Question

As shown in the figure below, a stick of length = 0.395 m and mass m -0.240 kg is in contact with a rough floor at one end and a frictionless bowling ball (diameter d = 15.00 cm) at some other point such that the angle between the stick and the floor is 0 - 30°. Determine the following. 30.0 (a) magnitude of the force exerted on the stick by the bowling ball 1.66 X Since the stick is in static equilibrium, we expect to use the first and second condition of equilibrium. Since we need to determine torques, before starting the problem, see if you can write an expression for the torque in terms the force producing the torque and the perpendicular distance between the line of action of this force and the point about which we are determining the torques. Determine a convenient point about which to dete ine torques. It will be to your advantage a careful figure which includes the stick, all dimensions and angles, all forces acting on the stick, the point about which you plan to take the torques and for each force. Now that all the preliminary work for the problem has been done, see if you can write an expression of the second condition of equilibrium that will allow you to determine the 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 0.5 X Now that you have determined N (the magnitude of the force exerted on the stick by the bowling ball from part(s) of the problem, see if you can write an expression of the first condition of equilibrium that will allow you to determine the 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 0.91 х Now that you have determined N, the magnitude of the force exerted on the stick by the bowling ball) from part(s) of the problem, see you can write an expression of the first condition of equilibrium that will allow you to determine the vertical component of the force exerted on the stick by the floor. N

Answer 1

V o Fy 1/2 Rcos e у. H L/2 30,4 mg х Fx w

from triangle xyz

cos 30 = r / L/2

r = L cos 30 / 2

and

from triangle xvw

L = H / sin 30

where , H = R + R cos 30

net torque about center of mass of stick

N L = mgr

N = mgr / L

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

where, R = 15/ 2 = 0.075 m

N = 0.240 * 9.8 * 0.395 * cos 30 sin 30 / 2 * 0.075 ( 1 + cos 30)

N = 1.437 Newtons

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Fx = N sin 30

Fx = 0.7186 N

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Fy = mg - N cos 30

Fy = 0.240 * 9.8 - 1.437 cos 30

Fy = 1.1075 N