# Statics problem ignoring horizontal reactions seems to require horizontal forces

I'm an industrial engineering student taking engineering mechanics, and I'm quite confused by this problem our proffessor gave us, the chair is supposely in equilibrium, yet in order to counter the moment created by the weight pressed on the chair upward y vector vertical force on leg be would need to be eade, and cos(b) is greater than cos(a) - and she says there's no friction to insure horizntal equilibrium through resistance, could someone please show me where I'm screwing up on this, - please help! I don't want the answer I just want to see what I 'm doing wrong and fi< it, I just want to know how to solve this problem...

• It is difficult to say where you may be having trouble without a better idea of what you have already tried. In your question you gave a couple of assumptions that you are using or have tried but where is your actual attempt to solve so that we can help redirect your effort? Commented Oct 23, 2018 at 2:56
• It's hard to guess what your problem is here, except that you might be trying to over-think the problem and making a mistake while doing that. But notice that this is not the usual type of "pin jointed frame" structure, because ADE and BCE are continuous members, not two members with a pin joint between them. Finding the forces in all the members is hard, but finding the reactions at the ground is easy. Commented Oct 23, 2018 at 8:25

Since the weight W acts between the points A and B, the chair will not tip over. So the chair is in equilibrium. Writing the following equations will let you solve for the two support reactions:

• sum of forces in Y direction = 0.
• sum of moments about point A = 0.

You should take moments around any one point, Say A. And because no data about the sections of the chair and their weight is given, you can ignore them.

No horizontal component of the weight and no friction tells us no horizontal reaction. So converting to centimeters we have the moment of weight around A

$$\begin{gather} M_{W\text{ about }A } = 32.5 \times 36\text{ cm} = 1152\text{ cmkg} \\ \sum M_A =0 \\ R_b\times 50\text{ cm} = M_{W\text{ about }A} = 1152\text{ cmkg} \\ 1152/50\text{ cm} = 23.4\text{ kg} = R_B \end{gather}$$ the vertical reaction of B and from there you subtract it from 32.5 to get the vertical reaction of A.

And there are no more reactions.

Using the principle of virtual work...

If the chair is rotated infinite infinitesimally about the point A, each point will move an amount vertically that is proportional to the horizontal distance from A. The only points that we are interested are B where the reaction force acts upwards and where W acts downwards.

$$(200+200+100) \ R_B + (-65/2) \ (60+200+100)=0$$ $$500 \ R_B = 11700$$ $$R_B = 117/5= 23.4$$

The same procedure can how be applied by rotating the chair an infinitesimal amount about the point B.

$$(200+200+100) \ R_A +(-65/2)\ (200-60)=0$$ $$500 \ R_A = 4550$$ $$R_A = 45.5/5 = 9.1$$