- #1

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Thanks for any help.

Here is the assignment question: http://members.shaw.ca/KingofGods/Phys24421.jpg

And here is my attempt: http://members.shaw.ca/KingofGods/Phys24422.jpg

- Thread starter KoGs
- Start date

- #1

- 106

- 0

Thanks for any help.

Here is the assignment question: http://members.shaw.ca/KingofGods/Phys24421.jpg

And here is my attempt: http://members.shaw.ca/KingofGods/Phys24422.jpg

- #2

Pyrrhus

Homework Helper

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- #3

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So we have a horizontal tension T6. This must be balanced out by the horizontal tension of T2 since we have no other horizontal forces, is this correct?

So T6 = T2cos60.

Now for the vertical force.

T6 applies no vertical force to vertex C. So we must have the normal force balancing the verticle force of T2.

or N = T2sin60.

but N = mg

and m = 0 doesn't it?

That's gonna make T2 = 0, and everything else = 0.

- #4

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- #5

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I talked to my prof today. He said the part of my equation with T2 should be using the angle of 120. That is, the T2cos(60) part should be T2cos(120) and the T2sin(60) part should be T2sin(120). Even after explaining to me why, for the life of me I cannot figure out what he is saying. Someone else want to give it a shot and explain to me why? All I did was drew a perpendicular line straight down to intersect line segment CD. And then simply using cos(60) = T2_x / T2. So T2_X = T2Cos(60). Where 60 is the angle of C. I also tried making a right angle triangle the other side on T2 (ie. outside the ABC triangle), and again I got T2Cos(60).

- #6

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Next sum the forces in the y direction which lets the Vertical component of T2 = that reaction, e.g. T2sin60=mg/2, so T2 = mg/2sin60

then the horizontal component of T2 will equal T6, e.g. T6 = T2 cos60 and so on.

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