Apparent Weight in Physics
These are my notes and examples on apparent weight in physics.
Apparent Weight
You usually encounter apparent weight when working force problems. One of the
first points to get straight is the difference between mass and weight. Mass is
the property that makes different objects accelerate differently when equal
forces are applied. operationally, mass is the m in the equation \(F=ma\). Mass
is measured in kilograms. Weight in equation form is \(W=mg\) and is the force
that something exerts. Weight is measured in newtons or pounds.
The first step in keeping all this straight is to remember that weight is force.
A scale that measures weight in pounds or newtons is just a force meter. Tension
in a rope is a force.
A person of 80 kg mass standing on the surface of the Earth is subject to the
force of gravity that acts between the 80 kg person and the mass of the earth.
This force is expressed as an acceleration due to gravity. On the earth, this
acceleration is \(9.8 m/s^2\). The force on the person, also called weight, is
\[F=80kg*9.8 m/s^2 = 784N\]
A force meter placed between the person and the earth would read 784 newtons.
To understand apparent weight, think of this analogy. Imagine a force meter
between you and the floor of a elevator. As you step on the elevator, the force
meter reads the same as when you are standing on the ground. As the elevator
accelerates upward, the force meter registers higher, reading maximum at maximum
acceleration.
As the acceleration decreases and the elevator assumes a constant
velocity(upward), the force meter reads the same as when the elevator was not
moving. This constant velocity condition is equivalent to forces in equilibrium,
so the only thing contributing to the reading of your force meter is the mg due
to the earth. As the elevator slows, by decelerating, the force meter reads
less than the force for constant velocity. As the elevator comes to a rest at a
higher level, the force meter again reads the same as when you were at ground
level.
Let's go back over this elevator ride and put in some numbers. At zero or
constant velocity, the force meter reads 784 N. If the elevator were accelerating
upward at \(2.0 m/s^2\), then the force meter would read:
\[F=784N+2.0 m/s^2(80kg)=944N\]
If the elevator were accelerating downward at \(3.0 m/s^2\), then the force
meter would read:
\[F=784N-3.0m/s^2(80)kg=544N\]
If the elevator were in free fall, then the force meter would read zero. The
acceleration acting on the elevator is the same as on the person.
Example 1
A 12 kg flower pot is hanging by a cord from the roof of an elevator. What is
the tension in the cord when the elevator is stationary and when it is
accelerating upward at \(3.0 m/s^2\)?
Solution:
When the elevator is stationary, the tension in the cord is:
\[T=12kg(9.8m/s^2)=118N\]
When the elevator is accelerating upward, the tension in the cord is increased
by an amount equal to ma:
\[T=118N+12kg(3.0m/s^2)=154N\]
Example 2
A rope fastened at the top is hanging over a cliff. What is the tension in the
rope with a 70 kg mountain climber sliding down the rope at a constant
acceleration of \(6.0 m/s^2\)?
Solution:
The tension in the rope is less than if the climber were at zero velocity or
sliding at constant velocity. In this instance, the \(6.0m/s^2\) must be
subtracted from the \(9.8 m/s^2\).
\[T=(9.8-6.0) m/s^2(79 kg)=266N\]
When in doubt whether to add or subtract the acceleration from g, look to the
extreme situation. In this problem, the climber sliding down the rope at
constant velocity would produce \(9.8m/s^2(70kg)=686N\) of tension in the rope.
If the climber were in free fall, or accelerating \(9.8m/s^2\), he or she would
produce no tension in the rope. So any acceleration down the rope should be
subtracted from the \(9.8m/s^2\).
Example 3
What is the tension in a rope with the 70 kg mountain climber accelerating up
the rope at a constant acceleration of \(1.0 m/s^2\)?
Solution:
In this case, the tension in the rope is increased by ma:
\[T=(9.8+1.0)m/s^2(70 kg)=756 N\]
Example 4
A 100 kg astronaut produces a force(weight) on the surface of the earth of 980
N. What force(weight) would the astronaut produce on the surface of the moon,
where g is about 1/6 the g on earth?
Solution:
The force or weight would be:
\[F=100 kg(9.8 m/s^2)1/6 = 163 N\]
Example 5
A 16 kg monkey wishes to raise a 20 kg mass by climbing(accelerating) up a rope
that passes over a pulley attached to the mass. How much must the monkey
accelerate up the rope to raise the mass?
Solution:
The mass produces a force or tension in the rope of:
\[T=20kg(9.8m/s^2)=196N\]
The mass of the monkey hanging on the rope produces a force of:
\[T=16kg(9.8m/s^2)=157N\]
To just balance the mass, the monkey must accelerate up the rope to produce a
force of:
\[(196-157)N=39N\]
So, \(39N=(16kg)a \rightarrow a=2.4m/s^2\)
At this acceleration, the tension in the rope is 196 N. To raise the mass, the
monkey must accelerate at a rate greater than \(2.4 m/s^2\). Any acceleration
greater than 2.4 will increase the tension in the rope by an amount equal to the
additional acceleration times the mass of the monkey.