# Mechanics: Elevator Accelerating Downward

• AN630078
In summary: So, if I amend my calculations to find the force on the man, would this be;ΣF= -ma= -mg+FN FN =mg - maThe force acting on the man=70*9.8-70*1.5Force=581 N ?Then to find the tension using the equation you have stated:(M+m)a=(M+m)g-Fcable(500+70)*1.5=(500+70)*9.8-FcableTherefore;Fcable=(M+m)g-(M+m)aFcable=(500+70)*9.8-(500+70)*1.5Fcable=4731 N (would
AN630078
Homework Statement
Hello, I have found a mechanics problem concerning an elevator accelerating downwards. I have tried to comprehensively answer the question but I find that I usually struggle with these questions, particularly whether acceleration is positive or negative etc.

A man of mass 70kg is travelling in a lift of mass 500kg (excluding the mass of the man) which is suspended from a vertical cable and is moving downwards with an acceleration of 1.5ms^-1. Take g=9.8ms^-2.
Find the force exerted on the man by the lift and the tension in the cable.
Relevant Equations
ΣF=ma
The acceleration of the elevator is downwards and therefore negative.
The overall acceleration of the man is downward with the the direction of the elevator meaning that ma is negative.
The external forces acting on the man are the force of gravity acting downwards (-W=-mg) and the supporting Normal Force FN applied upward on the man by the elevator.
So ΣF= -ma= -mg+FN
FN =mg - ma
The force acting on the man=70*9.8-70*-1.5
Force=791 N

(Is it correct to denote the acceleration as negative in this calculation i.e. -1.5ms^-1?)

To find the tension in the cable;
Force= ma is in direction of motion i.e. downwards. Also mg is being applied in the downward direction. The direction of tension is always along the string and away from the body i.e. here in the upward direction.

Balancing the forces:
T=mg+ma
T=m(g+a)

Would the mass be mass of man+mass of elevator = 70kg+500kg=570kg

Then; T=570(9.8*-1.5)
T=-8379 N

I apologise clearly I am rather confused here, I am trying to improve my understanding of mechanics problems I just struggle to visualise them to understand them fully. I have drawn free body diagram for this scenario to help me picture what is happening but I am still struggling a little. I would be very grateful of any help

Acceleration is a vector, reason for which it has a direction.
A negative scalar value means deceleration instead.
While descending, both car and person accelerate first, then deccelerate.

The problem refers to the first phase.
“... moving downwards with an acceleration of 1.5ms^-1” means that the downwards velocity increases 1 m/s per each second; hence, the value of that acceleration should be considered to be positive.

It is easiest to take the downward direction as positive, so the acceleration of gravity and the acceleration of the elevator are both positive, and the upward normal force of the elevator is negative: ##ma=mg-FN##. This gives the same answer you would have gotten if you did the arithmetic correctly: 581 Newtons.

In the second part, if the man and the elevator are being accelerated downward, the acceleration is again positive downward and the upward force of the cable is negative, and this time you get

##(M+m)a=(M+m)g-F_{cable}##

AN630078 said:
T=m(g+a)
[...]
Then; T=570(9.8*-1.5)
Looks like a problem in transcription.

If you find yourself multiplying one acceleration by another, it is a good bet that you've goofed. If you track units properly, you can detect such goofs.

Lnewqban said:
Acceleration is a vector, reason for which it has a direction.
A negative scalar value means deceleration instead.
While descending, both car and person accelerate first, then deccelerate.

The problem refers to the first phase.
“... moving downwards with an acceleration of 1.5ms^-1” means that the downwards velocity increases 1 m/s per each second; hence, the value of that acceleration should be considered to be positive.
So in finding the force acting on the man, one should consider the value of acceleration is to be positive.
So, would this be;
ΣF= -ma= -mg+FN
FN =mg - ma
The force acting on the man=70*9.8-70*1.5
Force=581 N ?

Then to find the tension in the cable, the acceleration downwards is positive therefore ma would be positive.
m= mass of elevator + mass of man = 570kg

F=mg-T
ma=mg-T
T=mg-ma
T=570*9.81-570*1.5
T=4731 N ?

Chestermiller said:
It is easiest to take the downward direction as positive, so the acceleration of gravity and the acceleration of the elevator are both positive, and the upward normal force of the elevator is negative: ##ma=mg-FN##. This gives the same answer you would have gotten if you did the arithmetic correctly: 581 Newtons.

In the second part, if the man and the elevator are being accelerated downward, the acceleration is again positive downward and the upward force of the cable is negative, and this time you get

##(M+m)a=(M+m)g-F_{cable}##
Thank you very much for your reply and explanation, it has really helped my understanding I believe!

So, if I amend my calculations to find the force on the man, would this be;
ΣF= -ma= -mg+FN
FN =mg - ma
The force acting on the man=70*9.8-70*1.5
Force=581 N ?

Then to find the tension using the equation you have stated:
(M+m)a=(M+m)g-Fcable
(500+70)*1.5=(500+70)*9.8-Fcable
Therefore;
Fcable=(M+m)g-(M+m)a
Fcable=(500+70)*9.8-(500+70)*1.5
Fcable=4731 N (would I state this as a negative value as the upward force of the cable is negative?)

jbriggs444 said:
Looks like a problem in transcription.

If you find yourself multiplying one acceleration by another, it is a good bet that you've goofed. If you track units properly, you can detect such goofs.
Yes, of course you are correct. I am sorry I had clearly overlooked my mistake, thank you for the tip, that multiplying one acceleration by another is erroneous!

AN630078 said:
Fcable=4731 N (would I state this as a negative value as the upward force of the cable is negative?)
There is a tricky little question of terminology here. You were not asked for the force of the cable on the elevator. That would be a vector and, given the sign convention of down=positive, that force would indeed be negative.

Instead, you were asked for the "tension in the cable". Tensions are always positive. Never negative. "You cannot push a rope".

The distinction between a "tension" and a "force" is eventually made clear in future courses where you may learn about the "stress tensor". For now, a "tension" is a condition in a cable -- how hard any two pieces are pulling on each other. The force is what the cable exerts on whatever it is attached to.

Chestermiller and Lnewqban
jbriggs444 said:
There is a tricky little question of terminology here. You were not asked for the force of the cable on the elevator. That would be a vector and, given the sign convention of down=positive, that force would indeed be negative.

Instead, you were asked for the "tension in the cable". Tensions are always positive. Never negative. "You cannot push a rope".

The distinction between a "tension" and a "force" is eventually made clear in future courses where you may learn about the "stress tensor".
Thank you again for your reply, I have indeed found it very enlightening and helpful! Yes, that is true you cannot push a rope. I will look into stress tensor, thank you for all of your help
Would the value I have calulated for the tension now be correct, if it is positive i.e. 4731 N ?

jbriggs444
AN630078 said:
Would the value I have calulated for the tension now be correct, if it is positive i.e. 4731 N ?
Yes. Looks good to me.

AN630078
jbriggs444 said:
Yes. Looks good to me.
Excellent, thank you again for your help I am very grateful for it

## 1. How does an elevator accelerate downward?

An elevator accelerates downward due to the force of gravity. When the elevator is at rest, the force of gravity is balanced by the normal force exerted by the elevator floor. However, when the elevator begins to move downward, the force of gravity becomes greater than the normal force, causing the elevator to accelerate downward.

## 2. What factors affect the acceleration of an elevator?

The acceleration of an elevator is affected by the mass of the elevator, the force of gravity, and the force of friction. The greater the mass of the elevator, the greater the force of gravity, and the greater the force of friction, the slower the acceleration will be. Additionally, the design and condition of the elevator's pulley system can also affect its acceleration.

## 3. How does the speed of an elevator change during its downward acceleration?

As an elevator accelerates downward, its speed increases. This is because the force of gravity is constantly pulling the elevator downward, causing it to gain speed. However, once the elevator reaches its maximum speed, it will continue to move at a constant velocity unless acted upon by another force.

## 4. How does an elevator's downward acceleration affect the passengers inside?

The downward acceleration of an elevator can cause passengers to feel a sensation of weightlessness or a slight increase in weight. This is because the force of gravity is acting on both the elevator and the passengers inside. As the elevator accelerates downward, the force of gravity on the passengers increases, causing them to feel heavier. However, once the elevator reaches a constant velocity, the passengers will feel their normal weight again.

## 5. How do safety mechanisms prevent an elevator from accelerating too quickly?

Elevators are equipped with safety mechanisms, such as speed governors, to prevent them from accelerating too quickly. These mechanisms monitor the speed of the elevator and will engage the brakes if the elevator exceeds a certain speed. Additionally, elevators are designed with a maximum acceleration rate to ensure the safety and comfort of its passengers.

• Introductory Physics Homework Help
Replies
42
Views
1K
• Introductory Physics Homework Help
Replies
4
Views
831
• Introductory Physics Homework Help
Replies
6
Views
2K
• Introductory Physics Homework Help
Replies
9
Views
3K
• Introductory Physics Homework Help
Replies
2
Views
276
• Introductory Physics Homework Help
Replies
13
Views
611
• Introductory Physics Homework Help
Replies
5
Views
2K
• Introductory Physics Homework Help
Replies
3
Views
1K
• Introductory Physics Homework Help
Replies
8
Views
3K
• Introductory Physics Homework Help
Replies
31
Views
2K