A stick with a mass of 0.217 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 21 cm, and the angle the stick makes with the horizontal is 30°. 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 upward component of the force.
for part a
sum FX=Gx(the Ground)-P(force at BBall) = 0 Gx=P
sum FY=Gy(the Ground)-Ws(stick weight) = 0 Gy=Ws
sum of the torques is zero
The Attempt at a Solution
.217kg*9.8m/s = 2.1266N
Ws=2.1266Ncos 30 = 1.8417 = -Wstick*Lstick
P = .435m*sin 30 = .2175
so sum of torques is -Wstick*Lstick+PLp
P = 2.1266N*.2175m(mid stick)*cos30 / .435m*sin 30 or .4006/.2175 = 1.8418N
is the diameter number extra info or am I missing something here? This is a first attempt
for me at a problem like this, I am probably off-base?