Question

Two sides of a step ladder BA and CA are 1.6 m long and hinges at A, A rope DE, 0.5 m long is tied half way up. A weight 40 kg is suspended from point F, 1.2 m from B along the ladder BA. assuming the floor to be frictionless and neglecting the weight of ladder find the tension in the rope and forces exerted by floor on ladder.

• A. T=350 N $R_1=250N$
  $R_2=150B$
• B. T=97 N $R_1=150N$
  $R_2=100N$
• C. T=97 N $R_1=250N$
  $R_2=150N$
• D. T=250 N $R_1=100N$
  $R_2=50N$

Answer 1

The correct option is B T=97 N $R_1=150N$

$R_2=100N$

$N_B=$ Force exerted on the ladder by the floor point $B$

$N_C=$ Force exerted on the ladder by the floor point $C$

$T=$ tension in the rope

$BA=CA=1.6m$

$DE=0.5m$

$BF=1.2m$

Man of the weight, $m=40kg$ drawing the perpendicular form $A$ on the floor $BC$,

This intersects $DE$ at mid-point $H$.

$\Delta ABJ$ and $\Delta AIC$ are similar $BI=IC$

Hence, $I$ is the mid-point of $BC$

$DE\parallel BC$

$BC=2\times DE=1m$

$AF=BA-BF$

$=0.4m⟶\left(i\right)$

$D$ is the mid-point of $AB$.

Hence, we can write

$AD=\left(\frac{1}{2}\right)\times BA$

$=0.8m⟶\left(ii\right)$

Using equations $\left(i\right)$ and $\left(ii\right)$ we get,

$FE=0.4m$

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

$\Delta AFG$ and $\Delta ADH$ are similar

$\therefore$ $FG/DH=AF/AD$

$FG/DH=0.4/0.8=1/2$

$FG=\left(1/2\right)\times DH$

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

In $\Delta ADH$,

$AH={({AD}^2-{DH}^2)}^{1/2}$

$={({0.8}^2-{0.25}^2)}^{1/2}$

$=0.76m$

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

$N_C+N_B=mg=392⟶\left(iii\right)$

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

$-N_B\times B_1+mg\times FG+N_C+C_1+T\times AG-T\times AG=0$

$-N_B\times 0.5+40\times 9.8+0.125+N_C\times 0.5=0$

$(N_C-N_B)\times 0.5=49$

$N_C-N_B=98⟶\left(iv\right)$

Adding equations $\left(iii\right)$ and $\left(iv\right)$, we get

$N_C=245N$

$N_B=147N$

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

$N_B\times B_1+mg\times FG+T\times AG=0$

$-245\times 0.5+40\times 9.8+0.125+T\times 6.76=0$

$T=96.7N$