Resolving reactant forces on supported beams.

[Adrian Haddock]

The questions States.

A uniform plank of weight 120N rests on two stools. a weight of 80N is placed on the plank, midway between the stools calculate

A) the force acting on stool A
B) the force acting on stool B
(Cant add a picture for some reason.)

Plank over length = 4.0m
from left hand edge to stool A = 0.5m
from right hand edge to stool B = 1.0m

The answers in the book state
A)48N
b)72N

I know the plank is in equilibrium so Σ M = 0 and Σ F = 0
I know a moment is F * d

My attempt.
I calculated the centre of mass via moments from the left hand side.

as in (m1*d1 + m2*d2)/m1+m2

so distance of the 80N mass is 1.75m
centre of the uniform plank is 120N at 2m.

put those into the above sum i get the effective centre to be at 1.9m. I think i can say this system is the same as a 200N force acting at 1.9m.

Is this correct?
Do I need to work out fro the over hangs?

The answers i get are Ra = 112N and Rb = 88N

by taking a 200N force acting at 1.9m and taking Moments about stool A...

 

[haruspex]

The answers i get are Ra = 112N and Rb = 88N

Not the other way around?

 

[Adrian Haddock]

Yes, quite right. Ra=88N and Rb = 112N

 

[haruspex]

Yes, quite right. Ra=88N and Rb = 112N

Looks good.
I would not have bothered finding the mass centre. Simpler just to take moments of all the components separately.

 

[Adrian Haddock]

I did that also, got the same answers.

If someone else agrees with me, I'll put the answers in the book down to an editting mishap and move on worth my life, same I can't put the hair I pulled out back into my head.

Thank you for your help.