# 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)(Hint: Consider the equilibrium of each side of the ladder separately.)

The diagram is shown below:

Given:

AB = AC = 1.6 m

DE = 0.5 m

BF = 1.2 m

Mass of the weight, m = 40 Kg

NB = Normal reaction at point B

NC = Normal reaction at point C

T = tension in the rope

Let us drop a perpendicular from A to imaginary line BC. This should bisect the line DE (ABC is Isosceles triangle). Let the point of intersection be H.

Let the point of intersection of perpendicular and BC be I.

ΔABI is similar to Δ AIC

So, BI = CI

Hence, I is the mid-point of BC.

DE||BC

BC = 2 × DE = 1m

And, AF = AB-BF = 0.4 m …(i)

Also, D is the mid-point of AB

So, we can write,

From (i) and (ii) we conclude that,

FE = 0.4 m

Hence, F is the mid-point of AD.

Now, FG||DH and F is the mid-point of AH.

G is the midpoint of AH.

So

FG/DH = 0.4 m/0.8 m

FG/DH = 0.5

FG = 0.125 m

AH = ((0.8m)2-(0.25m)2)1/2

AH = 0.76 m

For static equilibrium of ladder, the upward force should be equal to the downward force.

Nc + NB = mg = 392 N …(iii)

For rotational equilibrium,

The net moment of all forces about A should be Zero,

-NB × BI + mg × FG + NC × CI + T × AG = 0

-NB × 0.5m + 40kg × 9.8ms-2 × 0.125m + NC × 0.5m = 0

(NC - NB) × 0.5 = 49

NC - NB = 98 N …(iv)

Adding equations (iii) and (iv), we get,

NC = 245 N

NB = 147 N

For rotational equilibrium of the side AB, consider the moment about A,

-NB × BI + mg × FG + T × AG = 0

-245 N × 0.5 m + 40Kg × 9.8 ms-2 × 0.125m + T × 0.76m = 0

T = 96.7 N

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