- #1

- 21

- 0

I got the tension in the rope which is 440N, I am having trouble finding the Direction.

° (counterclockwise from x-axis) ?

I tried 1/cos of 164/440...its wrong

]

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter ny_aish
- Start date

- #1

- 21

- 0

I got the tension in the rope which is 440N, I am having trouble finding the Direction.

° (counterclockwise from x-axis) ?

I tried 1/cos of 164/440...its wrong

]

- #2

lightgrav

Homework Helper

- 1,248

- 30

how did you get 440 N?

you _should_ keep vertical Forces and horizontal Forces separate .

you _should_ keep vertical Forces and horizontal Forces separate .

- #3

- 21

- 0

i will look thru my notes and see how i came to it.

- #4

- 21

- 0

So x- direction....-447N*cos(12) + T2*cos(θ2) = 0...So T2*cos(θ2) = 437.2 [eqn 1]

and in the y- direction 447*sin(12) + T2*sin(θ2) - 146 = 0..

so T2*sin(θ2) = 146 - 447*sin(12) = 53.1 [eqn 2]

Now divide eqn 2 by eqn 1 giving sin(θ2)/cos(θ2) = 53.1/437.1

or tan(θ2) = 0.1214....so θ2 = arctan(0.1214) = 6.92o

Now plug back into eqn 1 to get T2...T2 = 437.1/cos(6.92) = 440N

Share:

- Replies
- 3

- Views
- 1K