Question

i have one question...

Answer 1

i dont have any idea>>> please guide me>>>

Answer 2

@Zarkon

Answer 3

Please help me>>>@Vincent-Lyon.Fr a little bit blur now..

Answer 4

F3 should probably be F1 + F2, but I can't read the wording of the problem.
Can't you type it?

Answer 5

What does "What if?" mean?

Answer 6

the question asking to find the F3 at B and along BC... So how can i calculate it.. did i must use formula of equilateral triangle..

Answer 7

F3 = F1 + F2 so that net torque is zero.

The distances of the forces' lines of action to O are the same for all three of them.

Answer 8

ok .. for formula of torque. is Fd = I(alpha) ... so how can i substitute this inside ???

Answer 9

if it has for x-axis and y-axis.. it,s mean sum of Fx and sum of Fy

Answer 10

???

I repeat:
Torque is Force times distance of the force's line of action to the pivot.
Using x and y is much more complicated.

Answer 11

so to find torque of F3=(F1+F2)(distance)....

Answer 12

The equation is very simple :

Let distance \(OD=d\) and anticlockwise be positive.

\(F_1.d + F_2.d-F_3.d=0\)
Hence
\(F_3=F_1+F_2\)

That's all you need!

Answer 13

so how if the torque of F3 change to another position

Answer 14

Re-write first equation to :

\(F_1.d + F_2.d-F_3.d\;' =0\)

And have a look at the picture to find what \(d\;'\) is.

Answer 15

Got to go now.

Answer 16

ok.. thank you. i understand already..
so the q just to approve torque zero = F3=F1+F2