Weight Of A Body / Object Inside A Lift
When a man stand in a lift or elevator, two forces act on the man.
The forces are:
I. The true weight of the man which act vertically downward (w = m*g)
II. The reaction force R of the floor of the lift on the man, which act upward.
We are going to look into different cases when the man is on the lift.
Case 1: when the lift is stationary or moving with a constant velocity.
When the lift is stationary, the downward force due to the weight of the man is equal to the upward reaction force of the floor of the lift. Therefore the lift is stationary or moving with constant velocity. Its acceleration is zero. ( a = 0 m/s² )
Therefore,
Weight of man = Reaction of floor = m*g
It means that there is no unbalance force acting on the man because the man’s weight is cancelled by the upward reaction force of the floor of he lift. If the man stands on a spring scale, the scale will record the man’s true weight.
Case2: when the lift accelerates upward with acceleration a:
If the lift accelerates upward with acceleration a, there is unbalance force acting on the man. This means that the upward reaction force of the floor of the lift on the man is greater than the downward force due to the weight of the man on the floor of the lift.
From Newton’s second law of motion,
Force F = m*a
Since the lift move upward, it means that floor Reaction is greater than man’s weight ( I.e R > W )
Effective force that move the lift is the difference between Man’s weight and floor reaction.
Effective force F = R – W
Therefore R – W = m*a
R – m*g = m*a
R = m*a + m*g
Factorize m: R = m ( a + g )
R = m ( a + g ) is also regarded as the apparent weight of the man when the lift is accelerating upward. The scale will record a value that is greater than the mans true weight. ( apparent weight ).
That is apparent weight W = R = m(a + g)
At this point, you can calculate R, m or a. This we shall see in the worked example below.
Case 3: If the lift is accelerating downwards with an acceleration, a:
If the lift accelerates downwards with acceleration a, it means that there is unbalancing force that is acting on the man. It means that the downward force due to the man’s weight is greater than the upward reaction force of the lift floor.
( W > R )
Man’s weight > Reaction force of lift floor
From Newton’s second law of motion,
Force F = m*a
Since the lift accelerates downwards, it means that the man’s weight is greater than the reaction force of the lift floor. ( W > R )
The effective force is the difference between the mans weight and the reaction force of the lift.
Effective force F = man’s weight – reaction force of lift floor
F = W – R
If we substitute for F in F = m*a, we will get
W – R = m*a
M*g – R = m*a
M*g – m*a = R
Factorize m: m ( g – a ) = R. ➡ R = m ( g – a )
At this point, you can calculate m, R or a by making it the subject of the formula. This we shall see in the worked example below.
R = m ( g – a ) can also be called apparent weight of the man.
W' = R = m { g – a ).
The man will fill lighter in weight. The scale will record a value the is smaller than the true weight of the man.
Case 4: when the lift is falling freely:
If the lift accelerate downward with acceleration that is equal to acceleration due o gravity, we say the lift is falling freely. His situation happen when the rope / cable of the lift cut.
From Newton’s second law of motion,
Force F = m*a
W > R ( the lift is falling )
Effective force F = W – R
Then. W – R = m*a
M*g – R = m*a
M*g - m*a = R
Factorize m: m( g - a ) = R
Recall that the lift is falling freely, then a = g .
Therefore,
R = m ( g – g ) = ( 0 ) = 0 N
We can see that R = 0. The man appears to have no weight ( weightless ). The man and the floor of the lift are not exerting force on each other. The scale of the weighing instrument reads zero.
Weightlessness In Satellites:
People and object is a satellite that is orbiting close to the earth experience weightlessness.
Exercises:
1. A ball of mass 6 kg moving with a velocity of 10 m/s collides with another ball of equal mass at rest. If the two balls off after the impact, calculate their common velocity.
2. A ball of mass 6 kg moving with a velocity of 10 m/s collides with a 2.0 kg mass ball moving in the opposite direction with a velocity of 5 m/s . After the collision, the two balls coalesce and move in the same direction. Calculate their common velocity.
3. A constant force of 5 N act for 7 seconds on a mass of 7 kg initially at rest. Calculate final momentum .
4. A foot baller applied a force of 39N when taking a penalty kick for a period of 0.24 second. If the mass of the ball is 0.045 kg, calculate the speed with which the ball took off.
5. A body of mass 100 g moving with a velocity of 10 m/s collides with a wall. Of after the collision, it moves with a velocity of 3 m/s in the opposite direction, calculate the change in momentum.
6. Define impulse. How is the rate of change of momentum related to the force that act on the body?
7. Explain she the velocity of a recoiling gun is lesser than that of the bullet shot out of he gun.
8. A 30 g bullet moving at 250 m/s hits a bag of sand and come to rest in 0.011 second. What is the momentum of the bullet just before hitting the bag? Find the average force that stopped the bullet.
9. State Newton’s laws of motion. Derive from one of the relation between the momentum produce in a body and the force applied on the body.
10. State the conservation law of linear momentum. A 250 g riffle lay on a smooth table when its suddenly discharges , firing a bullet of 0.05 g with a speed of 450 m/s . calculate the recoil speed of the gun.
11. Distinguish between mass, weight, momentum and inertia. Describe an experiment to demonstrate that momentum is conserved in a collision.
12. An object of mass 1.3kg falls a distance of 5 m to a horizontal surface and rebounds to a vertical height of 3 m . calculate the change in momentum.
13. A man whose mass is 85 kg stand on a spring weighing machine inside a lift. What is the reading of the weighing machine when:
I) When the lift starts to ascend with an acceleration of 2.0 m/s²?
II) The lift is moving with a uniform velocity?
III) The lift is coming to rest with a retardation of 4.0 m/s²
14. Explain the principle underlying the propulsion of rocket.
15. A rocket of mass 5800kg containing a propellant gas of 2000kg is to be launched vertically. If the fuel is consumed at a steady rate of 70kg/s, calculate the least velocity of the exhaust gases if the rocket and content will just lift off the launching pad immediately after firing.
16. Distinguish between impulse and momentum. How are they related? Calculate the momentum of : (i) The moon of mass 7x10^²²kg travelling with a velocity of 1.25 m/s. (ii) a tanker of mass 7x10^⁴ kg travelling at a speed of 6 m/s. (iii) an aircraft of mass 3.5 x 10^³ kg travelling at a speed of 400 m/s
17. Distinguish between elastic collision and inelastic collision. (ii) inertial mass and weight. From Newton’s law of motion show that F = ma. ( relationship between mass, force and acceleration ).
18. A 15 kg monkey hang from a cord suspended from the ceiling of a lift. The cord can withstand a tension of 200N and brakes as the lift accelerates. What was the lift minimum acceleration ( in magnitude and direction) g = 10.0 m/s².
19. A tractor of mass 5 x10³ kg s used to tow a car f mass 2.5 x10³ kg. The tractor moves with a speed of 3.0 m/s just before the towing rope become taut . calculate the (I) the speed of the tractor immediately the rope becames taut. (ii) loss kinetic energy of the system just after the car has started moving. (iii) impulse in the rope when it jerk the car into motion.
When a man stand in a lift or elevator, two forces act on the man.
The forces are:
I. The true weight of the man which act vertically downward (w = m*g)
II. The reaction force R of the floor of the lift on the man, which act upward.
We are going to look into different cases when the man is on the lift.
Case 1: when the lift is stationary or moving with a constant velocity.
When the lift is stationary, the downward force due to the weight of the man is equal to the upward reaction force of the floor of the lift. Therefore the lift is stationary or moving with constant velocity. Its acceleration is zero. ( a = 0 m/s² )
Therefore,
Weight of man = Reaction of floor = m*g
It means that there is no unbalance force acting on the man because the man’s weight is cancelled by the upward reaction force of the floor of he lift. If the man stands on a spring scale, the scale will record the man’s true weight.
Case2: when the lift accelerates upward with acceleration a:
If the lift accelerates upward with acceleration a, there is unbalance force acting on the man. This means that the upward reaction force of the floor of the lift on the man is greater than the downward force due to the weight of the man on the floor of the lift.
From Newton’s second law of motion,
Force F = m*a
Since the lift move upward, it means that floor Reaction is greater than man’s weight ( I.e R > W )
Effective force that move the lift is the difference between Man’s weight and floor reaction.
Effective force F = R – W
Therefore R – W = m*a
R – m*g = m*a
R = m*a + m*g
Factorize m: R = m ( a + g )
R = m ( a + g ) is also regarded as the apparent weight of the man when the lift is accelerating upward. The scale will record a value that is greater than the mans true weight. ( apparent weight ).
That is apparent weight W = R = m(a + g)
At this point, you can calculate R, m or a. This we shall see in the worked example below.
Case 3: If the lift is accelerating downwards with an acceleration, a:
If the lift accelerates downwards with acceleration a, it means that there is unbalancing force that is acting on the man. It means that the downward force due to the man’s weight is greater than the upward reaction force of the lift floor.
( W > R )
Man’s weight > Reaction force of lift floor
From Newton’s second law of motion,
Force F = m*a
Since the lift accelerates downwards, it means that the man’s weight is greater than the reaction force of the lift floor. ( W > R )
The effective force is the difference between the mans weight and the reaction force of the lift.
Effective force F = man’s weight – reaction force of lift floor
F = W – R
If we substitute for F in F = m*a, we will get
W – R = m*a
M*g – R = m*a
M*g – m*a = R
Factorize m: m ( g – a ) = R. ➡ R = m ( g – a )
At this point, you can calculate m, R or a by making it the subject of the formula. This we shall see in the worked example below.
R = m ( g – a ) can also be called apparent weight of the man.
W' = R = m { g – a ).
The man will fill lighter in weight. The scale will record a value the is smaller than the true weight of the man.
Case 4: when the lift is falling freely:
If the lift accelerate downward with acceleration that is equal to acceleration due o gravity, we say the lift is falling freely. His situation happen when the rope / cable of the lift cut.
From Newton’s second law of motion,
Force F = m*a
W > R ( the lift is falling )
Effective force F = W – R
Then. W – R = m*a
M*g – R = m*a
M*g - m*a = R
Factorize m: m( g - a ) = R
Recall that the lift is falling freely, then a = g .
Therefore,
R = m ( g – g ) = ( 0 ) = 0 N
We can see that R = 0. The man appears to have no weight ( weightless ). The man and the floor of the lift are not exerting force on each other. The scale of the weighing instrument reads zero.
Weightlessness In Satellites:
People and object is a satellite that is orbiting close to the earth experience weightlessness.
Exercises:
1. A ball of mass 6 kg moving with a velocity of 10 m/s collides with another ball of equal mass at rest. If the two balls off after the impact, calculate their common velocity.
2. A ball of mass 6 kg moving with a velocity of 10 m/s collides with a 2.0 kg mass ball moving in the opposite direction with a velocity of 5 m/s . After the collision, the two balls coalesce and move in the same direction. Calculate their common velocity.
3. A constant force of 5 N act for 7 seconds on a mass of 7 kg initially at rest. Calculate final momentum .
4. A foot baller applied a force of 39N when taking a penalty kick for a period of 0.24 second. If the mass of the ball is 0.045 kg, calculate the speed with which the ball took off.
5. A body of mass 100 g moving with a velocity of 10 m/s collides with a wall. Of after the collision, it moves with a velocity of 3 m/s in the opposite direction, calculate the change in momentum.
6. Define impulse. How is the rate of change of momentum related to the force that act on the body?
7. Explain she the velocity of a recoiling gun is lesser than that of the bullet shot out of he gun.
8. A 30 g bullet moving at 250 m/s hits a bag of sand and come to rest in 0.011 second. What is the momentum of the bullet just before hitting the bag? Find the average force that stopped the bullet.
9. State Newton’s laws of motion. Derive from one of the relation between the momentum produce in a body and the force applied on the body.
10. State the conservation law of linear momentum. A 250 g riffle lay on a smooth table when its suddenly discharges , firing a bullet of 0.05 g with a speed of 450 m/s . calculate the recoil speed of the gun.
11. Distinguish between mass, weight, momentum and inertia. Describe an experiment to demonstrate that momentum is conserved in a collision.
12. An object of mass 1.3kg falls a distance of 5 m to a horizontal surface and rebounds to a vertical height of 3 m . calculate the change in momentum.
13. A man whose mass is 85 kg stand on a spring weighing machine inside a lift. What is the reading of the weighing machine when:
I) When the lift starts to ascend with an acceleration of 2.0 m/s²?
II) The lift is moving with a uniform velocity?
III) The lift is coming to rest with a retardation of 4.0 m/s²
14. Explain the principle underlying the propulsion of rocket.
15. A rocket of mass 5800kg containing a propellant gas of 2000kg is to be launched vertically. If the fuel is consumed at a steady rate of 70kg/s, calculate the least velocity of the exhaust gases if the rocket and content will just lift off the launching pad immediately after firing.
16. Distinguish between impulse and momentum. How are they related? Calculate the momentum of : (i) The moon of mass 7x10^²²kg travelling with a velocity of 1.25 m/s. (ii) a tanker of mass 7x10^⁴ kg travelling at a speed of 6 m/s. (iii) an aircraft of mass 3.5 x 10^³ kg travelling at a speed of 400 m/s
17. Distinguish between elastic collision and inelastic collision. (ii) inertial mass and weight. From Newton’s law of motion show that F = ma. ( relationship between mass, force and acceleration ).
18. A 15 kg monkey hang from a cord suspended from the ceiling of a lift. The cord can withstand a tension of 200N and brakes as the lift accelerates. What was the lift minimum acceleration ( in magnitude and direction) g = 10.0 m/s².
19. A tractor of mass 5 x10³ kg s used to tow a car f mass 2.5 x10³ kg. The tractor moves with a speed of 3.0 m/s just before the towing rope become taut . calculate the (I) the speed of the tractor immediately the rope becames taut. (ii) loss kinetic energy of the system just after the car has started moving. (iii) impulse in the rope when it jerk the car into motion.
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