Step Ladder on a rough survace - torques help

[Bruno1983]

Homework Statement

A step ladder is moddelled as two model rods OA and Ab smoothly hinged at A. The other ends of the rods rest on a rough horizontal floor. The mass of rod AB is 4m and that of rod OA is m. The andle each rod makes with the vertical is theta. The origin is taken at O and the unit vectors i and j are defined as shown.

Here is a picture of the question
http://i93.photobucket.com/albums/l75/selah83/Q1.jpg

Homework Equations

Torque = position vector * Force

The Attempt at a Solution

Ok a) is drawing the force diagram Iv got that down I think

B) Express each for in terms of i and j (Iv got that)
F1 = Friction acting on O = lF1l i
F2 = Friction acting on B = - lF2l i
N1 = Normal Reaction of OA = lN1l j
N2 = Normal Reaction of AB = lN2l j
W1 = Weight of rod OA = -lw1l j
W2 = Weight of rod AB = -lw2l j

but is says hence obtain two scalar equations relating to the magnitude of the forces when the system is in equilibrium. and then find the torque of each force about O.

I really don't understand this any help much appreciated!

 

[tiny-tim]

Welcome to PF!

Hi Bruno1983! Welcome to PF!

(have a theta: θ )

A step ladder is moddelled as two model rods OA and Ab smoothly hinged at A. The other ends of the rods rest on a rough horizontal floor. The mass of rod AB is 4m and that of rod OA is m. The andle each rod makes with the vertical is theta. The origin is taken at O and the unit vectors i and j are defined as shown.
…
B) Express each for in terms of i and j (Iv got that)
F1 = Friction acting on O = lF1l i
F2 = Friction acting on B = - lF2l i
N1 = Normal Reaction of OA = lN1l j
N2 = Normal Reaction of AB = lN2l j
W1 = Weight of rod OA = -lw1l j
W2 = Weight of rod AB = -lw2l j

but is says hence obtain two scalar equations relating to the magnitude of the forces when the system is in equilibrium. and then find the torque of each force about O.

In this context, scalar means "not vector".

A vector equation would be ai + bj + ci + dj = 0, or (a,b) + (c,d) = 0

That can be decomposed into two scalar equations:

a + c = 0 and b + d = 0

Similarly you could decompose (a,b,c) x (d,e,f) = (u,v,w) into three scalar equations