# Mechanics statics equilibrium/trig question

## Homework Statement

The tension in the tow rope pulling the car in Newtons is 2 500N. Determine the minimum length of the rope l, between A and B, so that the tension in either AB or AC does not exceed the tension of 3 200 N.

Fy = 0
Fx = 0

## The Attempt at a Solution

Well so far I got an answer, which i have been told by my lecturer is not right.

I used the equations of equilibrium and solved for theta, afterwards i used trigonometry and solved for the side AB, which is the wanted length.

However, the answer is not correct.

Also, i put Fab and Fac to be 3200, as the maximum tensions. ( as it says in the question)

Any suggestions?

Thank you

#### Attachments

• question 5 help!.doc
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I do not know if your work is included in the document, but the submitted document is pending approval.

Hello mathmate,

The answer i get is not completely opposite to what the lecturer is looking for, it is just that it is not the right one.

Apparently there are more than one answers.

I summed all the forces in x direction, y direction, looking at the diagram(attached) as a 2D body.
When i sum all the points at y, i get
T=Fab sin theta + Fac sin 30 -------1
and when i sum all the forces at x, i get
Fab cos theta= Fac cos 30 -------- 2

From there i can get the unknown angle,

sin theta= T-Fac sin 30/Fab ------3
cos theta= Fac cos 30/Fab ------4

diving 3 by 4, i get that tan theta = T - Fac sin 30/Fab cos 30, from which i got the angle theta to be 19.12. ( I assumed that both Fac and Fab are 3 200, but they are not since the angles are not the same)

and afterwards, i used the trig and got the length of l.

But, i have been told by my lecturer that it is incorrect.

He mentioned something that we can change the value of theta, and also that we don't have to say that Fac and Fab are 3 200, that is only the maximum tension that can be allowed.

What is your point of view on all of this? I know it is a bit messy.

If you got any queries, feel free to post if forward.

Thanks for ur time mate.

If you want quick answers, namely before the attachment is approved, you will need to give a textual description of the diagram, or post the diagram elsewhere and include a link to the diagram.

Without seeing the diagram or a textual description, I do not (yet) understand what do A, B and C refer to. How many ropes are there? You need to define theta and l.
Is the tow rope AB, or AC, or both?

Before we go too far, I note that "maximum tension of 3200 N" doee not imply Fac or Fab=3200N, it means that the maximum permissible value is 3200N. It would generally not be correct to assume they take up the maximum value.

Haha sorry man, I just realised what you were talking about.

here is the full pic of that question.

By the way i got the answer of 79. ill post my full working tomorrow, but that is pretty much all i had done.

#### Attachments

• q5.bmp
172.6 KB · Views: 522
Your teacher is right, because unless theta is 30 degrees, the two tensions cannot be equal, as can be deduced from your equation 2.
In this situation, I would make two calculations.
First, assume Fac=3200, and calculate Fab. If Fab > 3200, this is not the solution.
You'd then need to assume Fab=3200 and calculate Fac, which should be less than 3200.
From the solution which satisfies Fac<=3200 and Fab<=3200, calculate theta.
The length cannot be known unless the length of AC is known, or some other additional piece of information not present in your link.
Give it a try and let me know what you get.

Note: if you have learned about the triangle of forces in equilibrium, the solution would be quite easy to obtain using this technique. On the other hand, your summation approach is OK too.

Nope, that is all i had to present in my FBD. Once i find the value of theta, i can solve for l(AB) by trig. either AB/sin 30= 1.2/sin (180 -theta - 30)= AC/sin theta
or tan theta=sin theta/cos theta, anyway by using trig functions

Also, that is exactly what i did for the first time when i got the answer to be 79, so don't think that will work.