Question

As shown in the figure below, a stick of length L = 0.460 m and mass m = 0.195 kg is in contact with a rough floor at one end and a frictionless bowling ball (diameter d = 19.00 cm) at some other point such that the angle between the stick and the floor is θ = 30°. Determine the following.

(a) magnitude of the force exerted on the stick by the bowling ball

(b) horizontal component of the force exerted on the stick by the floor

(c) vertical component of the force exerted on the stick by the floor

Answer 1

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)

a)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 = 0.195kg, L = 0.460 m, d = 0.19m, g = 9.81 m/s^2
F1 = 1.07 N

b) Total vertical force on the stick = 0
Therefore, F2v + F1 cos(30 deg) = M g
F2v = Mg - F1 cos(30 deg)
= 0.195 * 9.81 - 1.07cos(30 deg)
= 0.98 N

C) Net horizontal force on the stick = 0
Therefore, F2h = F1 sin(30 deg) = 1.07* sin(30 deg) = 0.535 N

Ans:
(a) 1.07 N
(b) 0.98 N
(c) 0.535 N

Hope this helps you.

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