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Minumum rope length

  • Thread starter Ry122
  • Start date
565
2
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.
 
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Answers and Replies

565
2
can someone please just tell me the equation that relates rope length to tension.
 
tiny-tim
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Hi Ry122! :smile:

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? :confused:
 
565
2
oops, I made a mistake. angle CBA is unknown and length BA is different from AC
 
565
2
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
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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. :confused:
 
565
2
a large mass is attached to the triangle at BC.
 
tiny-tim
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Sorry … this is no good.

Will you please type out the whole original question?
 
565
2
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
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So where does the 30º come into it? Or is that gone too?

Is angle CAB unknown as well as angle CBA?
 
565
2
angle CBA = theta
angle BCA = 30 degrees
BA = l
BC = 1.2m
 
15
0
can anyone help me with this?
 
15
0
30bp6c4.png
 

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stewartcs
Science Advisor
2,177
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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
 
15
0
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
Science Advisor
2,177
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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
 
15
0
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
Science Advisor
2,177
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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:

[tex] F_{ac}cos(30) - F_{ab}cos(\theta) = 0 [/tex]

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

CS
 
15
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ah true
that's because they have opposite directions right?

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

is that correct?
 
stewartcs
Science Advisor
2,177
3
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:
[tex] F_{ac}sin(30) + F_{ab}sin(\theta) + 3000 = 0 [/tex]

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
 
15
0
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
 
stewartcs
Science Advisor
2,177
3
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
 
15
0
ok so
[tex]3000 + F_{ac}sin(30) - F_{ab}sin(\theta) = 0 [/tex]


and now i got two equations:

[tex]3000 + F_{ac}sin(30) - F_{ab}sin(\theta) = 0 [/tex]

and

[tex]F_{ac}cos(30) - F_{ab}cos(\theta) = 0 [/tex]

if i make them equal to each other, it seems that i have too many unknowns
 
stewartcs
Science Advisor
2,177
3
ok so
[tex]3000 + F_{ac}sin(30) - F_{ab}sin(\theta) = 0 [/tex]
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
 
15
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yep, my bad...that was supposed to be - Fac(sin(30))
 

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