Excercises Question 7.22 - Page Number 179 of NCERT Text

Asked by midhunkrishna | 8th Oct, 2009, 08:31: PM

Expert Answer:

Pls. refer your text book for the diagram.

Condition for equilibrium: Net force=0 and net Torque=0


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,  R=245 and S=147

Answered by  | 25th Oct, 2009, 04:25: PM

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