Two uniform 75.9-g marbles 1.74 cm in diameter are stacked as shown in the figur
ID: 1506727 • Letter: T
Question
Two uniform 75.9-g marbles 1.74 cm in diameter are stacked as shown in the figure (Figure 1) in a container that is 3.00 cm wide. https://session.masteringphysics.com/problemAsset/1260581/3/YF-11-55.jpg 1. Find the force that the container exerts on the marble at the point of contact A. 2. Find the force that the container exerts on the marble at the point of contact B. 3. Find the force that the container exerts on the marble at the point of contact C. 4. What force does each marble exert on the other? Been stuck on this question for hours. I know that the force at B is 1.49N. Help me out pleaseExplanation / Answer
Taking the two marbles as the system, following forces act on them :
their total weight, 2*0.0759*9.8 N downwards.
horizontal reactive forces at A and C.
Reactive force that the container exerts on the marble at the point of contact B, upwards
2*0.0759*9.8 N = 1.49 N
Consider the upper marble. Three forces act on it.
its weight, 0.745 N downwards
Horizontal reactive force, F' at C
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) (3 - 1.74) = 63 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 = 1.74 cm
horizontal line = 3 - 1.74 = 1.26 cm
vertical line = [(1.74)^2 - (1.26)^2] = 2.1483
Vertical component of F
= F * (2.1483/1.74) = 1.2347F
This is equal to the weight of the upper marble
1.2347F = 0.745
F = 0.745 / 1.2347
F = 0.6034 N
Reaction at C,
F' = F * (0.97) /(1.74)
F' = 0.6034 * (1.26) /(1.74)
F' = 0.4370 N
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