Question

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.) Question 7.22 System Of Particles And Rotational Motion

Answer 1

See the figure ,
Length of each side of ladder
AB = AC = 1.6 m

Let the sides of the ladder subtend an angle ∅ with horizontal and forces exerted by the floor on the ladder at point B and C be Nb and Nc respectively.

Length of rope (DE) = 0.5 m
Weight suspended ( w) = 40kg-wt
= 40×9.8 N = 392N-------(1)
distance (BF) = 1.2 m
Distance (AF) = AB-BF = 1.6-1.2=0.4 m
Here,
angle ABC = ADE = ACB = AED = ∅
Let A' be the midpoint of the rope .
DA' = 0.5/2 = 0.25m
DF' = F'A' = (1/2)DA' = 0.125m
By translation Equilibrium,
Nb + Nc = w = 392 N----(2) { from (1)

Taking moment of forces about point A for side AB,
Nb×BC' = w×F'A' + T×AA'
BC' = ABcos∅
AA' = ADsin∅ use this above ,
Nb×ABcos∅ = w×0.125 + T×ADsin∅
[Cos∅ = DF/DF' = 0.125/0.4=0.3125
∅= 72.8° ]
Nb× 1.6× cos72.8° = 392×0.125+T×0.8×sin72.8°
After solving ,
0.5Nb = 0.764T + 49 ------(3)
Now, moment of forces taking about point A, for side AC .
Nc× CC' = T× AA'
CC' = AC cos∅
AA' = AE sin∅
Nc× AC cos∅ = T× AE sin∅
Nc × 1.6× cos72.8° = T×0.8×sin72.8°
0.5Nc = 0.764T ----------(4)
From eqns (3) and (4)
Nb - Nc = 98 ------(5)
From eqns (2) and (5)
2Nb = 490
Nb = 490/2 = 245N
From eqn (1) = Nc = 147N
Now, put this in eqn (4)
T = 0.5×1.47/0.764
= 96.7 N.

Answer 2

Answer:

So this is the answer plz do me brainlest answer