﻿ 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.) from Physics System of Particles and Rotational Motion Class 11 Manipur Board

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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 = 9.8 m/s2)

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

The given question is illustrated in the figure below:

NB = Force exerted on the ladder by the floor point B

NC = Force exerted on the ladder by the floor point C

= 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 triangles.

∴       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 =  × 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.

= (0.82 - 0.252)1/2

=  0.76 m

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

Nc + N = mg  = 392                      … (iii)

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

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

-NB × 0.5 + 40 × 9.8 × 0.125 + NC × 0.5  =  0

(NC - NB) × 0.5 = 49

NC - NB = 98                                      ...(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 × 0.5 + 40 X 9.8 × 0.125 + T × 0.76  =  0

∴                                                    T = 96.7 N.

T is the required tension in the rope.

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