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In order to define apparent weight, the apparent weight of an accelerating object is the vector sum of its true weight and the negative of all the forces that generate the object's acceleration. The apparent weight of a mass represented by â€˜mâ€™ is its real weight minus its mass times the acceleration of the frame (vector addition). Your true weight is induced by gravity - it is the force exerted on you through gravity; generally the earth's gravity.

Letâ€™s learn how to calculate the apparent weight. Your apparent weight is the sum of your real weight and a fictitious force related to your acceleration.

Usually, an object's apparent weight is its mass multiplied by the vector difference between the acceleration of the object and the gravitational acceleration. This definition means that apparent weight is a vector that can make a move in any direction, not just vertically.

Thus, apparent weight formula; a = dv/dt

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You might already be aware that the earth attracts everybody towards its centre with some force of attraction known as the gravity force. Because of this force of gravity, all bodies experience their weight. Given that â€˜mâ€™ is the mass of a body then its weight is â€˜mgâ€™ (vertically downwards) which is equivalent to the force of gravity.

But if the body is on a plane that is accelerated up or down, then the force deployed on the plane by the body (i.e., the weight of the body) alters while the gravity force remains the same. In such an instance, a person perceives his altered weight (apparent weight).

It implies real weight = true weight. True or real weight simply is weight. So, what is your true weight? It's simply the mg. Mass multiplied by gravity.

Apparent weight is represented by WA. It's described as;

WA = N

In which, â€˜Nâ€™ denotes the normal force in the direction opposite to the direction of gravity. That is simply away from the centre of the earth. You may be standing and someone may be attempting to push you horizontally. That normal force of reaction doesn't count. Only the vertical Normal Force is countable.

So assume that you jump from the top of the building since your cat died. You are falling and your '(True) Weight' is simply mg. Your apparent weight will be 0. Because there is no normal force exerted on you presently (Of course the ground will apply normal force when you finally reach it).

Now assume that you are standing in an elevator at rest. The (â€˜Trueâ€™) weight, of course, is mg. But the apparent weight is also mg. Since you are at rest, N = mg.

Elevator moving with constant speed: N = mg.

Assume that the magnitude of an elevator's acceleration is: |a|.

Elevator going up and slowing down: N = mg âˆ’ m|a|.

Elevator going up, and increasing speed: N = mg + m|a|.

Elevator going downwards and slowing down: N = mg + m|a|.

Elevator going up, and growing speed: N = mg âˆ’ m|a|.

Note: Apparent weight is the weight you 'feel'. When you are falling, you feel weightlessness. Thus, the Apparent Weight is 0. When in an elevator going upwards with growing speed, you feel heavier. Therefore, more is the Apparent Weight!

Letâ€™s check about the apparent weight of a man in a lift or elevator.

Assume that a boy of mass â€˜mâ€™ is standing on a weighing machine placed in a lift. The actual weight of the man will be â€˜mgâ€™. This weight bears down on the machine. The machine also expends a reactionary force â€˜Râ€™ on the boy in an upward direction where â€˜R = Wâ€™ (Newtonâ€™s 3rd Law). Hence, two forces are acting on the boy; reactionary force â€˜Râ€™ and gravity force â€˜mgâ€™. Seeing that, the two forces are in opposite directions, the net force on the boy is given by:

F = mg â€“ R (downwards)

The person is at rest (no acceleration) thus, the net force on him must be zero, that is

F = mg â€“ R = 0

R = mg

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If something is pulled with an ongoing force (and there is nothing dragging it back) it will continuously speed up! So, changing force and speed are associated.

Have you ever observed how you feel lighter when an elevator first begins going down, and you feel heavier when it slows again and moves upwards steadily?

That is due to the fact that the speed requires to be changing in order to affect the apparent weight!

If the elevator is moving at a constant speed then there is no difference in the apparent weight in comparison to when it is sitting still. Why is it so? This is because it takes force to make an object move faster (or slower). If something moves at a constant speed you will not feel any extra force.

That is why you can easily sit in a moving car or train, and everything appears normal (unless the driver speeds up, slows down or puts the brakes on).

FAQ (Frequently Asked Questions)

Q1. How is the Apparent Weight of an Object Calculated? What is the Formula for that?

Answer: Apparent weight generally refers to the weight of the object when suspended freely in water. Here, an upward force (known as buoyancy) acts up. This is the weight of the water displaced by the volume of the object.

Thus, Wt = M*g.

Where,

Wt = true Weight

M = mass of the body

g = acceleration due to gravity = (9.8 m/s^{2} at sea level)

However, when weighing an object resting in air, it is also subjected to an up buoyancy force, because of the weight of air displaced by the object (note that the air weighs about 1.2 kg /m^{3} x g). For objects of much higher density than air, however, this up force is small and can be overlooked.