A sqrt{34},m long ladder weighing 10,kg leans | vedclass

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(C) Let the length of the ladder be $L = \sqrt{34}\,m$ and the base distance be $b = 3\,m$.The height of the ladder on the wall is $h = \sqrt{L^2 - b^

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$\frac{3}{\sqrt{109}}$

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(C) Let the length of the ladder be $L = \sqrt{34}\,m$ and the base distance be $b = 3\,m$.The height of the ladder on the wall is $h = \sqrt{L^2 - b^2} = \sqrt{34 - 9} = \sqrt{25} = 5\,m$.Let $	heta$ be the angle the ladder makes with the floor. Then $\cos 	heta = \frac{b}{L} = \frac{3}{\sqrt{34}}$ and $\sin 	heta = \frac{h}{L} = \frac{5}{\sqrt{34}}$.Thus,$\cot 	heta = \frac{\cos 	heta}{\sin 	heta} = \frac{3}{5}$.Let $N_1$ be the normal force from the floor and $f$ be the friction force from the floor. Let $N_2$ be the normal force from the frictionless wall.For translational equilibrium: $\sum F_x = 0 \implies f = N_2$ and $\sum F_y = 0 \implies N_1 = mg$.For rotational equilibrium about the base: $N_2 	imes L \sin 	heta = mg 	imes \frac{L}{2} \cos 	heta$.$N_2 = \frac{mg}{2} \cot 	heta = \frac{mg}{2} 	imes \frac{3}{5} = \frac{3mg}{10}$.The reaction force of the floor is $F_f = \sqrt{N_1^2 + f^2} = \sqrt{(mg)^2 + (N_2)^2} = \sqrt{(mg)^2 + (\frac{3mg}{10})^2} = mg \sqrt{1 + \frac{9}{100}} = mg \sqrt{\frac{109}{100}} = \frac{mg}{10} \sqrt{109}$.The reaction force of the wall is $F_w = N_2 = \frac{3mg}{10}$.The ratio $\frac{F_w}{F_f} = \frac{3mg/10}{(mg/10)\sqrt{109}} = \frac{3}{\sqrt{109}}$.

$A$ $\sqrt{34}\,m$ long ladder weighing $10\,kg$ leans on a frictionless wall. Its feet rest on the floor $3\,m$ away from the wall as shown in the figure. If $F_{f}$ and $F_{w}$ are the reaction forces of the floor and the wall,then the ratio of $F_{w}/F_{f}$ will be:
(Use $g=10\,m/s^{2}$)

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$A$ $\sqrt{34}\,m$ long ladder weighing $10\,kg$ leans on a frictionless wall. Its feet rest on the floor $3\,m$ away from the wall as shown in the figure. If $F_{f}$ and $F_{w}$ are the reaction forces of the floor and the wall,then the ratio of $F_{w}/F_{f}$ will be:(Use $g=10\,m/s^{2}$)