Magnitude of the Acceleration of an Elevator when it first begins to move

In summary, the problem involves a person standing on a bathroom scale in a motionless elevator, which reads 0.88 of their regular weight when the elevator begins to move. The magnitude of the acceleration of the elevator can be calculated using the formula a = (F_n - F_g)/m, where F_n is the scale reading, F_g is the gravitational force, and m is the person's mass. The correct calculation should be a = (0.88mg - mg)/m.
  • #1
xsorealx
1
0

Homework Statement



A person stands on a bathroom scale in a motionless elevator. When the elevator begins to move, the scale briefly reads only 0.88 of the person's regular weight.

Calculate the magnitude of the acceleration of the elevator.


Homework Equations



w=mg
F=Ma

The Attempt at a Solution



Fn-Fg=ma
A=Fn-Fg/m
A=9.8-9.8/.88

I know that answer is definitely wrong because it can't be 0.

I am completely stuck on this problem, any help would be appreciated, thanks!
 
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  • #2
xsorealx said:

Homework Statement



A person stands on a bathroom scale in a motionless elevator. When the elevator begins to move, the scale briefly reads only 0.88 of the person's regular weight.

Calculate the magnitude of the acceleration of the elevator.


Homework Equations



w=mg
F=Ma

The Attempt at a Solution



Fn-Fg=ma
A=Fn-Fg/m
A=9.8-9.8/.88

I know that answer is definitely wrong because it can't be 0.

I am completely stuck on this problem, any help would be appreciated, thanks!
xsorealx, welcome to PF!
Your algebra is off, and so is your calculation for the variables. It's [tex] a = (F_n - F_g)/m [/tex], where [tex]F_g = mg[/tex], and [tex]F_n[/tex] is the scale reading.
 
  • #3


I would approach this problem by first clarifying the given information and assumptions. The statement mentions that the person is standing on a motionless elevator before it begins to move. This implies that the elevator is not accelerating yet, and therefore the person's weight should be equal to their regular weight (Fn = Fg). However, the scale reads 0.88 of the person's regular weight when the elevator starts to move. This suggests that there is a downward force acting on the person, causing a decrease in their weight.

Assuming that the scale is accurate and the person's weight remains constant, we can use the equation F = ma to determine the acceleration of the elevator. We know that the downward force acting on the person is equal to their regular weight minus the reading on the scale (F = Fn - Fg = mg - 0.88mg). Plugging this into the equation, we get:

F = ma
mg - 0.88mg = ma
0.12mg = ma

Solving for acceleration (a), we get:

a = 0.12g

Therefore, the magnitude of the acceleration of the elevator when it first begins to move is 0.12 times the acceleration due to gravity (g). This means that the elevator is accelerating downwards with a magnitude of approximately 1.176 m/s^2.

It is important to note that this calculation assumes that the scale is accurate and the person's weight remains constant. In reality, there may be other factors at play such as the person's movements or the elevator's initial velocity, which could affect the reading on the scale. However, based on the given information, this is the best estimation of the magnitude of acceleration for the elevator.
 

Related to Magnitude of the Acceleration of an Elevator when it first begins to move

1. What is the magnitude of the acceleration of an elevator when it first begins to move?

The magnitude of acceleration of an elevator when it first begins to move depends on various factors such as the weight of the elevator, the speed it is traveling, and the type of elevator. However, on average, elevators typically have an acceleration of 0.5 meters per second squared (m/s²) when they first start moving.

2. How is the magnitude of acceleration of an elevator calculated?

The magnitude of acceleration of an elevator can be calculated using the formula a = Δv/Δt, where a is acceleration, Δv is the change in velocity, and Δt is the change in time. To calculate the change in velocity, you can use the formula v = u + at, where v is the final velocity, u is the initial velocity (usually 0 for elevators), a is the acceleration, and t is the time.

3. Does the direction of the acceleration of an elevator matter?

Yes, the direction of the acceleration of an elevator does matter. Acceleration is a vector quantity, which means it has both magnitude and direction. In elevators, the direction of acceleration is typically upwards or downwards, depending on the direction the elevator is moving in.

4. How does the magnitude of acceleration of an elevator affect its passengers?

The magnitude of acceleration of an elevator can affect its passengers in different ways. If the elevator has a high acceleration, passengers may feel a greater force pushing them in the direction of the acceleration, making them feel heavier. On the other hand, if the acceleration is low, passengers may feel weightless or experience a slight floating sensation.

5. How can the magnitude of acceleration of an elevator be controlled?

The magnitude of acceleration of an elevator can be controlled by adjusting the speed and the starting and stopping mechanisms of the elevator. Modern elevators also use advanced control systems to ensure a smooth and comfortable ride for passengers, with acceleration and deceleration designed to minimize discomfort. Additionally, elevators are regularly inspected and maintained to ensure safe and efficient operation.

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