# Minumum rope length

http://users.on.net/~rohanlal/Q10.jpg [Broken]
The rope connected to the triangle is pulling away at 3000N
i need to determine the minimum length of ac so that the tension in either ab or ac does not exceed 4000N. AC and BC are the same length.

Last edited by a moderator:

Related Advanced Physics Homework Help News on Phys.org
can someone please just tell me the equation that relates rope length to tension.

tiny-tim
Homework Helper
Hi Ry122!

I don't understand … your diagram says BC = 5m, and angles BCA and CBA are 30º.

So AC is fixed, isn't it?

What am I missing?

oops, I made a mistake. angle CBA is unknown and length BA is different from AC

Is the tension in the rope pulling away (3000N) equal to the tension in BC? if so ill know how to answer the question

tiny-tim
Homework Helper
Is the tension in the rope pulling away (3000N) equal to the tension in BC?
No, certainly not.

You have to use Newton's second law: sum of the forces in any particular direction is zero.

But I still don't understand what stops the triangle from rotating.

a large mass is attached to the triangle at BC.

tiny-tim
Homework Helper
Sorry … this is no good.

Will you please type out the whole original question?

sure
The tension in the tow rope pulling the car in Newtons is 3000N. Determine the minumum length of the rope l, between A and B, so that the tension in either AB or AC does not exceed 4000N

tiny-tim
Homework Helper
So where does the 30º come into it? Or is that gone too?

Is angle CAB unknown as well as angle CBA?

angle CBA = theta
angle BCA = 30 degrees
BA = l
BC = 1.2m

can anyone help me with this?

#### Attachments

• 11.6 KB Views: 274
Last edited:
stewartcs
can anyone help me with this?
You need to show your attempt first.

HINT: Start by breaking the forces down into x and y components, then solve the equations for the desired unknown. Initially, the desired unknown will be theta. Once you have theta, you can use a little trig to find the length, L, which is what you truly desire.

CS

I get:
Sum of Fx
0 = Fab cos (theta) + Fac cos (30)

Sum of Fy
0 = Fab sin (theta) + Fac cos (30) - 3000N

Would Fab and Fac equal max. tension (4000N)? otherwise, I'm completely stuck

stewartcs
I get:
Sum of Fx
0 = Fab cos (theta) + Fac cos (30)

Sum of Fy
0 = Fab sin (theta) + Fac cos (30) - 3000N

Would Fab and Fac equal max. tension (4000N)? otherwise, I'm completely stuck
Check your FBD again, your component equations do not look correct. Once you solve the x-component equation you will be able to substitute that result into the y-component equation, which will leave only one unknown, theta (along with the assumption that one of the lines is at the max allowable tension of 4000N).

CS

I've been staring at those equations for quite a while now...
all i can see is that:

In the x-axis:
Fab cos(theta) = -Fac cos(30)

and in the y-axis:
3000N = Fab sin(theta) + Fac sin(30)

I know through the sin rule that:

sin (theta)/ Fac = sin (30)/Fab

but I don't know how the unknown length, l, can fit in

stewartcs
I've been staring at those equations for quite a while now...
all i can see is that:

In the x-axis:
Fab cos(theta) = -Fac cos(30)

and in the y-axis:
3000N = Fab sin(theta) + Fac sin(30)

I know through the sin rule that:

sin (theta)/ Fac = sin (30)/Fab

but I don't know how the unknown length, l, can fit in
I’ll get you started on your x-components:

$$F_{ac}cos(30) - F_{ab}cos(\theta) = 0$$

Do you see why this is so? If you do, then try the y-components and lets see what you get.

CS

ah true
that's because they have opposite directions right?

for the y-components:
$$F_{ac}sin(30) + F_{ab}sin(\theta) + 3000 = 0$$

is that correct?

stewartcs
ah true
that's because they have opposite directions right?
Correct!

Just remember that it is the x-components that are pointing in different directions.

for the y-components:
$$F_{ac}sin(30) + F_{ab}sin(\theta) + 3000 = 0$$

is that correct?
Not quite. The same applies for the vertical axis. In what direction is T shown in the problem? Now what about the y-component of the force (tension) Fab and Fac?

HINT: Draw a right triangle and resolve the resultant (Fab and Fac) into components to find the direction (remember vectors are drawn tip to tail).

CS

ah ok, so:
$$F_{ac}sin(30) + F_{ab}sin(\theta) - 3000 = 0 [tex] T in the problem opposes the y-components of the Fab and Fac stewartcs Science Advisor ah ok, so: [tex] F_{ac}sin(30) + F_{ab}sin(\theta) - 3000 = 0 [tex] T in the problem opposes the y-components of the Fab and Fac The normal convention is the other way around (i.e. up is positive and down is negative). I would recommend using that. CS ok so [tex]3000 + F_{ac}sin(30) - F_{ab}sin(\theta) = 0$$

and now i got two equations:

$$3000 + F_{ac}sin(30) - F_{ab}sin(\theta) = 0$$

and

$$F_{ac}cos(30) - F_{ab}cos(\theta) = 0$$

if i make them equal to each other, it seems that i have too many unknowns

stewartcs
$$3000 + F_{ac}sin(30) - F_{ab}sin(\theta) = 0$$