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.

Solution:

The given situation can be shown as:

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

NC = 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.

$\triangle \mathrm{ABI}$ and $\triangle \mathrm{AlC}$ are similar

$\therefore \mathrm{BI}=\mathrm{IC}$

Hence, $I$ is the mid-point of $B C$.

$D E \| B C$

$B C=2 \times D E=1 \mathrm{~m}$

$\mathrm{AF}=\mathrm{BA}-\mathrm{BF}=0.4 \mathrm{~m} \ldots(i)$

$D$ is the mid-point of $A B$.

Hence, we can write:

$\mathrm{AD}=\frac{1}{2} \times \mathrm{BA}=0.8 \mathrm{~m}$  $\ldots(i i)$

Using equations (i) and (ii), we get:

FE = 0.4 m

 

Hence, F is the mid-point of AD.

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

$\triangle \mathrm{AFG}$ and $\triangle \mathrm{ADH}$ are similar

$\therefore \frac{\mathrm{FG}}{\mathrm{DH}}=\frac{\mathrm{AF}}{\mathrm{AD}}$

$\frac{\mathrm{FG}}{\mathrm{DH}}=\frac{0.4}{0.8}=\frac{1}{2}$

$\mathrm{FG}=\frac{1}{2} \mathrm{DH}$

$=\frac{1}{2} \times 0.25=0.125 \mathrm{~m}$

In $\triangle A D H$ :

$\mathrm{AH}=\sqrt{\mathrm{AD}^{2}-\mathrm{DH}^{2}}$

$=\sqrt{(0.8)^{2}-(0.25)^{2}}=0.76 \mathrm{~m}$

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

$N_{\mathrm{C}}+N_{\mathrm{B}}=m g=392 \ldots$ (iii)

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

$-N_{\mathrm{B}} \times \mathrm{BI}+m \mathrm{~g} \times \mathrm{FG}+N_{\mathrm{C}} \times \mathrm{Cl}+T \times \mathrm{AG}-T \times \mathrm{AG}=0$

$-N_{\mathrm{B}} \times 0.5+40 \times 9.8 \times 0.125+N_{\mathrm{C}} \times(0.5)=0$

$\left(N_{\mathrm{C}}-N_{\mathrm{B}}\right) \times 0.5=49$

$N_{\mathrm{C}}-N_{\mathrm{B}}=98$ $\ldots(i v)$

Adding equations (iii) and (iv), we get:

$N_{\mathrm{C}}=245 \mathrm{~N}$

$N_{\mathrm{B}}=147 \mathrm{~N}$

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

$-N_{\mathrm{B}} \times \mathrm{BI}+m \mathrm{~g} \times \mathrm{FG}+T \times \mathrm{AG}=0$

$-245 \times 0.5+40+9.8 \times 0.125+T \times 0.76=0$

$0.76 T=122.5-49$

$\therefore T=96.7 \mathrm{~N}$