Question #123738

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

Expert's answer

"\\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}\n\\mu_2N_2\\nu + N_1 = mg, \\\\ N_2=\\mu_1N_1\\nu, \\\\\nN_1 = \\dfrac{mg}{2} + N_2\\tan\\alpha.\n\\end{cases}" "\\begin{cases}\n\\mu_1\\mu_2N_1\\nu^2 + N_1 = mg, \\\\ N_2=\\mu_1N_1\\nu, \\\\\nN_1 = \\dfrac{mg}{2} + \\mu_1N_1\\nu\\tan\\alpha.\n\\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}."

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