A stick with a mass of 0.213 kg and a length of 0.435 m rests in contact with a

Question

A stick with a mass of 0.213 kg and a length of 0.435 m rests in contact with a bowling ball and a rough floor, as shown in the figure below. The bowling ball has a diameter of 22.58 cm, and the angle ? the stick makes with the horizontal is 39?. You may assume there is no friction between the stick and the bowling ball, though friction with the floor must be taken into account. (a) Find the magnitude of the force exerted on the stick by the bowling ball. N (b) Find the horizontal component of the force exerted on the stick by the floor. N (c) Repeat part (b) for the upward component of the force. N

Explanation / Answer

Let A = point of contact of the stick with the ground, B = point of contact of the stick with the ball, C = the center of gravity of the stick, O = center of the ball r = radius of the ball, d = diameter of the ball, L = length of the stick, M = mass of the stick, F1 = force on the stick by the ball, F2v = vertical component of force on the stick by the floor, F2h = horizontal component of force on the stick by the floor OB is perpendicular to the stick. OB makes 30.0 deg with vertical. Height of B above the ground is h = OB cos(30 deg) + r h = r cos(30 deg) + r h = r(1 + cos 30 deg) AB = h/(sin 30deg) AB = r(1 + cos 30 deg)/(sin 30deg) AB = 2 r(1 + cos 30 deg) AB = d(1 + cos 30 deg) Torque of the stick around A = 0 Therefore, Mg * L/2 * cos(30 deg) = F1 * AB Mg * L/2 * cos(30 deg) = F1 * d(1 + cos 30 deg) F1 = Mg * L/2 * cos(30 deg) / [d(1 + cos 30 deg)] Substitute M = 0.202 kg, L = 0.441 m, d = 0.2289 m, g = 9.81 m/s^2 F1 = 0.886 N Total vertical force on the stick = 0 Therefore, F2v + F1 cos(30 deg) = M g F2v = Mg - F1 cos(30 deg) = 0.202 * 9.81 - 0.886 cos(30 deg) = 1.21 N Net horizontal force on the stick = 0 Therefore, F2h = F1 sin(30 deg) = 0.886 * sin(30 deg) = 0.443 N Ans: (a) 0.886 N (b) 1.21N (c) 0.443 N

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