A stick with a mass of 0.213 kg and a length of 0.425 m rests in contact with a

Question

A stick with a mass of 0.213 kg and a length of 0.425 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 22.07 cm, and the angle ? the stick makes with the horizontal is 29?. 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.

Explanation / Answer

Length of stick between floor and ball is r/Tan(/2) = (22.07/2)/Tan(29/2) = 42.67 cm

Horizontal distance between ball and point where stick touches the ground is also r/Tan(/2) = 42.67 cm.

Let the reaction force from ball on stick is R.

Wei ght of stick will act from its centre.

Taking moments of the forces acting on the stick w.r.t the point where it touches the ground,

R*0.4267 = (0.213*9.81)*((0.425/2) Cos29)

(a) This gives, R = 0.91 N

(b) Let the horizontal and vertical reaction forces acting on the stick from the floor be Fh and Fv respectively.

Resolving R in horizontal and vertical components,

Rh = R Sin 29 = 0.44 N (towards left)

Rv = R Cos 29 = 0.8 N

Balancing horizontal forces acting on the stick,

Rh+Fh = 0

So, Fh = -0.44 N (acting towards right)

(c) Balancing vertical forces acting on the stick,

Rv+Fv = 0.213*9.81

So, Fv = 1.3 N (upwards)

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.