Tension on Rope in Elevator: Find the Right Formula

[Viraam]

Homework Statement

An empty elevator of mass 2.7×10³ kg is going down by a cable at an acceleration of 1.2 m/s². What is the tension in the rope pulling the lift?

Homework Equations

$T=m(g-a)$

The Attempt at a Solution

$T=m(g-a) \\ \Rightarrow T = 2.7 *10^3 (9.8-1.2) \\ \Rightarrow T= 2.7 *10^3 \times 8.6 = 23.22*10^3 N$
Is my solution correct. Have I used the right formula?

Actually when an elevator goes down it moves in the direction of gravitational pull therefore why isn't the formula $T= m(g+a)$. In stark contrast, when the elevator goes up it is moving in the direction against gravity hence shouldn't the formula then be $T=m(g-a)$

 

[andrevdh]

Yes. That looks fine. Truncate your answer to two significant digits though.

 

[Qwertywerty]

Homework Statement

An empty elevator of mass 2.7×10³ kg is going down by a cable at an acceleration of 1.2 m/s². What is the tension in the rope pulling the lift?

Homework Equations

$T=m(g-a)$

The Attempt at a Solution

$T=m(g-a) \\ \Rightarrow T = 2.7 *10^3 (9.8-1.2) \\ \Rightarrow T= 2.7 *10^3 \times 8.6 = 23.22*10^3 N$
Is my solution correct. Have I used the right formula?

Your working is fine . However , what formula are you referring to here ?

 

[PeroK]

Homework Statement

An empty elevator of mass 2.7×10³ kg is going down by a cable at an acceleration of 1.2 m/s². What is the tension in the rope pulling the lift?

Homework Equations

$T=m(g-a)$

The Attempt at a Solution

$T=m(g-a) \\ \Rightarrow T = 2.7 *10^3 (9.8-1.2) \\ \Rightarrow T= 2.7 *10^3 \times 8.6 = 23.22*10^3 N$
Is my solution correct. Have I used the right formula?

What makes you doubt your answer?

 

[Viraam]

What makes you doubt your answer?

Actually when an elevator goes down it moves in the direction of gravitational pull therefore why isn't the formula $T= m(g+a)$. In stark contrast, when the elevator goes up it is moving in the direction against gravity hence shouldn't the formula then be $T=m(g-a)$

 

[Qwertywerty]

Actually when an elevator goes down it moves in the direction of gravitational pull therefore why isn't the formula $T= m(g+a)$. In stark contrast, when the elevator goes up it is moving in the direction against gravity hence shouldn't the formula then be $T=m(g-a)$

The easiest way to go about such a question is use F_net = ma .

 

[Viraam]

The easiest way to go about such a question is use F_net = ma .

But $F_{net}= ma$ is used when the elevator is stationary and $a=g$

 

[Viraam]

Yes. That looks fine. Truncate your answer to two significant digits though.

Actually I am not very well able to decipher what you are trying to interpret. Do you mean that I should write the answer as $23220 N$

 

[jbriggs444]

Actually when an elevator goes down it moves in the direction of gravitational pull therefore why isn't the formula $T= m(g+a)$. In stark contrast, when the elevator goes up it is moving in the direction against gravity hence shouldn't the formula then be $T=m(g-a)$

Put yourself in the position of the elevator machinery. You are holding up the cable which is holding up the car. If the car is just sitting there, you are contending only with gravity. Clearly the tension you must put on the cable must be equal to mg.

Now suppose that you want to allow the elevator to accelerate downward. Would you increase or decrease the tension you put on the cable?

 

[Qwertywerty]

Actually I don't understand what you mean. Could you
But $F_{net}= ma$ is used when the elevator is stationary and $a=g$

To what he meant - write it as 23 × 10³ N .

To my post - F_net = ma is Newton's second law ( for constant mass systems ) - applicable in all cases .

Hope this helps .

 

[Viraam]

Put yourself in the position of the elevator machinery. You are holding up the cable which is holding up the car. If the car is just sitting there, you are contending only with gravity. Clearly the tension you must put on the cable must be equal to mg.

Now suppose that you want to allow the elevator to accelerate downward. Would you increase or decrease the tension you put on the cable?

I would decrease the tension I suppose. Ohh! So is that why we use the formula $T= m(g-a)$ when the elevator accelerates downwards?

 

[Viraam]

To what he meant - write it as 23 × 10³ N .

To my post - F_net = ma is Newton's second law ( for constant mass systems ) - applicable in all cases .

Hope this helps .

I didn't get you. How can we use the $F_{net}=ma$ here. Thanks

 

[jbriggs444]

I would decrease the tension I suppose. Ohh! So is that why we use the formula $T= m(g-a)$ when the elevator accelerates downwards?

Bingo!

 

[andrevdh]

Answer in post number 10 (Qwertywerty).
The question gives values with two significant digits for both the mass
and the acceleration, thus two significanrt digits in the final answer.

 

[Viraam]

Bingo!

Thanks a lot!

 

[Viraam]

Answer in post number 10 (Qwertywerty).
The question gives values with two significant digits for both the mass
and the acceleration, thus two significanrt digits in the final answer.

Ok! Thanks

 

[Qwertywerty]

Acceleration of lift is ' a ' downwards . Forces acting on the elevator are gravity , and tension .

Therefore ,
mg - T = ma .

 

[Viraam]

Acceleration of lift is ' a ' downwards . Forces acting on the elevator are gravity , and tension .

Therefore ,
mg - T = ma .

Ohh Ok! Thanks

 

[NickAtNight]

why isn't the formula $T= m(g+a)$.

1) draw your Forces !

2) equation for each force is. $F = m*a$

3) the two forces are gravity and acceleration

4) tension comes from the forces pulling on the rope!

<------ ============== ------>
tension
Tension = F1 + F2

In this case, Forces are in the same direction so tension is lessened.

------>. ============== ------>
F1 = ma. F2 = mg

Tension = - F1 + F2

 

[Chestermiller]

Viraam:

If you had drawn a free body diagram for the elevator, you would not be experiencing all this uncertainty. Draw one, and you will see what I mean.

Chet

 

[Chestermiller]

1) draw your Forces !

2) equation for each force is. $F = m*a$

No. Only the net force on a body is equal to its mass times its acceleration, not each force. Certainly contact forces are not always equal to ma (unless they are the only force).

3) the two forces are gravity and acceleration

Gravitational attraction is a force, but there is no such thing as acceleration force.

4) tension comes from the forces pulling on the rope!

No. Tension is one of the forces acting on the elevator.

Please refrain from giving misinformation like this in your posts. Further misinformation will receive warning points and eventual ban from PF.

Chet

 

[jbriggs444]

3) the two forces are gravity and acceleration

I think you mean to write "gravity and tension".

 

[Alex Chen]

I didn't get you. How can we use the $F_{net}=ma$ here. Thanks

The formula Fnet=ma should always be your starter equation in questions like these. They are applicable to all cases like someone previously said, and it will cause less confusion than using the formula you did. Draw a free body diagram, which will show you the net forces, in this case, mg - T = ma. Then you can find what you're looking for.

 

[NickAtNight]

I think you mean to write "gravity and tension".

No, I meant the force from gravity and the force from acceleration.

The problem can be viewed as a 3 force problem with the forces being from gravity, from acceleration and the resulting force that we are calculating being the tension in the cable. That was the way they formulated the equation for him. $F_t = F_g + F_a$

 

[NickAtNight]

.

 

[NickAtNight]

How can we use the $F_{net}=ma$ here. Thanks

Like so

$$ \sum_{net} F = F_t + F_g = F_a = m \times a $$
where $F_t$ is the Force from the tension in the string
... $F_g$ is the Force from the acceleration due to gravity
...$F_a$ is the Net Force that can be calculated from the acceleration of the body

this equation can be rearranged to solve for the Tension on the string by moving the Force from gravity to the other side.
$$ F_t = F_a - F_g $$

We can then replace the Force with m*a and m*g respectively

$$ F_t = (m \times a) - (m \times g) $$

and then simplify the equation by removing the mass to the front.

$$ F_t = m \times (a - g) $$

Notice that you only need to use one equation. You have to decide which way acceleration is positive, up or down, and then apply the correct sign to both the acceleration and the gravity terms.

Now my equation appears backwards from theirs (your equation in post 1). The Tension on the string and the gravity are in opposite directions. They defined their Force tension as positive and their gravity Force as positive. If you define acceleration up as positive, the gravity used would be -9.8 N/m2. So upward acceleration increases the Force and downward acceleration decreases the force.

 

[NickAtNight]

There are two limits on these equations.

What happens if the Tension on the cable goes to 0?

Can we get a negative Tension in the cable? alternatively stated, can you push an object with a rope?

What happens if the Tension on the cable exceeds the maximum tension for the cable so that the cable snaps?

 

[Chestermiller]

There are two limits on these equations.

What happens if the Tension on the cable goes to 0?

Can we get a negative Tension in the cable? alternatively stated, can you push an object with a rope?

What happens if the Tension on the cable exceeds the maximum tension for the cable so that the cable snaps?

Are these questions you are asking as a test for the OP, or are they questions that you really don't know the answers to?

Chet

 

[Viraam]

There are two limits on these equations.

What happens if the Tension on the cable goes to 0?

Can we get a negative Tension in the cable? alternatively stated, can you push an object with a rope?

What happens if the Tension on the cable exceeds the maximum tension for the cable so that the cable snaps?

The tension is 0 when the elevator goes downwards with an acceleration equal to that of acceleration due to gravity. $T = m (g-a) \\T= m (9.8-9.8) = 0$
I am not sure if this is write but when the tension is 0 then no force is pulling on the elevator.
Yeah, negative acceleration when the acceleration of the elevator is greater than acceleration due to gravity. $a> g$
I don't understand how to push an object with a rope.
Finally, I guess the rope should snap if it exceeds the maximum tension. Is that the reason why the maximum number of people who can stand on the lift is mentioned.
Please correct me

 

[Viraam]

Are these questions you are asking as a test for the OP, or are they questions that you really don't know the answers to?

Chet

What is OP?

 

[Viraam]

When the lift accelerates upwards with an acceleration of $a$ then Tension $T$ is given by $m (g+a)$.
In this case should the acceleration due gravity be taken as negative that is $-9.8$ . No right.

 

[Viraam]

Viraam:

If you had drawn a free body diagram for the elevator, you would not be experiencing all this uncertainty. Draw one, and you will see what I mean.

Chet

Thanks, I got it.

 

[NickAtNight]

$T = m (g-a) \\T= m (9.8-9.8) = 0$
Please correct me

That is correct.

1) The tension in the cable can only go to 0. If we want the elevator to go down faster, we would have to attach a cable to the bottom of the elevator and pull it. With no tension, the elevator falls to the bottom of the shaft.

2) If the maximum tension for the cable is exceeded, then the cable snaps and the elevator falls to the bottom of the shaft.

3) If we raise and lower the elevator by letting out or pulling in cable at a fixed velocity (say 2 ft/second - so it takes 4 seconds to go up one 8 ft floor and 40 seconds to go up 10 - 8 foot floors {89 ft}, what is the acceleration of the elevator for most of the journey (excluding the very start and the end of the trip)?

Remember: acceleration is the rate of change of velocity.

4) Would you get on an elevator that has a significant acceleration (other than for a joy ride?)

5) Do you know where the value of g comes from?

 

[haruspex]

What is OP?

OP can mean the Original Post (post #1 in the thread) or Original Poster (you, in this case).

 

[Viraam]

That is correct.

1) The tension in the cable can only go to 0. If we want the elevator to go down faster, we would have to attach a cable to the bottom of the elevator and pull it. With no tension, the elevator falls to the bottom of the shaft.

2) If the maximum tension for the cable is exceeded, then the cable snaps and the elevator falls to the bottom of the shaft.

3) If we raise and lower the elevator by letting out or pulling in cable at a fixed velocity (say 2 ft/second - so it takes 4 seconds to go up one 8 ft floor and 40 seconds to go up 10 - 8 foot floors {89 ft}, what is the acceleration of the elevator for most of the journey (excluding the very start and the end of the trip)?

Remember: acceleration is the rate of change of velocity.

4) Would you get on an elevator that has a significant acceleration (other than for a joy ride?)

5) Do you know where the value of g comes from?

3) Is the acceleration 0. When velocity is constant then acceleration is zero
$u = v \ \ \ \ \ \therefore a = \frac{v-v}{t}=0$
4) I didnt understand what you mean by significant acceleration? (do you mean appreciable acceleration)