A stick with a mass of 0.234 kg and a length of 0.426 m rests in contact with a
ID: 1441309 • Letter: A
Question
A stick with a mass of 0.234 kg and a length of 0.426 m rests in contact with a bowling ball and a rough floor. 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.a) Find the magnitude of the force exerted on the stick by the bowling ball. b) Find the horizontal component of the force exerted on the stick by the floor. c) Repeat part B for the vertical component of the force
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 = 0.234 kg, L = 0.426 m, d = 0.206 m, g = 9.81 m/s^2
F1 = 1.1015 N
Total vertical force on the stick = 0
Therefore, F2v + F1 cos(30 deg) = M g
F2v = Mg - F1 cos(30 deg)
= 0.234 * 9.81 – 1.1015 cos(30 deg)
= 1.19 N
Net horizontal force on the stick = 0
Therefore, F2h = F1 sin(30 deg) = 1.1015 * sin(30 deg) = 0.550 N
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.