Tension of a string attached to an instrument inside rocket

  • #1
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Homework Statement


Suppose the rocket is coming in for a vertical landing at the surface of the earth. The captain adjusts the engine thrust so that rocket slows down at the rate of 2.40m/s2 . A 6.00-kg instrument is hanging by a vertical wire inside a space ship.

I don't know if there are any relevant equations for this...[/B]

2. The attempt at a solution

So I figured that since the tension on the string in one direction is the weight and the force of gravity, and in the other direction it is the force of the captain slowing it down, I thought you could multiply 6 by 9.8, and then subtract 2.40.

How am I supposed to do it?
 

Answers and Replies

  • #2
Simon Bridge
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Use free body diagrams... remember you are not in an inertial frame: you probably have examples in your notes involving elevators.
 
  • #3
haruspex
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multiply 6 by 9.8, and then subtract 2.40
You've got the right sort of idea, but seem to have confused acceleration and force.
What type of entity are you calculating from 6 * 9.8? What type of entity is the 2.4? Considering the units of each may help.
 
  • #4
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So I was supposed to find the force that the wire exerts on the instrument. So if I think about the just the wire, it should be equal to the amount of force that the instrument is actually pulling on the wire. That would include the weight of the instrument and gravity, and then I would think about how the rocket is slowing down...

So since gravity is affected, I would subtract 2.4 from 9.8 and get 7.4. Then could I multiply that by 6 to get the total force?
 
  • #5
haruspex
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I would subtract 2.4 from 9.8 and get 7.4. Then could I multiply that by 6 to get the total force?
Yes, though it feels like you are to some extent guessing. It would be more satisfactory to express it terms of net force and resulting acceleration, as Simon indicates.
If T is the tension in the wire, what is the net force on the instrument?
What is the acceleration of the instrument?
 
  • #6
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Um, the net force on the instrument would be...the wire pulling up, gravity pulling down, and the rocket's acceleration pulling up...but I'm honestly not sure. (I tried the answer 44.4, but it didn't work)
 
  • #7
Simon Bridge
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The rocket pulling up is the tension force... different names for the same thing.
There are only two forces on the instrument.
 
  • #8
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I'm afraid I still do not understand what to do. I feel like I have everything that I need to know, but I just don't know how to put it together
 
  • #9
haruspex
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I'm afraid I still do not understand what to do. I feel like I have everything that I need to know, but I just don't know how to put it together
There are two forces, and you listed them. As Simon says, the acceleration of the rocket is not another force. The instrument doesn't 'know' anything about the rocket - it just experiences the forces that act directly on it.
How fast and in what direction is the instrument accelerating? (is its acceleration different from the rocket's?)
 
  • #10
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Well, the acceleration should not be any different...
 
  • #11
haruspex
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Well, the acceleration should not be any different...
Right. So write out the (vertical) ##\Sigma F = ma## equation for the instrument.
 
  • #12
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So F (or tension?) should equal (6.0)(7.4)? I only am thinking about gravity and then the rocket slowing its ascent
 
  • #13
haruspex
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So F (or tension?) should equal (6.0)(7.4)? I only am thinking about gravity and then the rocket slowing its ascent
That's we're trying to show (but doing it properly).
You have two vertical forces (one unknown as yet), a mass and an acceleration . Write the equation connecting them.
 
  • #14
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One unknown as yet...?

mass=6.0
acceleration=7.7
force=44.4?
I am lost
 
  • #15
haruspex
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One unknown as yet...?
I'm trying to get you to solve the original question in a reliable manner, rather than as merely quite a good guess. So the unknown is the tension in the wire, T.
mass=6.0
Yes.
acceleration=7.7
You just agreed that the acceleration of the instrument is the same as the acceleration of the rocket. What is the rocket's acceleration, and in what direction? Here's what you are told:
rocket slows down at the rate of 2.40m/s2

force=44.4?
There are two forces on the instrument. The tension force is the unknown, T. The other is gravity. What is the force of gravity on the instrument?
 
  • #16
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Something was about to click..and then it went away. The force of gravity is 9.8, is it not? (I ask because I am getting my signs confused now as I go through my assignment). The acceleration of the rocket is going down, so it is negative...

so would the acceleration of the rocket be gravity (-9.8) minus the slowing (2.4)? I am getting jumbled

Thank you for teaching me. I end up learning rather than just getting the answer, and that's good because I have an exam on Thursday
 
  • #17
haruspex
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Something was about to click..and then it went away. The force of gravity is 9.8, is it not? (I ask because I am getting my signs confused now as I go through my assignment). The acceleration of the rocket is going down, so it is negative...

so would the acceleration of the rocket be gravity (-9.8) minus the slowing (2.4)? I am getting jumbled

Thank you for teaching me. I end up learning rather than just getting the answer, and that's good because I have an exam on Thursday
It's important to decide which way is positive in your analysis - and stick to it. Mostly people take up as positive, making g -9.8m/s2, so let's do that here.
What force does gravity exert on the mass (get the sign right)?
Is the force from the wire positive or negative?
Is the rocket's acceleration positive or negative?
 
  • #18
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Okay, so gravity is -9.8. The force from the wire is positive, and the acceleration of the rocket is negative
 
  • #19
haruspex
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Okay, so gravity is -9.8. The force from the wire is positive, and the acceleration of the rocket is negative
Two right out of three. Which way is the rocket moving? After a little time, is it moving that way faster or more slowly?
 
  • #20
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OH. the rocket is going down, but the acceleration is positive because it gets faster as it goes?
 
  • #21
haruspex
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OH. the rocket is going down, but the acceleration is positive because it gets faster as it goes?
No, the acceleration is positive because it is going down more slowly.
Acceleration is the rate of change of velocity.
If the velocity at time 0 is -10m/s and at a later time is -5 m/s then its change in velocity is (-5)-(-10) = +5.
 
  • #22
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oh. it was slowing down in the negative direction
 
  • #24
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sorry no. I still don't know where I am going
 
  • #25
haruspex
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sorry no. I still don't know where I am going
You have two forces, the tension and the force due to gravity.
Tension we will write as T.
What is the gravitational force on the instrument? (Include the sign.)
Do you understand what ##\Sigma F## means in the equation ##\Sigma F = ma##?
Can you write an algebraic expression for ##\Sigma F## for the instrument, using the answers to my questions above?
 
  • #26
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ΣF=it would be -44.4 and -9.8?
 
  • #27
haruspex
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ΣF=it would be -44.4 and -9.8?
No. When i ask a list of questions, please try answer each question in turn. I construct them to form a logical sequence.
I'll say it again:
- There are two forces
- One is the unknown force T. You cannot take it to be 44.4 anything, we don't know yet. Just leave it as T.
- The other force is the force due to gravity. It is not -9.8 anything. -9.8m/s2 the acceleration due to gravity. Do not mix up forces and accelerations.
 
  • #28
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The gravitational force is -9.8m/s^2
I think that ΣF=ma is the addition of all the forces being equal to the mass of the object multiplied by the acceleration of the rocket
I think that ΣF would be T+g
 
  • #29
haruspex
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The gravitational force is -9.8m/s^2
I think that ΣF=ma is the addition of all the forces being equal to the mass of the object multiplied by the acceleration of the rocket
I think that ΣF would be T+g
You're getting closer, but you are still confusing force and acceleration.
g = -9.8m/s2 is the acceleration due to gravity of an object in free fall. It is not a force. It is the result of the force exerted by gravity.
The gravitational force on an object of mass m is mg.
 
  • #30
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So the total force exerted on the instrument is T, and then mg, the force of gravity. ΣF=T+mg
 
  • #31
haruspex
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So the total force exerted on the instrument is T, and then mg, the force of gravity. ΣF=T+mg
Good. Now fill in the other side of the ΣF=ma equation.
 
  • #32
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Wait...

I got it

So I put T-mg=ma, or T-(6)(9.8)=(6)(2.4) and then I got my answer.

Yay ^^
 
  • #33
haruspex
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Wait...

I got it

So I put T-mg=ma, or T-(6)(9.8)=(6)(2.4) and then I got my answer.

Yay ^^
Good. It is now also well with your algebra.
 

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