Reaction in a Moving Lift
Reaction and Apparent Weight in a Moving Lift
When a person is standing on the floor of a stationary lift, their weight acts vertically downward and applies a force of mg on the floor. In response, the floor exerts an equal and opposite normal reaction force on the person. This reaction is what gives the sensation of weight and is referred to as the apparent weight of the person.
However, in a moving lift, the apparent weight changes depending on the acceleration or deceleration of the lift. The nature of the normal reaction in different conditions is discussed below:
1) Lift moving upward with uniform acceleration (a):
Reaction: R = m(g + a)
Apparent weight increases in this case.
2) Lift moving downward with uniform acceleration (a):
Reaction: R = m(g - a)
Apparent weight decreases in this case.
3) Lift moving with constant velocity (a = 0):
Reaction: R = mg
Apparent weight is equal to the actual weight.
4) Lift in free fall (a = g):
Reaction: R = m(g - g) = 0
Apparent weight becomes zero (weightlessness).
5) Lift accelerating downward with a > g:
Reaction: R = -m(a - g)
This results in a condition of negative weight or extreme weightlessness (not physically sustainable).
Some Important Problems
1) A 1 kg mass is hanging from the ceiling of a lift by a wire that can withstand a maximum tension of 14.7 N. What acceleration of the lift will cause the wire to break?
- a) More than 4.9 m/s² upward
- b) More than 4.9 m/s² downward
- c) Less than 4.9 m/s² upward
- d) Less than 4.9 m/s² downward
Answer: a) More than 4.9 m/s² upward
2) A lift is moving upward with an acceleration of 1 m/s². A 6 kg mass is hanging from the ceiling. What is the tension in the string?
- a) 60 N
- b) 66 N
- c) 54 N
- d) 42 N
Solution: T = m(g + a) = 6 × (10 + 1) = 66 N
Answer: b) 66 N
3) A spring balance is attached to the ceiling of a lift. A man hangs his bag from it. When the lift is stationary, the reading is 49 N. If the lift accelerates downward at 5 m/s², what will be the reading?
- a) 24 N
- b) 74 N
- c) 15 N
- d) 49 N
Solution: m = 49 / 10 = 4.9 kg ≈ 5 kg
R = m(g - a) = 5 × (10 - 5) = 25 N ≈ 24 N
Answer: a) 24 N
4) A lift is descending with 4 m/s² acceleration. The height from the floor to the ceiling is 3 m. A 10 g nail falls from the ceiling. What impulse (in kg·m/s) does it impart on the floor? (g = 10 m/s²)
- a) 0.01
- b) 0.02
- c) 0.03
- d) 0.06
Solution:
Relative acceleration = g - a = 6 m/s²
s = ½ at² ⇒ 3 = ½ × 6 × t² ⇒ t = 1 s
v = at = 6 × 1 = 6 m/s
m = 10 g = 0.01 kg
Impulse = mv = 0.01 × 6 = 0.06 kg·m/s
Answer: d) 0.06
0 Comments
Post a Comment