A uniform ladder of length L and weight 230 N rests against a vertical wall. The

Question

A uniform ladder of length L and weight 230 N rests against a vertical wall. The coe?cient of static friction between the ladder and the ?oor is 0.42, as is the coe?cient of friction between the ladder and the wall. What is the smallest angle the ladder can make with the ?oor without slipping? Answer in units of degrees.

Explanation / Answer

Let us take angle made by ladder with the wall be theta ? let us take point in contact with wall be A The point where the whole ladder weight acting at the centre of the ladder be B The point in contact with the floor be C There are two reaction forces one in vertical direction and horizontal direction Let us assume the ladder is moving towards right so the point C moves towards right so friction force(u*N2) at C will be in negative x direction and there will be a normal reaction(N2) at C acting in upward direction As point C move towards right A will be moving downwards friction force(u*N1) acting at A will be i upward direction at A and there will be normal reaction applied at the wall in positive x direction (N1) At B weight will be acting downward W Balancing in Forces in X and Y directions and balancing moments... w= (u * N1) + N2 N1 = u *N2 (W*(l/2)*cos ? ) = (N1 * L * sin ?) + (u * N1 * L cos ?)-----------------> (equation 1a) solving for N2 and N1 W = (u*u)N2 + N2 W= N2(1+u^2) N2 = (W/(1+u^2)) N1 = u *N2 N2 = (u*W/(1+u^2)) substituting in equation 1a dividing equation 1a by cos ? w*(L/2) = N1 * L *tan ? + u*N1*L w*(L/2) = N1 * L (tan ? + u ) W/(2*N1) = (tan ? + u ) substituting value of N1 (W(1+(u^2))/(2*u*W)) = (tan ? + u ) ((1+(u^2))/(2*u)) - u)) = tan ? ((1-(u^2))/(2*u))) = tan ? substituting value of u=0.42 solving for ?=44.43

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