Downward acceleration in an elevator

[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.

Homework Equations

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?

 

[tms]

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

 

[SammyS]

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.

Homework Equations

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: F_net = 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?

 

[racecar12]

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?

 

[SammyS]

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 .

F_NET = 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 . F_NET = N - W .

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

 

[racecar12]

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

 

[tms]

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.

 

[SammyS]

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 F_NET 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 ?

 

[Chestermiller]

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.

 

[FaraDazed]

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.