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The problem is as follows:

Two homogeneous spheres are at equilibrium by the action of an horizontal force of minimal magnitude $F= 25\,N$, as shown in the picture from below. The sphere which is in the bottom has a weight of $10\sqrt{3}$ and the upper sphere has a weight of $25\sqrt{3}$. Using this information, find the magnitude of the contact force excerted by sphere in the top over the sphere in the bottom. Assume the friction is negligible.

Sketch of the problem

The alternatives in my book are as follows:

$\begin{array}{ll} 1.&80\,N\\ 2.&75\,N\\ 3.&35\,N\\ 4.&20\,N\\ 5.&50\,N\\ \end{array}$

I often struggle in these kinds of problems. Can someone help me with the FBD of this thing?. I've attempted to draw all the forces which I could spot and these are drawn in the figure from below.

Sketch of the solution

However in the end that's where I'm stuck. I don't know how to relate all these forces in a resultant or a way to solve what it is being asked. Can someone help me here?. The problem where I'm struggling the most is the reaction from the floor in the ball from the bottom. Wouldn't this cancel with the weight of the ball in the bottom and from the top?.

I'm assuming what it is being asked is the reaction from orange which is pointing to the upper right corner in my drawing. But is this correct?. Can someone help me with the easiest way to solve this without geting tangled with too many equations?.

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  • $\begingroup$ I suggest asking this on the physics stack exchange.. $\endgroup$ Commented Jul 10, 2020 at 3:32
  • $\begingroup$ Home work? The contact force between spheres is 50N. They are at an angle of $\frac{\pi}{3}$. $\endgroup$
    – Cesareo
    Commented Jul 10, 2020 at 7:48

1 Answer 1

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One way to look at this, consider the system of both balls as one. Since this system as a whole is in equilibrium, the external forces acting on this sytem must all cancel out.

Using the left direction as positive x, and the up as positive y, we have

$$\vec {F} + \vec{N}_{wall, blue} + \vec{N}_{floor, orange} + \vec{W} = \vec{0}$$

$$25\hat{x}-N_{wb} \hat{x} + N_{fo}\hat{y} - 35\sqrt{3}\hat{y} = 0 \hat{x} + 0\hat{y}$$

Using this you can just solve for both the normal forces with the wall and the floor. Can you proceed from there?

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