Thread: 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

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.

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

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?

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 :/

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

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

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

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?

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?