Two spheres, each of mass m = 170 g, are placed in a glass beaker as shown below

Question

Two spheres, each of mass m = 170 g, are placed in a glass beaker as shown below. The dotted line in the figure makes a 45? angle with respect to the hori- zontal and crosses through the center of mass of the two spheres. Assume the walls of the beaker are frictionless.

(a) Determine the magnitude of the three forces in the diagram, P?1, P?2, and P?3.

(b) Calculate the magnitude of the forces that the spheres apply on each other.

Problem #3 Two spheres, each of mass m = 170 g, are placed in a glass beaker as shown below. The dotted line in the figure makes a 45° angle with respect to the hori zontal and crosses through the center of mass of the two spheres. Assume the walls of the beaker are frictionless. 45 9 (a) Determine the magnitude of the three forces in the diagram, P·A, and As. (b) Calculate the magnitude of the forces that the spheres apply on each other.

Explanation / Answer

Here it needs to be understood the the top sphere's weight is balanced by the vertical component of the normal force being exerted on it by the lower sphere.

Part a.) Now, we will form equations for each of the spheres along the vertical and horizontal direction and then resolve to determine the required values. Let us assume a force of N acts between the two spheres.

For lower sphere:

Horizontal forces would give: P3 = NCos45

Vertical forces would give: P2 = NSin45 + Mg

For the top sphere:

Horizontal forces yield: NCos45 = P1

Vertical forces give: NSin45 = Mg

or N = Mg2, using which in the equations above we get:

P1 = N Cos45 = Mg = 0.17 x 9.81 = 1.6677 N

Also, P2 = NSIn45 + Mg = Mg/2 + Mg = 2.847 N

P3 = Ncos45 = 1.179 N

Part b.) From the above relation we also have N = Mg2 = 2.358 N

Therefore the required force acting between the spheres is 2.358 N

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