Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store

Apparent Weight

ffImage
Last updated date: 25th Apr 2024
Total views: 345k
Views today: 8.45k
hightlight icon
highlight icon
highlight icon
share icon
copy icon

Definition of Apparent Weight

Accordingly, the apparent weight of an object that is accelerating is equal to the vector sum of its true weight and the negative of all of its acceleration forces. 'M' represents mass minus mass times frame acceleration multiplied by the real weight of that mass. It is gravity that determines your true weight- it is the force exerted upon you by gravity, which is usually the earth's gravity.

 

Formula for Apparent Weight

The apparent weight can be calculated using the formula below. You appear to weigh more than your actual weight if you add to it the effect of your acceleration.

A weight is usually measured as the vector difference between an object's acceleration and gravity's acceleration multiplied by its mass. Appearance weight can therefore be defined as a vector with a range of movement, not only vertically.

As a result, the apparent weight formula is; a = dv/dt

 

Real vs Apparent Weight

Real weight = true weight, so it follows that real weight = true weight. Do you know what your real weight is? It's simply the mg. It's the mass multiplied by gravity.

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

WA = N 

As shown, 'N' represents the normal force in the direction opposite to the direction of gravity. Basically, it means the opposite of the direction of the earth's center. It is possible that someone is pushing you horizontally while you are standing. Normally, this response is not taken into account. Normal Forces are a one-dimensional force.

Consider that you have just lost your cat and you are jumping from the top of the building. You are falling and your '(True) Weight' is simply mg. You appear to weigh zero. Due to the absence of normal force currently exerted on you by the ground (assuming you finally reach it, the ground will apply normal force).

If you are at rest in an elevator, imagine you are standing there. The (‘True’) weight, of course, is mg. Nevertheless, the apparent weight is mg as well. When at rest, N = mg.

The elevator moves at a constant speed: N = mg.

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

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

 

Changing Speed

When something is continuously pulled (with nothing dragging it back), it will speed up continuously! There is an association between force and speed. 

When an elevator first begins to descend, you feel lighter, whereas when it slows down again and moves upward steadily, you feel heavier. 

The reason for that is because the apparent weight must change if the speed changes!

There is no difference in apparent weight in an elevator that is moving at a constant speed in comparison with an elevator that is stationary. How can this be? An object must be moved with force in order to move more quickly or slower. The force that is applied is not increased if something moves at a constant speed.

In a moving car or train, you can sit comfortably and everything seems normal except when the driver accelerates, decelerates or puts the brakes on.

 

Apparent Weight in Lift

Here is how much the apparent weight of a man is in a lift or elevator.

In a lift, a young boy with a mass of 'm' stands on a weighing machine. The weight of the man will be 'mg'. As a result, the machine is burdened by this weight. The machine also expends a reactionary force ‘R’ on the boy in an upward direction where ‘R = W’ (Newton’s 3rd Law). Consequently, the boy is under the influence of both reactionary and gravitational forces. Considering the two forces are in opposing directions, the net force on the boy can be calculated as follows:

F = mg – R (downwards)

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

F = mg – R = 0

R = mg

FAQs on Apparent Weight

1. How do you define apparent weight?

  • A property of objects that characterizes their apparent weight is their apparent weight. 

  • In situations where an object is partially submerged in a liquid, or fully submerged in a liquid, there can be an apparent different weight from its actual weight, which is determined by the upthrust from the liquid, opposed to gravity.

  • Objects or people riding in elevators are another example of weight. As things move up in the elevator, they begin exerting forces in the opposite direction (downward).

  • The apparent weight then changes because of a downward force.

2. What is the formula for calculating the apparent weight of an object? How do you calculate that?

Generally, the apparent weight refers to the weight of the object when suspended in water. Buoyancy is responsible for this upward force. Weight calculated by dividing the volume of the object by the volume of water displaced.

Thus, Wt = M*g.

Where,

Wt = true Weight

M = mass of the body

g = acceleration due to gravity = (9.8 m/s2 at sea level)

Whenever we weigh an object which is resting in air, the object is also subject to up buoyancy forces due to the displaced air weight (the weight of the air is about 1.2 kg/m3 x g). This up force, however, can be ignored for objects that are much denser than air.

3. Is Gravity responsible for weight gain?

Many people are already aware that there is an attraction force known as gravity that draws everyone towards the center of the earth. The consequences of gravity are felt in the bodies of everyone. Because mass is measured in kilograms, the weight of a body is measured in grams (vertically downwards), so the weight is equal to gravity.

 

On the other hand, 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) changes while the gravity force remains the same. A person will perceive his apparent weight as being altered in such a case.