# Find tension of a string on the ceiling of an elevator

1. Apr 29, 2006

### bavenger

good day all i've no idea how to work this out

this is the problem
A 5.5-kg object is suspended by a string from the ceiling of an elevator that is accelerating downward at a rate of 2.5m/s^2 What is the tension in the string?

i presume tension is the force thats on the string?
so i did: 5.5(kg) * (9.8(grav) - 2.5) = 40.15N
is this correct?

the equation i was given is: Fnet = T + Fg
which i rearranged to get
T = Fnet - Fg
what the hell does that mean? lol
could someone tell me a general rule for tension?

2. Apr 29, 2006

### Kurdt

Staff Emeritus
Generally tension is just the balancing of the weight (weight is measured in newtons i.e. force) of an object suspended by a wire or a rope. It follows from newtons third law equal yet opposite reactions. If a rope is stationary and just supporting the weight of the object then the tension must be equal to that weight for it to remain in that position and therefore we can write T=fg. Obviously this is extended to the equation you have but be careful of the signs. try drawing a diagram to convince yourself of the sign of the net force.

3. Apr 29, 2006

### bavenger

so f = ma right? so it could be re-written as
T= m*a*g
which in most cases is
T= m*g^2...which doesnt look right.

i presume you meant T=mg which is what i did

i have a harder question now. what if it's not just 1 object haging off a fixed space there is a nother problem that goes like this.

____object1___|
|
|
|
|
object 2

where object 1 is being pulled --> that way with 44N of force. i know object 1 is 3.6 kg and object 2 is 1.6 kg. there is string connecting both of them and we dont have to worry about friction. what i've tried to do is work out the tension on the first string so
T1= fg
T1=15.68N

and then i thought i'll have to add the tension from object 1 also so i did this
T2= mg
T2= 31.36
but it's on a table so i worked out what the acceleratoin due to the 44N was
F= ma so
a= 13.75 (44N/3.2Kg)
so i changed T2 to equal
T2= ma
T2= 44 which is the force ?????????? i'm completely confused as how to balance the equation for object 2. what am i doing wrong?

4. Apr 29, 2006

### Staff: Mentor

You got the right answer, but let's review it so you understand what you're doing.

It's an application of Newton's 2nd law, Fnet = ma.

The forces on the object are: its weight (mg) acting down and the tension force of the string (T) acting up. So the net force, Fnet = T - mg. (Note that I use a convention that up is positive and down is negative.)

Thus Newton's 2nd law tells us:
Fnet = ma
T - mg = ma

Solving for T gives: T = mg + ma

When you plug the numbers in, realize that a = -2.5m/s^2 since the acceleration is downward and thus negative. As Kurdt points out, signs are important.

5. Apr 29, 2006

### Kurdt

Staff Emeritus
Sorry I was using your notation for weight which is Fg, I forgot the capital.

6. Apr 29, 2006

### Kurdt

Staff Emeritus
With regards to the second question where is the rope on object one? Is it pointing to the right or at some angle from the horizontal?

7. May 3, 2006

### bavenger

yeah sorry it was a little ambiguous...i'm not even sure if thats the right word i think it's at a right angle but it didnt specify. there is a weight hanging off a table connected to an object of double that weight that has a force applied to it in the opposite direction (ie it is applying tension to the rope thing)

8. May 3, 2006

### Hootenanny

Staff Emeritus
The first thing you must realise is that the tension is constant throughout the string, therefore the is only one tension acting on each particle. Using newton's law (F = ma) can you formulate an equation for each particle seperately, using T to represent the tension, you should obtain two equations? You didn't state in you original problem whether you need to find the tension or the accleration?

~H

Last edited: May 3, 2006
9. May 3, 2006

### Staff: Mentor

I'm unclear about the problem description. Is this one mass on a table connected to another mass hanging off the table (via a pulley, perhaps)?

In any case, as Hoot says a massless rope has but a single tension throughout. So the tension force on each mass will be the same. And, since the masses are connected, they will have the same acceleration (at least the same magnitude).

10. May 3, 2006

### bavenger

thats correct Doc Al one mass that weighs 3.2kg (m1)is on the table connected to another mass of 1.6kg(m2) hanging off the table cia a pulley. a force of 44N is acting on the 3.2 kg mass

so F=ma (m1)
44 = 3.2 * a
a = 13.75 ms^-2

the acceleration of m2 is gravity so 9.8 downwards.

so the Net acceleration will be 13.75 - 9.8 = 3.95 correct?

so m1 will be acceleratin along the table at 3.95 ms^-2 and m2 will be doing the same acceleratoin.

but the question asked what the tension on the string will be

so does that mean that F = ma
so F = (3.2 + 1.6) * 3.95?
so the tension on the string would be 18.96N

11. May 4, 2006

### Hootenanny

Staff Emeritus
I think your reasoning is eronous. When calculating the accleration of each particle you have to take into account the tension of the string. I think you should start by formulating two equations using netwon's second law, one for each particle. I'll start;

For the 3.2kg particle, the tension in the string is acting in the opposite direction of motion, however, the 44N force is acting in the same direction as the motion, therefore;

$$F = ma$$
$$44 - T = 3.2a$$

Can you now formulate and equation to represent the motion of the 1.6kg particle?

~H