Question

As shown in the figure below, a stick of length L = 0.495 m and mass m = 0.235 kg is in contact with a rough floor at one end and a frictionless bowling ball (diameter d = 17.00 cm) at some other point such that the angle between the stick and the floor is e 30º. Determine the following. 30.0° (a) magnitude of the force exerted on the stick by the bowling ball N (b) horizontal component of the force exerted on the stick by the floor N (c) vertical component of the force exerted on the stick by the floor N

Answer 1

N Z o Fy 1/2 Rcose X h L/2 30.0° Y WI

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from triangle OXY

cos 30 = r / L/2

r = L cos 30 / 2

and

from triangle OZW

L = H / sin 30

where , H = R + R cos 30

net torque about center of mass of stick

torque = force * lever arm

where lever arm - perpendicular distance from line of action of force

N L = mgr

N = mgr / L

N = mgL sin 30 cos 30 / 2R ( 1 + cos 30)

where, R = 17/ 2 = 0.085 m

N = 0.235 * 9.8 * 0.495 * cos 30 sin 30 / 2 * 0.085 ( 1 + cos 30)

N = 1.556 N

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Fx = N sin 30

Fx = 0.778 N

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Fy = mg - N cos 30

Fy = 0.235 * 9.8 - 1.556cos 30

Fy = 0.955 N