**Question 7.22:**

*g*= 9.8 m/s

^{2})

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

__Solution__:*N*

_{B}= Force exerted on the ladder by the floor point B

*N*

_{C}= Force exerted on the ladder by the floor point C

*T*= Tension in the rope

BA = CA = 1.6 m

DE = 0. 5 m

BF = 1.2 m

Mass of the weight,

*m*= 40 kg

Draw a perpendicular from A on the floor BC. This intersects DE at mid-point H.

ΔABI and ΔAIC are similar

∴BI = IC

Hence, I is the mid-point of BC.

DE || BC

BC = 2 × DE = 1 m

AF = BA – BF = 0.4 m … (

*i*)

D is the mid-point of AB.

Hence, we can write:

AD = (1/2) X BA = 0.8 m ...(ii)

Using equations (

*i*) and (

*ii*), we get:

FE = 0.4 m

Hence, F is the mid-point of AD.

FG||DH and F is the mid-point of AD. Hence, G will also be the mid-point of AH.

ΔAFG and ΔADH are similar

∴ FG / DH = AF / AD

FG / DH = 0.4 / 0.8 = 1 / 2

FG = (1/2) DH

= (1/2) X 0.25 = 0.125 m

In ΔADH:

AH = (AD

^{2}- DH

^{2})

^{1/2}

= (0.8

^{2}- 0.25

^{2})

^{1/2}= 0.76 m

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

*N*

_{c}+

*N*

_{B}=

*m*g = 392 … (

*iii*)

For rotational equilibrium of the ladder, the net moment about A is:

-N

_{B}X BI + mg X FG + N

_{C}X CI + T X AG - T X AG = 0

-N

_{B}X 0.5 + 40 X 9.8 X 0.125 + N

_{C}X 0.5 = 0

(N

_{C}- N

_{B}) X 0.5 = 49

N

_{C}- N

_{B}= 98 .....(iv)

Adding equations (

*iii*) and (

*iv*), we get:

N

_{C}= 245 N

N

_{B}= 147 N

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

-N

_{B}X BI + mg X FG + T X AG = 0

-245 X 0.5 + 40 X 9.8 X 0.125 + T X 0.76 = 0

∴ T = 96.7 N

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