Finding the tension force and torque

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Homework Help Overview

The discussion revolves around finding the tension force and torque in a physics problem involving a rope and a trap door. Participants are exploring the relationships between tension, weight, and angles in the context of mechanics.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the calculation of tension force and torque, with attempts to derive expressions based on given parameters. Questions arise regarding the correctness of these expressions and the relevance of angles involved.

Discussion Status

Some participants have offered guidance on adjusting calculations and interpretations of the problem. There is an ongoing exploration of different approaches, particularly regarding the assumptions made about the setup and the variables involved.

Contextual Notes

There are indications of confusion regarding the problem's wording and the assumptions about the physical setup, particularly concerning the roles of angles α and θ. Some participants express uncertainty about the relevance of certain components in their calculations.

Jessica8956
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1. Finding tension force and torque (details attached)

http://imgur.com/cZLG9xP




2. please see attachment

http://imgur.com/cZLG9xP




3.
For part 1). I get

​[itex]{​T} = -\sin \alpha(i) +\cos \alpha(j)[/itex]

is this correct?

(Where T is the tension force)


For part 2). I am stuck .

What I did was get

​[itex]{W} = -mgj[/itex]

Then get the position vector as

​[itex]\frac{1}{2}2L\sin \alpha(i) \sin \alpha(j)[/itex]

Simplifies to

​[itex]L\sin \alpha(i) \sin \alpha(j)[/itex]

So the answer is

​[itex]-mgL\sin \alpha(k)[/itex]


Is this correct or am I completely off? Also will the end answer contain ​[itex]\alpha[/itex] or [itex]\theta[/itex]?

Thanks (not sure if I can chat here but great forum and hello as I'm new here)

 
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no attachment.
 
SteamKing said:
no attachment.

Sorry about that, I have added the questions now :redface:
 
For part 1) you should multiply by W/2.

Part 2) is a little confusing since they say "for the rope and trap door" which seems non-sensible to me in context. If that part could be dropped and they were strictly looking for "the torque caused by the weight of the door about O" then I would change your answer from α to θ.
 
paisiello2 said:
For part 1) you should multiply by W/2.

Part 2) is a little confusing since they say "for the rope and trap door" which seems non-sensible to me in context. If that part could be dropped and they were strictly looking for "the torque caused by the weight of the door about O" then I would change your answer from α to θ.

Thanks for the reply :)

So part one becomes

[itex]{​T} = W(-\sin \alpha(i) +\cos \alpha(j))/2[/itex]

and part two

​[itex]-mgL\sin \theta(k)[/itex]

How does that look?
 
Hi Jessica8956. http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif

Unfortunately, neither answer looks right to me. You don't know the textbook answers, do you?

Let's focus on (b). Can you show your working?
 
Last edited by a moderator:
NascentOxygen said:
Hi Jessica8956. http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif

Unfortunately, neither answer looks right to me. You don't know the textbook answers, do you?

Let's focus on (b). Can you show your working?

Thanks NascentOxygen

We had a supply teacher who gave us questions like the one I am stuck on. They only have a list of multiple choice answers to choose from. Sadly that teacher was only there for a week and my regular teacher said to not worry about it as it is not relevant to what we're doing :rolleyes:



My workings for part (b) are


What I did was get

​[itex]{W} = -mgj[/itex]

Then get the position vector as

​[itex]\frac{1}{2}2L\sin \alpha(i) \sin \alpha(j)[/itex]

Simplifies to

​[itex]L\sin \alpha(i) \sin \alpha(j)[/itex]

So the answer is

​[itex]-mgL\sin \alpha(k)[/itex]
 
Last edited by a moderator:
That is the position vector to what? I can't see there should be an ##\alpha## involved, nor do I see a ##\sin()##. Please explain how you determined this.
 
NascentOxygen said:
That is the position vector to what? I can't see there should be an ##\alpha## involved, nor do I see a ##\sin()##. Please explain how you determined this.

I thought that was the position vector for W

As I thought the line OA was ​[itex]\sin \alpha(i) \sin \alpha(j)[/itex]

and as the length was 2L it was half that as I assumed it was in the middle.

I'm a bit lost with this so I hope that makes sense :blushing:
 
  • #10
For part 1), intuitively I would expect the tension force T to be greater than W/2. To confirm we can look at limiting cases:

-when α is close to zero then T should be pretty close to W/2; so far your formula looks like it might work
-when α is close to 90° then T should be much much bigger than W/2; now your formula doesn't look too healthy

Can you rework your formula to meet the above limiting cases? You might want to explicitly do an equilibrium of joint A.
 
  • #11
Jessica8956 said:
I thought that was the position vector for W

As I thought the line OA was ​[itex]\sin \alpha(i) \sin \alpha(j)[/itex]

and as the length was 2L it was half that as I assumed it was in the middle.

I'm a bit lost with this so I hope that makes sense :blushing:

I suggest that you start over, abandoning trying to use i and j. (If necessary, you can at the end resolve your final answer into vert and horiz components.)

Just work it out using trigonometry, not vectors. See how you go with that. Try (b) first, it looks easier than (a).
 
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