A stick with a mass of 0.244 kg and a length of 0.406 m rests in contact with a

Question

A stick with a mass of 0.244 kg and a length of 0.406 m rests in contact with a bowling ball and a rough floor, as shown in the figure(Figure 1) . The bowling ball has a diameter of 20.6 cm , and the angle the stick makes with the horizontal is 30.0 . You may assume there is no friction between the stick and the bowling ball, though friction with the floor must be taken into account.

Part A

Find the magnitude of the force exerted on the stick by the bowling ball.

Part B

Find the horizontal component of the force exerted on the stick by the floor.

Part C

Repeat part B for the vertical component of the force.

Explanation / Answer

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 = 0.244 kg, L = 0.406 m, d = 0.206 m, g = 9.81 m/s^2
F1 = 1.0947 N

Total vertical force on the stick = 0
Therefore, F2v + F1 cos(30 deg) = M g
F2v = Mg - F1 cos(30 deg)
= 0.244 * 9.81 - 1.0947 cos(30 deg)
= 1.4455 N

Net horizontal force on the stick = 0
Therefore, F2h = F1 sin(30 deg) = 1.0947 * sin(30 deg) = 0.5487 N

Ans:
(a) 1.0947 N
(b) 1.44 N
(c) 0.5487 N

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