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. Â
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.Â
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.Â
Apparent weight and true weightÂ
True weight of a body is basically defined as the
force exerted by the earth on it.Â
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.Â
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.
- Force due to gravity of weight of the body i.e. W = mgÂ
- Upward normal contact force i.e. N exerted by the platform of weighing machine on the bodyÂ
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.Â
Apparent weight of a man in a liftÂ
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.Â
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.Â
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.Â
Case 1: Apparent weight of a man in a lift when lift is moving up and acceleratingÂ
Let us consider that a man is standing inside a lift
over the spring scale as displayed here in following figure.Â
Let us think that lift is moving up with an
acceleration a. Â
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.Â
∑ Fy = m x ay
R – mg = ma
R = m (g + a)Â
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.Â
Spring scale will display the weight of man equivalent
to the reaction force exerted on the man by the surface of spring scale.Â
Therefore, apparent weight of a man in a lift, going
upward with an acceleration, will be higher than his actual weight or true
weight.Â
Case 2: Apparent weight of a man in a lift when lift is moving with a constant velocity i.e. acceleration is zeroÂ
Let us consider that a man is standing inside a lift
over the spring scale as displayed here in following figure.Â
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. Â
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. Â Â
∑ Fy = m x ay
R – mg = 0
R = mg  Â
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.Â
Spring scale will display the weight of man equivalent
to the reaction force exerted on the man by the surface of spring scale.Â
Therefore, apparent weight of a man in a lift, going
with a constant velocity, will be equal to his actual weight or true weight.Â
Case 3: Apparent weight of a man in a lift when lift is moving down and acceleratingÂ
Let us consider that a man is standing inside a lift
over the spring scale as displayed here in following figure.Â
Let us think that lift is moving down with an
acceleration a. Â
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.Â
∑ Fy = m x ay
R – mg = - ma
R = m (g - a)Â
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.Â
Spring scale will display the weight of man equivalent
to the reaction force exerted on the man by the surface of spring scale.Â
Therefore, apparent weight of a man in a lift, going downward
with an acceleration, will be lower than his actual weight or true weight.Â
Case 4: Apparent weight of a man in a lift when lift cable is snappedÂ
Let us consider that a man is standing inside a lift
over the spring scale as displayed here in following figure.Â
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. Â
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.Â
∑ Fy = m x ay
R – mg = - mg
R = 0Â
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. Â Â
Spring scale will display zero as the apparent weight
of man.Â
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.Â
Further we will find out another concept in
engineering mechanics i.e. Lami's theorem with the help of our next post. Â
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right hand side of page for further and continuous update from www.hkdivedi.com. Â
Reference:Â Â
Engineering Mechanics, By Prof K. Ramesh Â
Image courtesy: Google  Â
Also read  Â
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