1 answer

As shown in the figure below, a stick of length L = 0.460 m and mass...

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

(i know A is 1.07)

(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

30.0

30.0

Answers

B о А. 30.0°

Given

a stick of length L = 0.460 m

mass M = 0.195 kg

diameter d = 19.00 cm = 0.19 m

C = the center of gravity of the stick,

F1 = force on the stick by the ball,

Fv = vertical component of force on the stick by the floor,

Fh = horizontal component of force on the stick by the floor

center of the ball & point of contact of the stick with the ball is perpendicular to the stick and makes 30.0 deg with vertical.

Height of point of contact of the stick with the ball above the ground is

h = OB*cos30 + r

h = r*cos30 + r

h = r(1 + cos30)

AB = h/(sin30)

AB = r*(1 + cos30)/(sin30)

AB = 2*r(1 + cos30)

AB = d*(1 + cos30)

Torque of the stick around A = 0

Mg*L/2*cos30 = F1*AB

Mg*L/2*cos30 = F1*d(1 + cos30)

F1 = Mg* L/2*cos(30) / [d(1 + cos30)]

F1 = 0.195*9.81* 0.460/2*0.866 / [0.19(1 + 0.866)]

F1 = 1.0747 N

F1 = 1.07 N

Total vertical force on the stick = 0

Therefore,

Fv + F1*cos 30 = M*g

Fv = M*g - F1*cos30

Fv = 0.195 * 9.81 – 1.0747*0.866

Fv = 0.98 N

Net horizontal force on the stick = 0

Fh = F1*sin 30

Fh = 1.0747 * sin 30

Fh = 0.537 N

.

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