As shown in the two sides of a step ladder BA and CA are 1.6m long and hinged at A. A rope DE, 0.5m is tied half way up. A weight 40kg is suspended from a point F, 1.2m 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 $g = 9.8m/s^2$) (Hint: Consider the equilibrium of each side of the ladder separately).
Exercise
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The given situation can be shown as:
$N _{ B }=$ 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
$B A=C A=1.6 m D E=0.5 m B F=1.2 m$ Mass of the weight,
$m =40 kg$ Draw a perpendicular from A on the floor BC .
This intersects DE at mid-point $H . \Delta ABI$ and $\triangle AIC$ are similar
$\therefore BI = IC$
Hence, I is the midpoint of $B C$. $D E \| B C B C$
$=2 \times D E=1 m A F=B A-B F=0.4 m \ldots \ldots$ (i)
$D$ is the mid-point of $A B$.
Hence, we can write, $AD =\left(\frac{1}{2}\right) \times BA =0.8 m \ldots (ii)$
Using equations (i) and (ii),
we get, $FE =0.4 m$ Hence, F is the mid-point of AD .
$FG \| DH$ and F is the mid-point of AD .
Hence, G will also be the mid-point of $AH . \triangle AFG$ and $\triangle ADH$ are similar
$\therefore \frac{ FG }{ DH }=\frac{ AF }{ AD } \frac{ FG }{ DH }=\frac{0.4}{0.8}=\frac{1}{2}=\left(\frac{1}{2}\right) \times 0.25=0.125 m \ln \Delta ADH , AH =\left( AD ^2- DH ^2\right)^{1 / 2}=\left(0.8^2-0.25^2\right)^{1 / 2}=$
0.76 m For translational equilibrium of the ladder,
the upward force should be equal to the downward force.
$N _{ C }+ N _{ B }$ $= mg =392$.....(iii)
For rotational equilibrium of the ladder, the net moment about A is,
$- N _{ B } \times BI + mg \times FG + N _{ C } \times$
$Cl + T \times AG - T \times AG =0- N _{ B } \times 0.5+40 \times 9.8 \times 0.125+ N _{ C } \times 0.5$
$=0\left(N_{ C }- N _{ B }\right) \times 0.5=49 N_{ C }- N _{ B }=98 \ldots..(iv)$
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 } \times BI + mg \times FG + T \times AG $
$=0-245 \times 0.5+40 \times 9.8 \times 0.125+ T \times 0.76=0 T$
$=96.7 N$
$N _{ B }=$ 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
$B A=C A=1.6 m D E=0.5 m B F=1.2 m$ Mass of the weight,
$m =40 kg$ Draw a perpendicular from A on the floor BC .
This intersects DE at mid-point $H . \Delta ABI$ and $\triangle AIC$ are similar
$\therefore BI = IC$
Hence, I is the midpoint of $B C$. $D E \| B C B C$
$=2 \times D E=1 m A F=B A-B F=0.4 m \ldots \ldots$ (i)
$D$ is the mid-point of $A B$.
Hence, we can write, $AD =\left(\frac{1}{2}\right) \times BA =0.8 m \ldots (ii)$
Using equations (i) and (ii),
we get, $FE =0.4 m$ Hence, F is the mid-point of AD .
$FG \| DH$ and F is the mid-point of AD .
Hence, G will also be the mid-point of $AH . \triangle AFG$ and $\triangle ADH$ are similar
$\therefore \frac{ FG }{ DH }=\frac{ AF }{ AD } \frac{ FG }{ DH }=\frac{0.4}{0.8}=\frac{1}{2}=\left(\frac{1}{2}\right) \times 0.25=0.125 m \ln \Delta ADH , AH =\left( AD ^2- DH ^2\right)^{1 / 2}=\left(0.8^2-0.25^2\right)^{1 / 2}=$
0.76 m For translational equilibrium of the ladder,
the upward force should be equal to the downward force.
$N _{ C }+ N _{ B }$ $= mg =392$.....(iii)
For rotational equilibrium of the ladder, the net moment about A is,
$- N _{ B } \times BI + mg \times FG + N _{ C } \times$
$Cl + T \times AG - T \times AG =0- N _{ B } \times 0.5+40 \times 9.8 \times 0.125+ N _{ C } \times 0.5$
$=0\left(N_{ C }- N _{ B }\right) \times 0.5=49 N_{ C }- N _{ B }=98 \ldots..(iv)$
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 } \times BI + mg \times FG + T \times AG $
$=0-245 \times 0.5+40 \times 9.8 \times 0.125+ T \times 0.76=0 T$
$=96.7 N$
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