In this example we will consider the classic static-equilibrium situation of a l

Question

In this example we will consider the classic static-equilibrium situation of a ladder leaning against a wall. Suppose a medieval knight is climbing a uniform ladder that is 5.0 m long and weighs 180 N. The knight, who weighs 810 N, stops a third of the way up the ladder (Figure 1). The bottom of the ladder rests on a horizontal stone ledge and leans across the castle's moat in equilibrium against a vertical wall that is frictionless because of a thick layer of moss. The ladder makes an angle of 53 degree with the horizontal, conveniently forming a 3-4-5 right triangle, (a) Find the normal and friction forces on the ladder at its base, (b) Find the minimum coefficient of static friction needed to prevent slipping, (c) Find the magnitude and direction of the contact force on the ladder at the base. SOLUTION SET UP (Figure 2) shows a free-body diagram for the ladder. Since it is described as "uniform," we can assume that its center of gravity is at its center, halfway between the base and the wall. The knight's weight pushes down on the ladder at a point one-third of the way from the base toward the wall. The contact force at the top of the ladder is horizontal and to the left because the wall is frictionless. The components of the contact force at the base are the upward normal force n_2 and the friction force f_s, which must point to the right to prevent slipping. SOLVE Part (a) The first condition for equilibrium, in component form, gives sigma F_x = f_s + (-n_1) = 0 sigma F_y = n_2 + (-810N) + (-180N) = 0 These are two equations for the three unknowns n_1, n_2, and f_s. The first equation tells us that the two horizontal forces must be equal and opposite, and the second equation gives n_2 = 990N. That is, the ledge pushes up with a force of 990N to balance the total (downward) weight (810 N +180N). We don't yet have enough equations, but now we can use the second condition for equilibrium. We can take torques about any point we choose. The smart choice is the point that will give us the fewest terms and fewest unknowns in the torque equation. With this thought in mind, we choose point B, at the base of the ladder. The two forces ft 2 and fs have no torque about that point. From (Figure 2), we see that the moment arm for the ladder's weight is 1.5 m, the moment arm for the knight's weight is 1.0 m, and the moment arm for n is 4.0 m. The torque equation for point B is thus sigma tauB = n_1(4.0 m) - (180N)(1.5 m) -(810N)(1.0 m) + n_2(0) + f_s(0) = 0 Solving for n_1, we get n_1 = 271N Now we substitute this result back into the equation sigma F_x = 0 to get f_s = 271N The static-friction force f_s cannot exceed mu s n_2, so the minimum coefficient of static friction to prevent slipping is (mu_s)_min = fs/n_2 = 271N/990N = 0.27 The components of the contact force at the base are the static-friction force fs = 271N and the normal force n_2 = 990N; the magnitude of F vector_B is F_B = squareroot (271N)^2 + (990N)^2 = 1030N and its direction (Figure 3) is theta = tan^-1 990N/271N = 75 degree REFLECT First, the contact force F vector_B at an angle of 75 degree is not directed along the length of the ladder, which is at an angle of 53 degree. If it were, the sum of the torques with respect to the point where the knight stands couldn't be zero. Second, as the knight climbs higher on the ladder, the moment arm and torque of his weight about B increase, thereby increasing the values of n_1, f_s, and (mu_s)_min- At the top, his moment arm would be nearly 3 m, giving a minimum coefficient of static friction of nearly 0.7. An aluminum ladder on a wood floor wouldn't have a value of mu_s this large, so present-day aluminum ladders are usually equipped with nonslip rubber feet. Third, a larger ladder angle (i.e., a more upright ladder) would decrease the moment arms of the weights of the ladder and the knight with respect to B and would increase the moment arm of n_1, all of which would decrease the required friction force. Some manufacturers recommend that their ladders be used at an angle of 75 degree. (Why not 90 degree) What is the magnitude of the contact force on the ladder at the base if the knight has climbed two-thirds of the way up the ladder? What is the direction of the contact force on the ladder at the base if the knight has climbed two-thirds of the way up the ladder?

Explanation / Answer

Assume the force due to the wall be Fw, the friction force to be Ff and the Normal force be N

Lets write the torque balannce out the base

1/2 *180 *5 cos (53) + 2/3 *810 * 5 * cos(53) = Fw *5 * sin(53)

Fw = 475 N

Using the force balance in the horizonntal direction we can say

Fw= Ff = 475N

Now using the force balance in the erticle diretion

N = 180 +810 = 990 N

Hence the net contact force will be 1097 N

At the angle of 64 degree from the horizontal

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