Question

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 D, 0.5 m is tied Halfway 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)

Answer 1

it is given that , AB = AC = 1.6 m, DE = 0.5 m and BF = 1.2 m.

Mass of the weight, m = 40 Kg
$N_B$ is the Normal reaction at point B
$N_C$ is the Normal reaction at point C
T is the tension in the rope.

Let's take 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.
so,  DE||BC [ from Thales theorem ]
 BC = 2 × DE = 1m
And, AF = AB-BF = 0.4 m .....…(i)
Also, D is the mid-point of AB
So, we can write,
AD = 0.5 × BA
⇒ AD = 0.8 m ….....(ii)
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.
 ΔAFG and ΔADH are similar,
So, FG/DH = AF/AD
⇒ FG/DH = 0.4 m/0.8 m
⇒ FG/DH = 0.5
⇒ FG = 0.125 m

Now, In ΔADH
AH² = (AD²- DH²) [ from Pythagoras theorem ]
⇒ AH² = (0.8m)²-(0.25m)²
⇒ AH = 0.76 m
For static equilibrium of ladder, the upward force should be equal to the downward force.
⇒ $N_C+N_B$= mg = 392 N ...…(iii)
For rotational equilibrium,
The net moment of all forces about A should be Zero,
⇒ -$N_B$ × BI + mg × FG + $N_C$ × CI + T × AG = 0
⇒ -$N_B$ × 0.5m + 40kg × 9.8m/s²  × 0.125m + $N_C$ × 0.5m = 0
($N_C-N_B)$) × 0.5 = 49
($N_C-N_B)$)= 98 N .....…(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,
⇒ -$N_B$ × BI + mg × FG + T × AG = 0
⇒ -245 N × 0.5 m + 40Kg × 9.8 m/s² × 0.125m + T × 0.76m = 0
⇒ T = 96.7 N