circles in a box

by skiboka33
Tags: circles
skiboka33 is offline
Mar22-04, 07:06 PM
P: 60
Two identical uniform spheres, each weighing 75 N are at rest on the bottom of a fixed rectangular container. The line of centers of the spheres makes an angle of 35* with the horizontal. Find the forces exerted on the spheres by the container bottom, the container sides as well as the force that each sphere exerts on the other.

ill try to draw a picture for you...

* O O *
* O O O O*
*O O O O*
*O O O O *
* O O *
*************** the circle on the right is above the circle on the left... if you drew a line between the two centers you would have a line 35 degrees to the horizontal...

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skiboka33 is offline
Mar22-04, 07:10 PM
P: 60
wow my picture didnt turn out too well, basically its two circles, the one on the right is above the one on the left ( they dont both fit flat on the bottom of the container )

| |
| --- |
| ( )|
| --- --- |
|( ) |
| --- |
--------------- i hope this one turns out better
skiboka33 is offline
Mar22-04, 07:11 PM
P: 60
nope, damn, i hope you get the idea

HallsofIvy is offline
Mar23-04, 10:49 AM
Sci Advisor
PF Gold
P: 38,898

circles in a box

I take it, then, that the container is too small for the balls to sit next to one another on the bottom but too large to hold them one above the other. Let's assume that the bottom sphere is pressed against the left wall and that the upper sphere is pressed against the right wall.

There are three forces acting on the upper sphere. First, the force of gravity, <0,-75> (vector form: horizont, vertical components). Second, the wall pressing against the sphere: <-a,0> (we don't yet know how hard it is pressing). Third the lower sphere pressing against the upper sphere. That, of course, occurs where the two spheres are tangent and so along the line between their centers: that is 35 degrees to the horizontal. Taking its magnitude to be "b", that will be <b cos(35), b sin(35)> (Be careful about the signs: the force of the lower ball on the upper is up and to the right so both components are positive. Of course, the "cos" and "sin" are from the right triangle formed.) The total force on the upper sphere is <0-a + b cos(35),-75+ 0+ b sin(35)> and, since the upper sphere is not moving, that must be 0: a= b cos(35) and
b sin(35)= 75.
That's surprising: that's enough to calculate a and b right there!
b= 75/sin(35)= 131 Newtons. And then a= 131 cos(35)= 107 Newtons.

Now let's look at the bottom sphere. It has four forces acting on it. First the force of gravity: <0, -75>. Second the force of the wall: <c,0> (it's fairly easy to see that c= a but let's not assume that). Third the force of the upper sphere pressing down on it: <-b cos(35),-b sin(35)>- exactly the force the lower sphere applies to the upper but oppositely directed, of course. Finally, there is the force of the bottom supporting the lower sphere: <0,d>.
The total force on the lower sphere is <c- b cos(35), -75- b sin(36)+ d> and, since the lower sphere is not moving, that must be 0. We must have c= b cos(35) and d= 75+ b sin(35). The "b" we already know: it is 131 Newtons.

We can see that, in fact, that c= b cos(35) is exactly the same as the previous a= b cos(35)- since the only horizontal force is coming from the two sides, those must be the same in equilibrium. We also know that b sin(35)= 75 (the weight of the upper sphere pressing on the bottom one) so c= 75+ b sin(35)= 75+75= 150 Newtons. That's obvious isn't it? The bottom sphere is pressing down on the base with the weight of both spheres.

Making sure we answer all questions: the container bottom is pushing up on the spheres with 150 Newtons, exactly the weight of the two. Each side is pressing in against a sphere with a= c= 107 Newtons and the two spheres are pressing against one another with b= 131 Newtons.

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