Finding the tension force and torque

In summary, in the conversation, the participants discuss finding tension force and torque for a problem involving a trap door attached to a rope. They discuss the formula for tension force and the position vector for the weight of the door. There is some confusion about the answer and the participants try to work through the problem using trigonometry.
  • #1
Jessica8956
8
0
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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  • #2
no attachment.
 
  • #3
SteamKing said:
no attachment.

Sorry about that, I have added the questions now :redface:
 
  • #4
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.
 
  • #5
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?
 
  • #6
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?
 
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  • #7
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 :uhh:



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:
  • #8
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.
 
  • #9
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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What is tension force?

Tension force is the pulling or stretching force that occurs when an object is being pulled from both ends. It is also known as the force of tension or the tensional force.

How is tension force measured?

Tension force can be measured using a spring scale or a force sensor. The force applied to the object is measured in Newtons (N) or pounds (lbs).

What factors affect tension force?

The tension force on an object can be affected by the mass of the object, the force applied, and the angle at which the force is applied. Other factors include the type of material and any external forces acting on the object.

What is torque?

Torque is the measure of the force that can cause an object to rotate around an axis. It is calculated by multiplying the force applied by the distance from the axis of rotation.

How is torque related to tension force?

Torque and tension force are related because they both involve the application of a force to an object. In the case of torque, the force is applied at a distance from the axis of rotation, while in the case of tension force, the force is applied to opposite ends of an object. The calculation of torque also takes into account the angle at which the force is applied, which can also affect tension force.

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