Find the tension in the rope and forces exerted by the floor

[deathnote93]

I'm new here, so please bear with me.

Homework Statement

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Verbatim from my textbook:
As shown in Fig.7.40, the two sides of a step ladder BA and CA are 1.6 m long and
hinged at A. A rope DE, 0.5 m is tied half way up. A weight 40 kg is suspended from
a point F, 1.2 m from B along the ladder BA. Assuming the floor to be frictionless
and neglecting the weight of the ladder, find the tension in the rope and forces
exerted by the floor on the ladder. (Take g = 9.8 m/s²)

Homework Equations

Condition for equilibrium: Net force=0 and net Torque(\tau)=0
\tau=rF sin\theta

The Attempt at a Solution

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Let the angle BAC be 2x, tension in string be T, hung weight of 40kg be W, and the reaction forces of the floor be R and S.

Then, for the equilibrium of AB(net torque=0):
0.4W*sinx + 0.8T*cosx = 4Rsinx which on simplification led to
T = 2/3 *tanx(4R-W) ------(1)

For the equilibrium of AC (Net torque=0)
0.6*T*cosx = 1.6*S*sinx
T = 8S/3 tanx --------(2)

For vertical equilibrium of ladders, W = R + S -------(3)

Equating 1 and 2,
R - S = W/4 --------(4)

Solving 3 and 4, I got R=245 and S=147 which are correct answers.

The next question is to find the tension, which is what I'm having trouble with. Using either the law of sines or the law of cosines gives me x=24.6 degrees, and substituting that into eqns 1 or two gives me the wrong answer for the tension - the correct answer according to the book is T=96.7N.

Any help is appreciated.

 

[willem2]

AE and BD are 0.8 long, not 0.6, so your value for x is wrong.

The values of R and S are independent both the length of the ladder and the length of
the rope BTW. This is easy to see by considering the torque around F on the entire ladder

 

[deathnote93]

That sucks, I don't usually do stupid things like that. Thank you very much.

Also - is there a shorter way to do this problem?