- #26

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SohCahToa

Sin=Opp/Hyp

Cos=Adj/Hyp

Tan=Opp/Adj

sohcahtoa

Sin=Opp/Hyp

Cos=Adj/Hyp

Tan=Opp/Adj

sohcahtoa

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- Thread starter jarmen
- Start date

- #26

- 27

- 0

SohCahToa

Sin=Opp/Hyp

Cos=Adj/Hyp

Tan=Opp/Adj

sohcahtoa

Sin=Opp/Hyp

Cos=Adj/Hyp

Tan=Opp/Adj

sohcahtoa

- #27

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so, cos(65)=adj/hyp

then hyp=565/cos(65) ?

then hyp=565/cos(65) ?

- #28

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- #29

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- #30

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i cant... But i just realized what u are saying.. Ok i get it

- #31

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so the triangles are on the outside of the ropes not inside... ok i understand

- #32

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- #33

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right. so, cos=adj/hyp

so hyp=565/cos(65) right?

so hyp=565/cos(65) right?

- #34

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so let T1 = hyp1

then cos(80)=565N/T1

let T2 = hyp2

then cos(65)=565N/T2

then cos(80)=565N/T1

let T2 = hyp2

then cos(65)=565N/T2

- #35

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I wish the pictures would work, as that would have made this so much easier, but hopefully that helped you

- #36

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so T1, 80*, =3253.7N

T2, 65*, =1336.9N

does this look right, seems like quite a bit of tension

T2, 65*, =1336.9N

does this look right, seems like quite a bit of tension

- #37

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i tried those answers but they were incorrect.

- #38

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no- I think we did something wrong. T1+T2=565N so you could say

T1=565-T2

T1=565-T2

- #39

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ok, then how do i find T1 or T2

- #40

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- #41

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ok, if both lengths of rope were equal

T1=Fg/2

T2=Fg/2

they are not but this gives me an idea...

T1=Fg/2

T2=Fg/2

they are not but this gives me an idea...

- #42

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ok...

- #43

Doc Al

Mentor

- 45,248

- 1,596

- Tension (T1) from left rope (acting at the angle of the rope)

- Tension (T2) from right rope (acting at the angle of the rope)

- Weight (W) acting down (which is given)

Since she's in equilibrium:

- The sum of the vertical force components must equal 0

- The sum of the horizontal force components must equal 0

That will give you two equations, which you can solve to find the two unknowns: T1 & T2.

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