# Downward acceleration in an elevator

• racecar12
In summary, a person weighing 0.9 kN rides in an elevator with a downward acceleration of 1 m/s^2. Using Newton's second law, the net force acting on the person is equal to their mass (approximately 9.183 kg) multiplied by the acceleration. The two forces acting on the person are their weight (mg) and the normal force exerted by the elevator floor (ma). Therefore, the magnitude of the normal force can be calculated by subtracting the weight from the net force, which is equal to the person's mass multiplied by the acceleration.
racecar12

## Homework Statement

Person weighing 0.9kN rides in an elevator that has downward acceleration of 1m/s^2. What is the magnitude of the force of the elevator floor on the person. Answer in units of kN.

F=ma

## The Attempt at a Solution

I drew a force body diagram and set y upward to be the positive y direction. The relevant forces acting on the person are then 1) -1m/s^2 and 2) -9.8m/s^2 due to gravity and 3) the normal force of the elevator floor that pushes up in the positive y direction. That's the one I am solving for.

Sum of the forces in the y direction have to equal each other, right?

I have three forces working in the y direction.

So F becomes N, the normal force of the elevator yielding: N=ma.

I'm guessing a then is just the sum of the two accelerations? So, -9.8 + -1 =-10.8

Then N = 0.9kN(-10.8)...N=-9.72kN?

You've ridden elevators. Do you fell heavier when the elevator starts down? Look again at the forces on the person.

racecar12 said:

## Homework Statement

Person weighing 0.9kN rides in an elevator that has downward acceleration of 1m/s^2. What is the magnitude of the force of the elevator floor on the person. Answer in units of kN.

F=ma

## The Attempt at a Solution

I drew a force body diagram and set y upward to be the positive y direction. The relevant forces acting on the person are then 1) -1m/s^2 and 2) -9.8m/s^2 due to gravity and 3) the normal force of the elevator floor that pushes up in the positive y direction. That's the one I am solving for.
Those are not forces. They're accelerations.

The two forces acting are
1) the upward for the floor exerts on the person

2) the (downward) force of gravity on the person.​

Sum of the forces in the y direction have to equal each other, right?
No, that's not right. The forces cancel out only id the acceleration is zero.

Newton's 2nd Law: Fnet = ma .

I have three forces working in the y direction.

So F becomes N, the normal force of the elevator yielding: N=ma.

I'm guessing a then is just the sum of the two accelerations? So, -9.8 + -1 =-10.8

Then N = 0.9kN(-10.8)...N=-9.72kN?

I tried it again...

There are 2forces acting in this problem. One is the weight force pushing down, which is W=mg.
The second force is the normal force pushing up from the floor of the elevator, N=ma.

W=mg
m=0.9kN
g= -9.8
W=-8.82N

N=ma
m=0.9kN
a=-1
N=-0.9N

Since the elevator is moving down, the total force acting on the person is W-N, which is -7.92N?

racecar12 said:
I tried it again...

There are 2forces acting in this problem. One is the weight force pushing down, which is W=mg.
The second force is the normal force pushing up from the floor of the elevator, N=ma.

W=mg
m=0.9kN
g= -9.8
W=-8.82N

N=ma
m=0.9kN
a=-1
N=-0.9N

Since the elevator is moving down, the total force acting on the person is W-N, which is -7.92N?
N ≠ ma .

FNET = ma.

There are two force acting on the person.

N, the normal force exerted by the elevator floor and W = mg.

Therefore, the net force is N - W . FNET = N - W .

That along with FNET = ma, should get you the answer.

I clearly don't understand a lot about physics, forces, etc.

If I'm understanding correctly, Fnet = N-W

Is Fnet supposed to equal zero?

I understand that W is equal to mg. the force of W then is -8.82kN. Is that right? I mean do I account for the downward direction of gravity in this situation?

I'm lost at what N equals, or even how to calculate it.

I could say Fnet-W = N, but I don't understand how to get Fnet then.

Fnet = ma.. a is equal to 1, because it was given to me in the problem. m is also given as 0.9kN.
so the net force is then -0.9, because the elevator moves in the downward or negative y direction?

Then -0.9- -8.82 = N? With that thought I'm still wrong.

If Fnet is zero, I get a different answer. N =-8.82kN

The problem tells you there is acceleration, so you know that the net force is not zero. You are given the mass and the acceleration, which is enough for you to calculate the net force. From that and the weight you can calculate the normal force.

racecar12 said:
I clearly don't understand a lot about physics, forces, etc.

If I'm understanding correctly, Fnet = N-W

Is Fnet supposed to equal zero?

I understand that W is equal to mg. the force of W then is -8.82kN. Is that right? I mean do I account for the downward direction of gravity in this situation?

I'm lost at what N equals, or even how to calculate it.

I could say Fnet-W = N, but I don't understand how to get Fnet then.

Fnet = ma.. a is equal to 1, because it was given to me in the problem. m is also given as 0.9kN.
so the net force is then -0.9, because the elevator moves in the downward or negative y direction?

Then -0.9- -8.82 = N? With that thought I'm still wrong.

If Fnet is zero, I get a different answer. N =-8.82kN
If FNET is zero, then the persons acceleration would be zero..

Look at the problem this way.

You are given that the acceleration of the elevator, and thus the acceleration of the person, is 1 m/s2 downward.

The person weighs 0.9 kN , so the person's mass is 900/9.8 ≈ 9.183 kg .

What net force is required to make an object of mass 9.183 kg, accelerate at 1 m/s2 ?

Draw a free body diagram of the person, and show the two forces acting on him, including their directions. The net force acting on the person is the vector sum of these two forces. If the person is accelerating, this net force, according to Newton's second law, must be equal to the person's mass times his acceleration. Capture this relationship in the form of an equation.

racecar12 said:
I understand that W is equal to mg. the force of W then is -8.82kN. Is that right?

How can a persons weight be minus something? You are given W in the question it said it was 0.9kN (which is equivalent to 900N).

You know what the acceleration due to gravity is, 9.8, and you said you know that W=mg. So from that you can work out the mass of the person.

Then you can use F=ma.

## 1. What is downward acceleration in an elevator?

Downward acceleration in an elevator is the rate at which the elevator speeds up as it moves downwards. This is caused by the force of gravity pulling the elevator towards the ground.

## 2. Why do we feel heavier when the elevator accelerates downwards?

When the elevator accelerates downwards, the force of gravity pulling us towards the ground increases, making us feel heavier. This is because our bodies are also accelerating at the same rate as the elevator.

## 3. How do elevators control downward acceleration?

Elevators control downward acceleration by using a counterweight system. This system balances the weight of the elevator car and its passengers, allowing for a smoother and more controlled descent.

## 4. What happens to downward acceleration as an elevator approaches the ground floor?

As an elevator approaches the ground floor, the downward acceleration decreases because the elevator is slowing down to come to a stop. This is known as deceleration or negative acceleration.

## 5. Can downward acceleration in an elevator be dangerous?

Downward acceleration in an elevator is typically not dangerous as elevators are designed to have a maximum acceleration rate that is safe for passengers. However, sudden or excessive acceleration or deceleration can cause discomfort or even injury, which is why elevators are highly regulated and maintained for safety.

Replies
10
Views
2K
Replies
42
Views
1K
Replies
4
Views
987
Replies
6
Views
2K
Replies
14
Views
2K
Replies
12
Views
3K
Replies
8
Views
3K
Replies
3
Views
2K
Replies
9
Views
2K
Replies
35
Views
4K