the magnitude of the force exerted by the block on the floor, in N.
In a separate stage with upward acceleration \(\displaystyle \frac{4m}{s^{2}}\), find the normal reaction in N.
Show the reasoning
(a)
Use the downward acceleration sign in the vertical force equation.
Upward-positive acceleration is −4, so \(\displaystyle R=Mg\)−Md=48 N.
Answer: \(48\).
(b)
Convert the measured support force using the stated scale calibration.
The calibrated display is \(\displaystyle \frac{R}{g}=\)\(\displaystyle \frac{24}{5}\) kg. The actual mass remains 8 kg.
Answer: \(\displaystyle \frac{24}{5}\).
(c)
Use Newton’s third law for the block-floor pair, keeping the recipient explicit.
The floor interaction partner has the same magnitude 48 N, directed downward on the floor.
Answer: \(48\).
(d)
Change the acceleration sign for the explicitly different stage.
Now a is positive, giving \(\displaystyle R=M(g+a)=112 N\).
Answer: \(112\).
Example 3 · Contact forces in a moving lift
A 9 kg block stays on the horizontal floor of a lift that is moving upward and slowing at \(\displaystyle \frac{1m}{s^{2}}\). Use \(\displaystyle g= \frac{10m}{s^{2}}.A\) scale is calibrated to display reaction divided by g as a mass reading.
Schematic force diagram for the selected particle; arrow lengths are not a force scale.
Find the normal reaction on the block in N.
Find the scale’s displayed mass in kg.
Find the magnitude of the force exerted by the block on the floor, in N.
In a separate stage with upward acceleration \(\displaystyle \frac{1m}{s^{2}}\), find the normal reaction in N.
Show the reasoning
(a)
Use the downward acceleration sign in the vertical force equation.
Upward-positive acceleration is −1, so \(\displaystyle R=Mg\)−Md=81 N.
Answer: \(81\).
(b)
Convert the measured support force using the stated scale calibration.
The calibrated display is \(\displaystyle \frac{R}{g}=\)\(\displaystyle \frac{81}{10}\) kg. The actual mass remains 9 kg.
Answer: \(\displaystyle \frac{81}{10}\).
(c)
Use Newton’s third law for the block-floor pair, keeping the recipient explicit.
The floor interaction partner has the same magnitude 81 N, directed downward on the floor.
Answer: \(81\).
(d)
Change the acceleration sign for the explicitly different stage.
Now a is positive, giving \(\displaystyle R=M(g+a)=99 N\).
Answer: \(99\).
Common mistakes
An upward-moving lift can have downward acceleration.
The scale reading can change while the actual block mass does not.
