Two identical, uniform, frictionless spheres

In summary: Fb2*sin(45)=WWhich would still meen the force from the left wall is W*sin(45). What am i doing wrong?In summary, the force acting on the spheres due to the container surfaces and one another is Fw*sin(45), where W is the weight of the spheres.
  • #1
juggalomike
51
0

Homework Statement



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

Homework Equations



Newtons 3rd law(not really an equation)

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:
Physics news on Phys.org
  • #2
Hello juggalomike,

What makes you say that the force in the x direction applied to ball 2 from ball 1 is Wcos(45o)?
 
  • #3
collinsmark said:
Hello juggalomike,

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

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
juggalomike said:
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.

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
collinsmark said:
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).

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
juggalomike said:
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

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

using your result for Fb2.
 
  • #7
collinsmark said:
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

using your result for Fb2.

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

W/sin(45)=Fb2=W/cos(45)?
 
  • #8
juggalomike said:
y component----Fb2=W/sin(45)
x component----Fb2=W/cos(45)

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

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
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.
 

1. What is the definition of "identical" in this context?

In this context, "identical" means that the two spheres have the same shape, size, and mass.

2. What does it mean for the spheres to be "uniform"?

When we say the spheres are "uniform", it means that the mass is evenly distributed throughout the entire surface of the sphere. In other words, there are no areas of varying density within the sphere.

3. Can you explain what a "frictionless" sphere is?

A frictionless sphere is one that does not experience any resistance or force when moving against another surface. This means that the spheres will continue to move in a straight line at a constant speed unless acted upon by an outside force.

4. What are the implications of having frictionless spheres?

The implications of having frictionless spheres are that they will behave according to the laws of motion without any external forces affecting their movement. This allows for simpler and more accurate calculations and predictions in physics experiments and models.

5. Are there any real-world examples of two identical, uniform, frictionless spheres?

While it is difficult to find objects that are completely frictionless in the real world, there are some examples that closely resemble the properties of the two identical, uniform, frictionless spheres in theory. One example is a pair of ball bearings rolling against each other on a perfectly smooth surface, where the weight and shape of the balls are identical and the surface has no friction.

Similar threads

  • Introductory Physics Homework Help
Replies
12
Views
2K
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
16
Views
4K
  • Introductory Physics Homework Help
Replies
7
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
349
  • Introductory Physics Homework Help
Replies
19
Views
3K
  • Introductory Physics Homework Help
Replies
5
Views
1K
  • Introductory Physics Homework Help
Replies
19
Views
5K
  • Introductory Physics Homework Help
Replies
8
Views
1K
  • Introductory Physics Homework Help
2
Replies
43
Views
2K
Back
Top