A stick with a mass of 0.244kg and a length of 0.416mrests in contact with a bow

Question

A stick with a mass of 0.244kg and a length of 0.416mrests in contact with a bowling ball and a rough floor, as shown in the figure(Figure 1) . The bowling ball has a diameter of 20.6cm , 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 and vertical components of the force.

Figure 1:

Explanation / Answer

Let:
O be the centre of the ball,
P be the point where the stick meets the floor,
Q be the point where the stick touches the ball,
m be the mass of the stick,
L be the length of the stick,
d be the diameter of the ball,
x be the distance from the lower end of the stick to its point of contact with the ball,
a be the inclination of the stick to the floor,
R be the upward reaction of the floor on the stick,
F the the friction force between the floor and the stick,
u be the coefficient of friction,
S be the normal reaction of the ball on the stick,
g be the acceleration due to gravity.

Resolving horizontally:
F = S sin(a) ...(1)

Resolving vertically:
mg = R + S cos(a) ...(2)

Friction constraint:
F / R <= u ...(3)

Moments about P:
mgL cos(a) / 2 = Sx ...(4)

Geometry of triangle OPQ:
d / (2x) = tan(a / 2) ...(5)

(a)
Eliminating x from (4) and (5):
mgL cos(a) = Sd / tan(a / 2)
S = mgL cos(a) tan(a / 2) / d
= 0.254 * 9.81 * 0.416 cos(30) tan(15) / 0.206
= 1.1676
= 1.17 N to 3sf.

(b)
From (1):
F = S sin(a)
= 1.1676 sin(30)
= 0.5838
= 0.584 N to 3sf.

(c)
From (2):
R = mg - S cos(a)
= 0.254 * 9.81 - 0.584 cos(30)
= 1.9859
= 1.99 N to 3sf.

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