As shown in Figure 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 halfway up. A weight 40 kg is suspended from a point F, 1.2 m from B along with 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.)

Asked by Abhisek | 1 year ago |  81

##### Solution :-

The above situation can drawn as :

Here,

NB = Force being applied by floor point B on the ladder

NC = Force being applied by floor point C on the ladder

T   = Tension in string.

BA= CA = 1.6 m

DE = 0.5 m

BF = 1.2 m

Mass of the weight, m = 40 kg

Now,

Make a perpendicular from A on the floor BC. This will intersect DE at mid-point H.

ΔABI and ΔAIC are similar

∴BI = IC

This makes I the mid-point of BC.

DE || BC

BC = 2 × DE = 1 m AF

BA – BF= 1.6 – 1.2 = 0.4 m . . . . . . . . . . . ( 1 )

D is the mid-point of AB.

Thus, we can write:

AD = ($$\dfrac{1}{2}$$) × BA  =  0.8 m    . . . . . . . .. . . . ( 2 )

Using equations ( 1 ) and ( 2 ), we get:

DF = 0.4 m

Thus, F is the mid-point of AD.

FG || DH and F is the mid-point of AD. This will make  G  the mid-point of AH.

$$\dfrac{FG}{DH}=\dfrac{AF}{AD}$$

$$\dfrac{FG}{DH}$$ = $$\dfrac{0.4}{0.8}$$$$\dfrac{1}{2}$$

FG = ($$\dfrac{1}{2}$$) DH

H is the midpoint of the rope.

Therefore, DH = $$\dfrac{0.5}{2}$$ = 0.25 m

= ($$\dfrac{1}{2}$$) × 0.25  =  0.125 m

$$AH= (AD^2 – DH^2)^\dfrac{1}{2}$$

$$= (0.82 – 0.252)^\dfrac{1}{2}= 0.76 m$$

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

NC+ N= mg = 392 N . . ... .  ( 3 ) [ mg = 9.8 x 40 ]

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

-NB× 0.5 + 392× 0.125 + NC × 0.5  =  0

(NC – NB) × 0.5 = 49

NC – NB = 98       . . . . . . . . . .. . . .. . .. . . ( 4 )

Adding equation ( 3 ) and equation ( 4 ), we get:

NC = 245 N

NY = 147 N

-NB × BI +  FG x mg + T × AG  =  0

-245× 0.5 + 392 × 0.125 + 0.76 x T  =  0

∴ T = 96.7 N.

Answered by Pragya Singh | 1 year ago

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