In Example 5.8, we investigated the apparent weight of a fish in an elevator. Now consider a
72.0 kg man standing on a spring scale in an elevator. Starting from rest, the elevator ascends,
attaining its maximum speed of 1.20 m/s in 0.80 s. It travels with this constant speed for the
next 5.00 s. The elevator then undergoes a uniform acceleration in the negative y direction for
1.50 s and comes to a rest. What does the spring scale register: a) before the elevator starts to
move, b) during the first 0.80 s, c) while the elevator is travelling at constant speed, and d)
during the time interval it is slowing down.
a) Before the elevator starts to move, the spring scale will register the man's weight, which is approximately equal to the weight of his struggles to get through Monday mornings.
b) During the first 0.80 s when the elevator is accelerating upwards, the spring scale will register a slightly higher weight than the man's actual weight. It's like trying to hold onto your snack in a roller coaster- things tend to feel a bit heavier when you're going up.
c) While the elevator is traveling at a constant speed for 5.00 s, the spring scale will register the man's actual weight. It's the moment of calm when you realize life isn't throwing any surprises at you, and you can just sit back and enjoy the ride without any extra baggage.
d) During the time interval when the elevator is slowing down, the spring scale will register a slightly lower weight than the man's actual weight. It's like when you see the weekend coming to an end and your excitement levels drop, making it feel like the weight of the world is on your shoulders.
To find the readings on the spring scale at different time intervals, let's analyze each situation step by step:
a) Before the elevator starts to move:
When the elevator is at rest, the man is also at rest. Thus, the spring scale will register the weight of the man, which is equal to his mass multiplied by the acceleration due to gravity (g).
So, the reading on the spring scale is:
Weight = mass * g
= 72.0 kg * 9.8 m/s^2
= 705.6 N
Therefore, the spring scale will register a reading of 705.6 N before the elevator starts moving.
b) During the first 0.80 s (while the elevator accelerates):
During this time interval, the elevator is accelerating upwards, and the man inside the elevator also experiences an upward force due to this acceleration.
To calculate the reading on the spring scale, we need to find the net force acting on the man, which is the difference between the upward force due to acceleration and the weight of the man.
The upward force due to acceleration can be calculated using Newton's second law:
Force = mass * acceleration
= 72.0 kg * (1.20 m/s^2 / 0.80 s)
= 108.0 N
The weight of the man remains the same as in part a) since the gravitational force hasn't changed.
So, the net force acting on the man is:
Net Force = Upward force due to acceleration - Weight
= 108.0 N - 705.6 N
= -597.6 N
The negative sign indicates that the net force is downward, causing a decrease in the apparent weight.
Therefore, the spring scale will register a reading of -597.6 N during the first 0.80 s.
c) While the elevator is traveling at a constant speed:
When the elevator is moving at a constant speed, its acceleration is zero. As a result, the net force acting on the man is also zero, and the spring scale will register the actual weight of the man.
Therefore, the spring scale will register a reading of 705.6 N while the elevator is traveling at a constant speed.
d) During the time interval it is slowing down:
During this time interval, the elevator is undergoing a negative acceleration. Similar to part b), we need to find the net force acting on the man.
The downward force due to deceleration can be calculated using Newton's second law:
Force = mass * acceleration
= 72.0 kg * (-1.50 m/s^2)
= -108.0 N
The weight of the man remains the same as in part a).
So, the net force acting on the man is:
Net Force = Downward force due to deceleration - Weight
= -108.0 N - 705.6 N
= -813.6 N
Again, the negative sign indicates that the net force is downward, causing a decrease in the apparent weight.
Therefore, the spring scale will register a reading of -813.6 N during the time interval it is slowing down.
To find what the spring scale registers at different times, we need to consider the forces acting on the man in the elevator.
a) Before the elevator starts to move:
When the elevator is at rest, the man experiences the force of gravity acting on him. The spring scale registers the force due to his weight, which can be calculated using the equation F = mg, where m is the mass of the man and g is the acceleration due to gravity (approximately 9.8 m/s²).
b) During the first 0.80 s:
When the elevator starts to ascend with an acceleration, the man experiences an additional upward force due to the acceleration. This force is equal to ma, where m is the mass of the man and a is the acceleration of the elevator. We can calculate the force using this equation and add it to the force due to his weight to find the total force on the spring scale.
c) While the elevator is traveling at constant speed:
When the elevator is moving at a constant speed, there is no net force on the man. The force due to his weight and the upward force due to the acceleration of the elevator cancel each other out. Therefore, the spring scale registers only the force due to his weight.
d) During the time interval it is slowing down:
When the elevator undergoes uniform acceleration in the negative y direction, the man experiences an additional downward force due to this acceleration. This force is equal to ma, but with a negative sign since it is in the opposite direction. We can calculate the force using this equation and subtract it from the force due to his weight to find the total force on the spring scale.
By calculating the forces acting on the man at different times, we can determine what the spring scale registers in each scenario.