A uniform ladder of length L rests against a smooth, vertical wall (Figure 8.1).

Question

A uniform ladder of length L rests against a smooth, vertical wall (Figure 8.1). The mass of the ladder is m, and the minimum angle at which the ladder docs not slip is = 51 degree. Find the coefficient of static friction between the ladder and the ground. ( Answ. 0.40 ) Two 80-N forces are applied to opposite comers of a rectangular plate, as shown in Figure 8.3. (a) Find the torque produced by this couple, (b) Show that the result is the same as if you determine the torque about the lower Ieft-corner.

Explanation / Answer

Given Length of the ladder is L Mass of the ladder is m The angle made by the ladder with the wall is 510 Let the frictional force between the ladder and the ground is f The height of the wall be d The normal force acting on the base of the ladder is n The force exerted by the wall at the top of the ladder is P Since the system is in equilibrium , the net force and the net torque acting on the system is zero. Fx = f -P = 0 f = P     ..... (1) And Fy = n - mg = 0 n = mg    ..... (2) Taking the torques around the base of the ladder is i = f + n + grav + p = 0      = 0 +0 - mg (L/2) cos51 + P (L) sin 51 = 0 P = (mg /2) (cos51 / sin51) -----(3) From eq 1 ,    P = f                       P = s (n) = s (mg)      (mg /2) (cos51 / sin51) =  s (mg)                                      s = 0.40              ----------------------------------------------------------------------------------------- Magnitude of the two forces , F = 80 N Angle made by the force F is 300 a) Torque produced by the couple is calculated by taking the torque     about the center of the rectangular plate      = F cos30 (-b/2) - F cos30 (+ b/2) - F sin 30 (a/2) + F sin 30 (-a/2)        = -0.8660 b F - 0.5 F a b) Computing the torque from the lower left corner is ,       = - a F sin 30 - bF sin 60          = - 0.5 F a - 0.8660 bF Thus, the result is same in both the cases a) Torque produced by the couple is calculated by taking the torque     about the center of the rectangular plate      = F cos30 (-b/2) - F cos30 (+ b/2) - F sin 30 (a/2) + F sin 30 (-a/2)        = -0.8660 b F - 0.5 F a b) Computing the torque from the lower left corner is ,       = - a F sin 30 - bF sin 60          = - 0.5 F a - 0.8660 bF Thus, the result is same in both the cases

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