A uniform ladder of mass (m) and length (L) leans against a frictionless wall, s

Question

A uniform ladder of mass (m) and length (L) leans against a frictionless wall, see figure. If the coefficient of static friction between the ladder and the ground is 0.41, what is the minimum angle (q) between the ladder and the floor at which the ladder will not slip?

A uniform ladder of mass (m) and length (L) leans against a frictionless wall, see figure. If the coefficient of static friction between the ladder and the ground is 0.41, what is the minimum angle (q) between the ladder and the floor at which the ladder will not slip?

Explanation / Answer

Now, the sum of these three forces must be exactly zero and the same for their torques. The former gives a set of equations RAx + RBx = 0 RAy - mg = 0 We immediately observe RAx = -RBx RAy = mg So there's only one unknown left. The torque, calculated with respect to the point A, is -2r*RBx*sin ? - r*mg*cos ? and this must be zero. Solving this, we get 2r RBx sin ? = -r mg cos ? RBx = -cos ? / (2sin ?) mg = -1/2*cot ?*mg Note the sign: we measure RBx to the right, so it is OK that we got it negative. We are about to finish. The x-component of the RA is 1/2*cot ?*mg, but this can't exceed the coefficient 0.41*the y-component, which is mg. In equations, 1/2*cot ?*mg < 0.41*mg 1/2*cot ? < 0.41 cot ? < 0.82 ? > 50°39'

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