1. ## Frame works help

Basic maths frameworks

How to find the reaction force on B in below frameworks?

http://oi56.tinypic.com/a13f6e.jpg

step by step explanation would be appreciated because I can't understand my freak tutorial !
thanks

2. Originally Posted by silvercats
Basic maths frameworks

How to find the reaction force on B in below frameworks?

http://oi56.tinypic.com/a13f6e.jpg

step by step explanation would be appreciated because I can't understand my freak tutorial !
thanks
hello silvercats!
it'd be better if you post the exact problem. is there an equilateral triangle hanging freely? have you shown the free body diagram there? this problem cannot be solved until you specify this.

3. Also at B you have listed two different coordinate systems. You have to choose one of them and then stick with it, else you will get confusing answers.

-Dan

4. Originally Posted by abhishekkgp
hello silvercats!
it'd be better if you post the exact problem. is there an equilateral triangle hanging freely? have you shown the free body diagram there? this problem cannot be solved until you specify this.

AB and BC bars are connected to B smoothly(no friction).A and C corners are fixed to 2a long two pints which is horizontal.is this ok?

5. Originally Posted by topsquark
Also at B you have listed two different coordinate systems. You have to choose one of them and then stick with it, else you will get confusing answers.

-Dan
That is how they are given it :/

6. To be clear,here is the answer they have given but i don't understand

AB ~ Y(2a cos 60)-x(2a Sin 60)-w(a cos 60) = 0
~ Y-(√3)x = w/2 ------- (1)
CD ~ y(2a cos 60)+x(2a Sin 60) + w(a cos 60)=0
~ Y+(√3)x=-w/x -----------(2)

and they solve it and I am like :O

7. Originally Posted by silvercats
To be clear,here is the answer they have given but i don't understand

AB ~ Y(2a cos 60)-x(2a Sin 60)-w(a cos 60) = 0
~ Y-(√3)x = w/2 ------- (1)
CD ~ y(2a cos 60)+x(2a Sin 60) + w(a cos 60)=0
~ Y+(√3)x=-w/x -----------(2)

and they solve it and I am like :O
Hmmmm....I had taken the x and y to be proposed coordinates. Are the x and y labeling reaction forces at B?

-Dan

8. Originally Posted by topsquark
Hmmmm....I had taken the x and y to be proposed coordinates. Are the x and y labeling reaction forces at B?

-Dan
yes reactions.I think they divided it to X and Y so we can easily divide forces of both

9. Originally Posted by silvercats
yes reactions.I think they divided it to X and Y so we can easily divide forces of both
the forces applied to AB at the point A by the rod AC has not been considered. why?
consider this force too in the FBD. It will have two components. same story will be there with point C on rod BC. again two more unknown forces are introduced.
we have a total of 6 unknowns now. x,y, two components of force on point A, and two components of force on point B.

you must be aware of the fact that we can have three equations of statics for a rigid body in two dimensions.Consider rod AB, these three equations can be:

1) summation of torques about point A is zero.
2)summation of torques about point B is zero.
3)vector sum of the forces acting on AB is zero.

similarly we will have 3 more equations when we consider rod BC's statics.
hence 6 equations and six unknowns.

Did this help?

10. Originally Posted by silvercats
To be clear,here is the answer they have given but i don't understand

AB ~ Y(2a cos 60)-x(2a Sin 60)-w(a cos 60) = 0
LHS is summation of torques about A and this is set equal to zero since AB is in equilibrium

~ Y-(√3)x = w/2 ------- (1)

CD ~ y(2a cos 60)+x(2a Sin 60) + w(a cos 60)=0
LHS is summation of torques about C and this is set equal to zero since BC is in equilibrium.

~ Y+(√3)x=-w/x -----------(2)

and they solve it and I am like :O
note that even though the FBD is not complete( see my previous post) yet these equations give right answer because the torques of the forces acting at point A( and C) are zero for rod AB(and BC).

did this help?