Find the force that the container exerts on the marble at the point of contact A.?

Question

Two uniform 66.4 g marbles 1.76 cm in diameter are stacked in a container that is 2.73 cm wide.

http://session.masteringphysics.com/problemAsset/1039128/5/YF-11-55.jpg

1) Find the force that the container exerts on the marble at the point of contact B.

2) Find the force that the container exerts on the marble at the point of contact C.

3) What force does each marble exert on the other?

Answer 1

Answer : 1) 0.3 N, 2) 0.18 N and 3) 0.1 N worked out as under.

1)

Taking the two marbles as the system, following forces act on them :

(1) their total weight, 2*0.0664*9.8 N downwards

(2) horizontal reactive forces at A and C

(3) Reactive force that the container exerts on the marble at the point of contact B, upwards = R

=> R

= 2*0.0664*9.8 N

= .30144 N

Consider the upper marble. Three forces act on it.

(1) its weight, 0.15072 N downwards

(2) Horizontal reactive force, F' at C

(3) Reactive force, F,  from the lower marble at the point of contact in a direction passing through the center of the upper marble. Horizontal distance between C and center of the upper marble = (1/2) (2.73 - 1.76) = 0.485 cm

Consider a right triangle having line joining the centers of the marbles as hypotenuse and horizontal line through the center of lower marble and a vertical line from the center of the upper marble as the other sides.

hypotenuse = 1.76 cm

horizontal line = 2.73 - 1.76 = 0.97 cm

vertical line = √[(1.76)^2 - (0.97)^2] = 1.47

Vertical component of F

= F * (1.47/1.76) = 0.835F

This is equal to the weight of the upper marble

=> 0.835F = 0.15072

=> F

= 0.15072 / 0.835

= 0.181 N (This is answer to Q 3)

Reaction at C,

F'

= F * (0.97) /(1.76)

= 0.181 * (0.97) /(1.76)

= 0.1 N