We were discussing the projectile motion - trajectory equation, definition and formulas and we have also seen the concept of Terminal velocity with the help of our previous posts.

Now, we will be interested further to understand a very important topic in engineering mechanics i.e. apparent weight of a man in a lift.

Today we will first understand the basics of apparent weight and true weight and further we will find out the apparent weight of a man in a lift for various situations.

### Apparent weight and true weight

True weight of a body is basically defined as the force exerted by the earth on it.

Let us think that a person whose mass is m and he is located at or near the earth surface, true weight or simply weight of the person will be mg. Where g is the acceleration due to gravity.

Weight of the body will be measured with the help of weighing machine. When a body will be placed over the surface of weighing machine platform, there will be acting basically two types of forces.
1. Force due to gravity of weight of the body i.e. W = mg
2. Upward normal contact force i.e. N exerted by the platform of weighing machine on the body

The upward normal contact force (N) exerted on the body will be termed as the apparent weight of the body. Value of the upward normal contact force will be dependent over the state of motion of the body.

### Apparent weight of a man in a lift

Now let us see the apparent weight of a man in a lift for the various state of motion of the lift. Please note that we are going to use here the Newton’s law of motion in order to secure the expression for the apparent weight of a man in a lift for the various state of motion of the lift.

I hope all of you are well aware with the lifts or elevators used in various multi-storey buildings or towers. If you have an opportunity to travel in a lift that travels longer distance such as lift of a 50 storey or 70 storey buildings, you will surely feel that there is acceleration, deceleration and steady state motion too.

So, we will analyse and find out the apparent weight of a man for each state of motion of the lift and we will also see here a case of snap of lift or elevator with the help of this post.

### Case 1: Apparent weight of a man in a lift when lift is moving up and accelerating

Let us consider that a man is standing inside a lift over the spring scale as displayed here in following figure.

Let us think that lift is moving up with an acceleration a.

Let us draw the free body diagram, as displayed above, considering the man as a particle. I will determine the apparent weight of man by using the newton’s law of motion and please note that I am not standing inside the lift. I am standing outside of the lift on ground and analysing the apparent weight of man in the lift.

∑ Fy = m x ay
R – mg = ma
R = m (g + a)

As we have already seen here that apparent weight is basically the reaction force exerted on the body and hence apparent weight will be higher as compared to the actual weight of the man.

Spring scale will display the weight of man equivalent to the reaction force exerted on the man by the surface of spring scale.

Therefore, apparent weight of a man in a lift, going upward with an acceleration, will be higher than his actual weight or true weight.

### Case 2: Apparent weight of a man in a lift when lift is moving with a constant velocity i.e. acceleration is zero

Let us consider that a man is standing inside a lift over the spring scale as displayed here in following figure.

Let us think that lift is moving with a constant velocity V. We will see here the apparent weight of man and we will find out here that what will be the reading of spring scale.

Let us draw the free body diagram, as displayed above, considering the man as a particle. I will determine the apparent weight of man by using the newton’s law of motion and please note that as lift is moving with a constant velocity and hence acceleration will be zero.

∑ Fy = m x ay
R – mg = 0
R = mg

As we have already seen here that apparent weight is basically the reaction force exerted on the body and hence apparent weight will be same as the actual weight of the man.

Spring scale will display the weight of man equivalent to the reaction force exerted on the man by the surface of spring scale.

Therefore, apparent weight of a man in a lift, going with a constant velocity, will be equal to his actual weight or true weight.

### Case 3: Apparent weight of a man in a lift when lift is moving down and accelerating

Let us consider that a man is standing inside a lift over the spring scale as displayed here in following figure.

Let us think that lift is moving down with an acceleration a.

Let us draw the free body diagram, as displayed above, considering the man as a particle. I will determine the apparent weight of man by using the newton’s law of motion and please note that I am not standing inside the lift. I am standing outside of the lift on ground and analyzing the apparent weight of man in the lift.

∑ Fy = m x ay
R – mg = - ma
R = m (g - a)

As we have already seen here that apparent weight is basically the reaction force exerted on the body and hence apparent weight will be lower as compared to the actual weight of the man.

Spring scale will display the weight of man equivalent to the reaction force exerted on the man by the surface of spring scale.

Therefore, apparent weight of a man in a lift, going downward with an acceleration, will be lower than his actual weight or true weight.

### Case 4: Apparent weight of a man in a lift when lift cable is snapped

Let us consider that a man is standing inside a lift over the spring scale as displayed here in following figure.

Let us think that lift cable is snapped and lift is free falling. Lift will free fall with an acceleration g i.e. acceleration due to gravity.

Let us draw the free body diagram, as displayed above, considering the man as a particle. I will determine the apparent weight of man by using the newton’s law of motion and please note that I am not standing inside the lift. I am standing outside of the lift on ground and analyzing the apparent weight of man in the lift.

∑ Fy = m x ay
R – mg = - mg
R = 0

As we have already seen here that apparent weight is basically the reaction force exerted on the body and hence apparent weight will be zero.

Spring scale will display zero as the apparent weight of man.

Therefore, we have seen here the basics of apparent weight and true weight of a body with the help of this post. We have also secured here the apparent weight of a man in a lift for various situations.

Further we will find out another concept in engineering mechanics i.e. Lami's theorem with the help of our next post.

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### Reference:

Engineering Mechanics, By Prof K. Ramesh