Apparent Weight

Apparent weights are properties of the object that correspond to the weights of the object. Whenever the gravitational forces acting on an object are equal but not balanced by the opposite normal forces, the apparent weight of the object is different from the weight of the object. Thus, the weight of an object becomes equal to the amount of gravity acting on it. Read on further to understand the whole concept. 

Astronauts in low earth orbit with an apparent weight of zero weigh about the same as when standing on the ground. This is because gravity is about the same on low earth orbit and on the ground. Objects on the ground are bound to be subjected to the normal force applied from the ground. Only on the boundaries of objects in contact with the ground, experience normal force. This power is transmitted to the body. The gravity of any part of the body is balanced by the stress acting on that part, which is the reason why  Astronauts feel weightless.  You can capture this effect of tension by defining the apparent weight of the object in terms of the normal force. 

The apparent weight may differ from the weight of the object that is “partially or completely submerged” and has “buoyancy” from the liquid acting against gravity. Another illustration of the concept is the weight of a person in an elevator or lift. As the elevator begins to rise, the object begins to exert a downward force. When using a scale, the downward force makes the object heavier and changes its apparent weight. 

The weight of an object in the escalator

 When a man is standing in a lift or elevator, two forces act on him. The units are as follows:

• A man’s real weight acting vertically in the downwards direction (w = m * g)
• Elevator floor reaction force R against men acting upwards.

Example 1: The lift is either motionless or travelling at a steady pace.  

The downward force of a man’s weight = the upward force of the elevator floor while the elevator is stationary. 

As a result, either the elevator is immobile or it is travelling at a consistent speed. 

It has no acceleration.

(A = 0m / s²)

Therefore

Human weight = ground reaction = m * g

 This means that there is no imbalanced force acting on the man, as the man’s weight is offset by the upward reaction force on the elevator floor.

Example 2:  During upward acceleration of an elevator, an unbalanced force acts on a man.

This means that the upward reaction force of a man’s elevator floor is greater than the downward force of the man’s weight on the elevator floor.

From the second law of Newton’s second law,

 Force F = m * a

 Because the elevator is rising, it means that the reaction force on the

the floor is greater than the person’s weight (that is, R> W).

The effective force required to displace the lift = difference between the weight of the people in it and the reaction force of the ground.

 Effective force F = R – W

 Therefore, R – W = m * a

 R – m * g = m * a

 R = m * a + m * g

 Coefficient m: R = m (a + g)

 R = m (a + g) is also used as the apparent weight of a man as the elevator accelerates upwards. The scale records measurements that are larger than the man’s actual weight. (Apparent weight).

apparent weight W = R = m (a + g)

Now, you can calculate R, m, or a. This can be confirmed in the following work example.

Example 3: When the elevator accelerates and accelerates downwards, it looks like this:

 If the elevator accelerates downward with acceleration a, it means that an imbalanced force is acting on the man, which means that the man’s weight (downward force) is greater than the reaction force of the floor (upward) of the lift W> R

 weight> buoyancy reaction        

According to newton’s 2nd  law, 

. Force F = m * a 

 Since the lift accelerates down, it means that the man’s weight is lesser than the reaction force of the lift. (W> R) 

The effective force is the difference between the man’s weight and the reaction force of the lift. 

 Effective force F = man’s weight – reaction force of lift 

 F = W – R 

F = m *a. W = m * g 

 M * g – R = m * a 

 M * g – m * a = R 

Factorize m m (g – a) = R. ➡ R = m (g – a) 

Now, calculate m, R. This is shown in the worked illustration below. 

R = m (g – a) can also be called the apparent weight of the man. 

 W’ = R = m{ g – a). 

The man will feel lighter in weight. The scale will record a value that is lower than the true weight of the man. 

Example 4 when the lift is falling freely 

The lift falls with the acceleration that’s equal to the acceleration due to o gravity.From Newton’s, 2nd law,  

. Force F = m * a 

 W> R (the lift is falling) 

 Effective force F = W – R 

 Also. W – R = m * a 

 M * g – R = m * a 

 M * g-m * a = R 

 Factorise m m (g-a) = R 

 Recall that the lift is falling freely, also a = g. 

 Thus, 

. R = m (g – g) = (0) = 0 N 

We can see that R = 0. The man appears to have no weight ( light). The man and the bottom of the lift aren’t applying force on each other. The scale reads zero. 

Conclusion

Apparent weight is a very interesting concept, and when visualised and understood correctly it can be a thrill to study physics. It is one of those topics that ignite curiosity and wonder in physics, and if grasped accurately can improve understanding of other concepts in physics as well. However, to be adept at these concepts and to lock them in memory, it is important to keep revising them frequently.