Answer to Question #123738 in Mechanics | Relativity for Brian

Question

A uniform ladder AB of weight 180N and length 8m, rest in equilibrium with its foot on a rough horizontal ground and its upper end against a rough vertical wall. The ladder makes an angle of 60 degrees with the horizontal ground. The coefficient of friction between the ladder and the wall is 1/4 and that between the ladder and the ground is 1/2. Find the normal reactions at A and B

Answer 1

$\theta = 60^\circ, \;\; \mu_1 = 0.5, \;\; \mu_2 = 0.25.$ The length $l = 8\,\mathrm{m},$ the weight $mg = 180\,\mathrm{N}.$ Let $N_1$ be the normal reaction at the foot of the ladder and $N_2$ be the normal reaction at the highest point of the ladder. We may write two equations of the force balance in projection on vertical and horizontal axes:

$F_{fr,2} + N_1= mg\,; \;\; N_2 = F_{fr,1}\,.$

Let us assume that the friction is proportional to the friction of rest, so $\mu_2N_2\nu + N_1 = mg, \;\; N_2=\mu_1N_1\nu.$

We may also write equation concerning torques using the top of ladder:

at the top: $mg\dfrac l2\cos\alpha = N_1l\cos\alpha - F_{fr,1}l\sin\alpha.$

Substituting $F_{fr,1}\,, \; F_{fr,2}$ from the balance of forces, we get

$mg\dfrac 12\cos\alpha =N_1\cos\alpha -N_2\sin\alpha$ . Therefore, $N_1 = \dfrac{mg}{2} + N_2\tan\alpha.$

So we obtain the system of equations

$\begin{cases} \mu_2N_2\nu + N_1 = mg, \\ N_2=\mu_1N_1\nu, \\ N_1 = \dfrac{mg}{2} + N_2\tan\alpha. \end{cases}$ $\begin{cases} \mu_1\mu_2N_1\nu^2 + N_1 = mg, \\ N_2=\mu_1N_1\nu, \\ N_1 = \dfrac{mg}{2} + \mu_1N_1\nu\tan\alpha. \end{cases}$

Then we get equation $\mu_1\mu_2\nu^2+2\mu_1\nu\tan\alpha-1=0, \;\;$ $\nu\approx 0.555.$

$N_1 = \dfrac{mg}{\mu_1\mu_2\nu^2+1} = 173\,\mathrm{N}.$

$N_2 = 0.5\cdot N_2\cdot 0.555 = 48\,\mathrm{N}.$