Tensions and reactions for a fixed length cable, pivoting member & a movable pulley.

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Member ABC is supported by a pin and bracket at C and by an inextensible cable of length
L  7.8 m that is attached at A and B and passes over a frictionless pulley at D. Assuming
P  450 N and   35 and neglecting the mass of ABC and the radius of the pulley, simplify
the problem as much as possible analytically and, for 0  s  6 m, use MS Excel and/or
MATLAB to:
(a) Calculate and plot the tension T in the cable as a function of s ;
(b) Calculate and plot the magnitude of reactions x C and y C at hinge C as a function of s ;
(c) Calculate and plot the magnitude of the reaction at hinge C as a function of s ;
(d) Calculate and plot the angle  of the reaction at hinge C (with respect to +x-axis) as a
function of s ;
(e) Determine the maximum tension max T and the corresponding value of s ;
(f) Determine the (absolute) maximum reactions x C & y C and the corresponding values of s ;
(g) Determine the (absolute) maximum reaction C and the corresponding value of s;
(h) If it exists, determine the value of s at which x y C  C and the corresponding value of y C .


upload_2014-10-13_20-27-39.png



I don't know how to approach this. I think I should look at it when s=0 because it is then a right triangle. I have been told that the member does not move and that only the pulley moves from s=0 to s=6. I continued on after doing some reading and looked at the moment about C and without in cluding tension its a lil over 2100 Nm. Also I broke up P into X and Y components, now the rope of known length is attached to body at 2 points. do i count it twice? If so how do I get the angle it forms at A??
 
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Answers and Replies

  • #2
NascentOxygen
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Hi mannyk29! http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif [Broken]

If the T-member did not rotate at all, the the rope would go slack as the pulley moves horizontally to the right. So I expect what you'll find is that the member pivots around C in order to keep the rope taut. Force P, tending to lift end A, ensures there is tension in the rope and this adjusts the shape of ΔABD to maintain a triangle having a fixed perimeter, fixed at whatever it is as you see it pictured now.
 
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  • #3
gneill
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Please note: I have adjusted the thread title to better reflect its content - gneill
 
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Thanlk you for the welcome! I can certainly agree with that. could you please show me how I would get the equations for this? I need to get an understanding because I also need to enter this into matlab once I solve it.
 
  • #5
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Hi mannyk29! http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif [Broken]

If the T-member did not rotate at all, the the rope would go slack as the pulley moves horizontally to the right. So I expect what you'll find is that the member pivots around C in order to keep the rope taut. Force P, tending to lift end A, ensures there is tension in the rope and this adjusts the shape of ΔABD to maintain a triangle having a fixed perimeter, fixed at whatever it is as you see it pictured now.
I double checked about that and I was told it doesn't move. However I was able to solve for tension at s=3.5 drawing a vertical line creating 2 right triangles. now my final issue is how do I write the tension in terms of S???? please help. my time limit for this is getting down to the wire
 
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  • #6
NascentOxygen
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You checked with your lecturer and were told that AB remains horizontal? I would have expected this to be made clear, either on the diagram or in the accompanying text.

If that is so, then what needs to happen is that the pulley moves vertically as well as horizontally. With AB remaining horizontal, geometry dictates that the pulley when above the middle of AB needs to be about 1 metre more distant from AB than when the pulley is near A or B.

If our observations commence with the pulley positioned above B, then as it moves towards A the member will be held stable up until a point is reached by the pulley which sees the beam begin to rotate. You could label this distance from B as Smax.

We'll have to hope someone else can help you with this, I'm afraid I've lost my way in the algebra for this arrangement.
 
  • #7
berkeman
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Thread has been closed by OP request.
 

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