Min Length of AC to Maintain Tension in Triangle: 4000N

[Ry122]

http://users.on.net/~rohanlal/Q10.jpg
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.

 

[Ry122]

can someone please just tell me the equation that relates rope length to tension.

 

[tiny-tim]

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?

 

[Ry122]

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

 

[Ry122]

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]

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.

 

[Ry122]

a large mass is attached to the triangle at BC.

 

[tiny-tim]

Sorry … this is no good.

Will you please type out the whole original question?

 

[Ry122]

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]

So where does the 30º come into it? Or is that gone too?

Is angle CAB unknown as well as angle CBA?

 

[Ry122]

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

 

[gozita73]

can anyone help me with this?

 

[gozita73]

 

[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

 

[gozita73]

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

 

[gozita73]

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 let's see what you get.

CS

 

[gozita73]

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

 

[gozita73]

ah ok, so:
F_{ac}sin(30) + F_{ab}sin(\theta) - 3000 = 0 <br /> <br /> T in the problem opposes the y-components of the Fab and Fac

 

[stewartcs]

ah ok, so:
F_{ac}sin(30) + F_{ab}sin(\theta) - 3000 = 0 <br /> <br /> T in the problem opposes the y-components of the Fab and Fac

<br /> <br /> The normal convention is the other way around (i.e. up is positive and down is negative). I would recommend using that.<br /> <br /> CS

 

[gozita73]

ok so
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]

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

The only positive term in the y-component equation should be T. Both, Fab and Fac should be negative since their y-components point downward. Make sense?

CS

 

[gozita73]

yep, my bad...that was supposed to be - Fac(sin(30))

 

[gozita73]

ok, so this is the part where I'm completely stuck because there are 3 unknowns...

would either Fac or Fab have the maximum tension?
that is...either
Fac = 4000N
or
Fab = 4000N?

 

[stewartcs]

ok, so this is the part where I'm completely stuck because there are 3 unknowns...

would either Fac or Fab have the maximum tension?
that is...either
Fac = 4000N
or
Fab = 4000N?

Now that you have both of your x and y component equations, you can solve the x-component equation for Fab. Then you can sub Fab into the y-component equation and solve for theta.

CS

 

[gozita73]

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

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

F_{ab} = \frac{F_{ac}cos(30)}{cos(\theta) }

therefore...subbing into the other eq.


F_{ac}sin(30) + \frac{F_{ac}cos(30)}{cos(\theta) }sin(\theta) = 3000

F_{ac}(sin(30) + cos(30)}tan(\theta))= 3000

 

[stewartcs]

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

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

F_{ab} = \frac{F_{ac}cos(30)}{cos(\theta) }

therefore...subbing into the other eq.


F_{ac}sin(30) + \frac{F_{ac}cos(30)}{cos(\theta) }sin(\theta) = 3000

F_{ac}(sin(30) + cos(30)}tan(\theta))= 3000

Looks good to me. Although it may be easier to solve for theta if you don't factor Fac out. That is to say start with the equation below and solve for theta:

F_{ac}sin(30) + F_{ac}cos(30)tan(\theta) = 3000

CS

 

[gozita73]

Fac = 4000N ?

 

[stewartcs]

Fac = 4000N ?

Yes, since that is the maxium force it can have per the problem statement.

CS

 

[gozita73]

cool thanks so much...
theta turns out to be around 16.1 degrees

doing the rest should be no problem for me

thanks again

 

[stewartcs]

cool thanks so much...
theta turns out to be around 16.1 degrees

doing the rest should be no problem for me

thanks again

Good job.

CS