How Do You Calculate the Force Between Two Masses?

[JoeyBob]

Homework Statement

See attached

Relevant Equations

Fnet=ma, F 1 on 2 = -F2 on 1

Answer is 7.00.

I don't really know where to start on this one. Because there is no friction, shouldn't there be no force pushing against the first mass? If I assume the next mass has a force that acts against the first, I'm not sure how to find it.

 

[Chestermiller]

Let's see your three free body diagrams (one for each mass), and, based on these free body diagrams, what are the corresponding force balances (one for each mass).

 

[JoeyBob]

Let's see your three free body diagrams (one for each mass), and, based on these free body diagrams, what are the corresponding force balances (one for each mass).

Well you can have Fnet1=Applied Force-Force of mass 2 on mass 1. Then Fnet2=Force mass 1 on mass 2-Force mass 3 on mass 2. And finally Fnet3=Force mass 2 on mass 3.

But none of these net forces will be the same, because although they have the same acceleration, they have different masses.

 

[berkeman]

Well you can have Fnet1=Applied Force-Force of mass 2 on mass 1. Then Fnet2=Force mass 1 on mass 2-Force mass 3 on mass 2. And finally Fnet3=Force mass 2 on mass 3.

But none of these net forces will be the same, because although they have the same acceleration, they have different masses.

Sure. Please post your 3 FBDs and 3 equations. Thanks.

 

[Chestermiller]

Well you can have Fnet1=Applied Force-Force of mass 2 on mass 1. Then Fnet2=Force mass 1 on mass 2-Force mass 3 on mass 2. And finally Fnet3=Force mass 2 on mass 3.

But none of these net forces will be the same, because although they have the same acceleration, they have different masses.

Well, you have the right idea, but those really aren't equations. I'm going to take the liberty of writing out the equations.

$$F_0-F_{12}=m_1a$$
$$F_{12}-F_{23}=m_2a$$
$$F_{23}=m_3a$$

OK so far?

 

[JoeyBob]

Well, you have the right idea, but those really aren't equations. I'm going to take the liberty of writing out the equations.

$$F_0-F_{12}=m_1a$$
$$F_{12}-F_{23}=m_2a$$
$$F_{23}=m_3a$$

OK so far?

Kind of. Shouldnt it be

$$F_0-F_{21}=m_1a$$ instead of

$$F_0-F_{12}=m_1a$$ ? ect.

Because the force of mass two is pushing against the applied force. It doesn't make sense to me that the force mass 1 has on 2 is pushing against the applied force.

Or is this an application of Newtons third law?

 

[Chestermiller]

Kind of. Shouldnt it be

$$F_0-F_{21}=m_1a$$ instead of

$$F_0-F_{12}=m_1a$$ ? ect.

Because the force of mass two is pushing against the applied force. It doesn't make sense to me that the force mass 1 has on 2 is pushing against the applied force.

Or is this an application of Newtons third law?

I’m calling it the magnitude of the force and applying the third law.

 

[JoeyBob]

I’m calling it the magnitude of the force and applying the third law.

Okay so it would be

$$F_0-F_{21}=m_1a$$

but wouldn't it change to

$$F_0+F_{12}=m_1a$$ using the third law because F(21)=-F(12)?

 

[Chestermiller]

Okay so it would be

$$F_0-F_{21}=m_1a$$

but wouldn't it change to

$$F_0+F_{12}=m_1a$$ using the third law because F(21)=-F(12)?

Let |F| be the magnitude of the contact force between the masses m1 and m2. Then the force exerted by m1 on m2 is $|F|\mathbf{i}$ and the force exerted by m2 on m1 is $F(-\mathbf{i})=-|F|\mathbf{i}$. So the force balance on m1 is $$F_0-|F|=m_1a$$And the force balance on m2 is $$|F|-|G|=m_2a$$where |G| is the magnitude of the contact force between the masses m2 and m3.

 

[JoeyBob]

Let |F| be the magnitude of the contact force between the masses m1 and m2. Then the force exerted by m1 on m2 is $|F|\mathbf{i}$ and the force exerted by m2 on m1 is $F(-\mathbf{i})=-|F|\mathbf{i}$. So the force balance on m1 is $$F_0-|F|=m_1a$$And the force balance on m2 is $$|F|-|G|=m_2a$$where |G| is the magnitude of the contact force between the masses m2 and m3.

Ok I think I get it. Then you just solve for F(12), which you can do by finding an equation for acceleration and then substituting a bunch.

 

[Chestermiller]

Ok I think I get it. Then you just solve for F(12), which you can do by finding an equation for acceleration and then substituting a bunch.

Add the three equations together and see what you get.

 

[JoeyBob]

Add the three equations together and see what you get.

Yeah a bunch of stuff cancels. and you can get the acceleration. Thanks for all the help.