# Two identical, uniform, frictionless spheres

1. Apr 12, 2010

### juggalomike

1. The problem statement, all variables and given/known data

Two identical, uniform, frictionless spheres, each of weight W , rest in a rigid rectangular container as shown in the figure. Find, in terms of a ratio to W, the forces acting on the spheres due to the container surfaces and one another, if the line of centers of the spheres makes an angle of 45° with the horizontal. Normal force from the left side of the container?

[PLAIN]http://img260.imageshack.us/img260/7907/1368b.jpg [Broken]

2. Relevant equations

newtons 3rd law(not realy an equation)

3. The attempt at a solution

Looking at the picture i believe the force from the left side of the container will be equal to the force in the x direction applied to ball 2 from ball 1, so i set the force to be W*cos45, however that is not correct.

Last edited by a moderator: May 4, 2017
2. Apr 12, 2010

### collinsmark

Hello juggalomike,

What makes you say that the force in the x direction applied to ball 2 from ball 1 is Wcos(45o)?

3. Apr 12, 2010

### juggalomike

I was originaly thinking of the ball is touching the other ball at a 45 degree angel then the components would be wcos45 and wsin45, however thinking about it the wall is also part of the equation, however i have no idea how to include that.

4. Apr 13, 2010

### collinsmark

Okay, let's think about that. And let's ignore the wall just for the moment -- one can come back to it later (and yes, the wall does fit into things. But you can come back to that later anyway).

Let's break up the problem into its x and y components. I'll concentrate only on the y component for now. Gravity exerts a force on the top ball with a magnitude of W in the down direction. Since nothing is accelerating, something else must be acting on that ball with a force of W in the up direction. What is exerting that force? And if that force in the up direction is only the y component of some other total force, what is the magnitude of that total force?

A good way to solve problems like this is to start by drawing a "free body diagram." Nothing is moving, so you know all the forces involved in the y direction must cancel each other (everything is in a state of equilibrium). The same is true for all the forces in the x direction. Once you figure out the y forces involved, you should be able to use your free body diagram to find the x forces (the x forces involve the balls and the walls too).

5. Apr 13, 2010

### juggalomike

alright i did as you said and for each ball here are the forces i calculated

ball1
x component----Fb2*cos(45)=Fw
y component----Fb2*sin(45)=W

ball2
x component----Fb1*sin(45)=Fw
y component----Fb1*cost45+W=Fn

Which would still meen the force from the left wall is W*sin(45). What am i doing wrong?

6. Apr 13, 2010

### collinsmark

Okay, let's just start with ball 1 for now. You can come back to ball 2 later.

You have stated,

y component----Fb2*sin(45)=W

Now solve for Fb2. (Note that if a x b = c, then ab x c. Instead the correct relationship is, a = c/b).

So what do you have for Fb2? Now find

x component----Fb2*cos(45)=Fw

7. Apr 13, 2010

### juggalomike

y component----Fb2=W/sin(45)
x component----Fb2=W/cos(45)

W/sin(45)=Fb2=W/cos(45)?

8. Apr 13, 2010

### collinsmark

Well, the left side the equation is correct (the y component).

W/sin(45)=Fb2

But the x component deals with the force from the wall, not W (at least not directly). So now use the equation you already derived to show the force from the wall, in terms of Fb2.

Fb2cos(45)=Fw,

where Fw is the force from the wall. Using the two equations, solve for Fw in terms of W.

9. Apr 13, 2010

### juggalomike

Thank you so much for the help, this was actualy an 8 part question(i only posted the 1st part). After ur last post i was able to figure out the other 7 parts within a few minutes.