Force Acting on a System of connected masses

[Feodalherren]

Homework Statement

Find a3. Assume that the system is placed on a frictionless table.

Assumptions: 45 degree angles and rigid connectors between masses.

Homework Equations

The Attempt at a Solution

So my first thought was just to sum all the forces on mass 1, find the reaction along 1-3 and use that to find a3, is that the correct approach? I.e. the reaction in beam 1-3 divided by the total mass of the system is the solution.

Sum of the forces in x:

mAx = F1 - R13 - R12 Sin45
mAy=R13cos45

were R12 is the reaction along beam 1-2 and R13 is the reaction along beam 1-3. But then I'm stuck, and I'm also not sure I'm approaching this correctly, perhaps I should use T=Iα somehow?

 

[Orodruin]

So my first thought was just to sum all the forces on mass 1, find the reaction along 1-3 and use that to find a3, is that the correct approach?

No. Mass 3 will also have forces acting on it from the other connectors.

Hint: The fact that mass 3 is the centre of mass of the construction makes this problem particularly simple.

 

[BvU]

Very nice exercise !

I.e. the reaction in beam 1-3 divided by the total mass of the system is the solution

I don't think so, and you too, considering your follow-up where you do use R12.

use T=Iα somehow

Good plan: the structure will accelerate to the right and will also undergo counter-clockwise angular acceleration. Where will the axis of the rotation be ?

 

[Feodalherren]

The axis of rotation is on mass 3.

 

[Orodruin]

Good plan: the structure will accelerate to the right and will also undergo counter-clockwise angular acceleration. Where will the axis of the rotation be ?

Hint 2: Rotation is irrelevant.

Also, the axis of rotation depends on the current velocity and angular velocity of the structure.

 

[Feodalherren]

Hmm... I don't know then. I can't use conservation of angular momentum since there's a force acting on the object... I'm lost, I need more help.

 

[Orodruin]

Can you give an expression for the total linear momentum of the structure? What is the time derivative of this momentum?

 

[Feodalherren]

Well the total linear momentum is just the total mass multiplied by the velocity of the COM but I don't know the velocity. The time derivative of the linear momentum is F=ma?

 

[Orodruin]

Well the total linear momentum is just the total mass multiplied by the velocity of the COM

Indeed.

I don't know the velocity.

Do you need to? You are not asked for the velocity of the mass 3.

The time derivative of the linear momentum is F=ma?

What does Newton 3 say?

 

[Feodalherren]

Indeed.Do you need to? You are not asked for the velocity of the mass 3.

That's why I didn't consider using linear momentum hmm...

What does Newton 3 say?

For every action there's an equal but opposite reaction.

I really don't see at all where I'm going with this, I'm scratching my head as I'm writing this.

 

[jbriggs444]

The time derivative of the linear momentum is F=ma?

The time derivative of the linear momentum can be determined using the formula F=ma, yes.

$\frac{dp}{dt}=\frac{d(mv)}{dt}=m\frac{dv}{dt}=ma=F$

What is the time derivative of the velocity of the center of mass?

 

[Feodalherren]

Same thing... The derivative of the velocity is the acceleration, the only thing that is different is the mass, which is just five times m0. I still don't get any Y-component for the acceleration.

 

[jbriggs444]

Same thing... The derivative of the velocity is the acceleration, the only thing that is different is the mass, which is just five times m0. I still don't get any Y-component for the acceleration.

What do you mean "same thing"? What is the first derivative of the velocity of the center of mass? Answer the question.

Why would you expect a y component for the acceleration?

Are you picturing the force as an invisible hand moving rightward and pushing on the structure? And then picturing the structure rolling up and off as the hand continues in a purely horizontal path? That would be an incorrect mental picture. Better to picture a smooth wall pushing at the indicated point without resisting any resulting upward or downward deflection. [Or take the formalist approach, close your eyes and see where the math takes you].

 

[Feodalherren]

What do you mean "same thing"? What is the first derivative of the velocity of the center of mass? Answer the question.

dVcom/dt=a3 ... That's what I mean with "same thing".

Why would you expect a y component for the acceleration?

Because of the way that the vector is drawn, it has a slight y-component to it.

Otherwise a3=F1/5*m0

 

[jbriggs444]

dVcom/dt=a3 ... That's what I mean with "same thing".

It would be good if you would have filled in your reasoning for that. Something along the lines of:

Point 3 is located at the center of mass of the structure. Accordingly, the velocity, position and acceleration of the center of mass are the same as the velocity, position and acceleration of point 3. The two are interchangeable.

By definition, acceleration is the first derivative of velocity. So the first derivative of the velocity of the center of mass is equal to the acceleration of point 3. That is to say that it is equal to "a3".​

That is all true and correct. But is not the answer I was hoping for. What is the acceleration of the center of mass in terms of the given's of the problem? What does F=ma tell you?

Because of the way that the vector is drawn, it has a slight y-component to it.

Yes, it does. Can you tell us by pure force of reason whether that detail in the drawing is correct or incorrect?

Otherwise a3=F1/5*m0

[Repairing formatting, that's $a_3=\frac{F_1}{5m_0}$. Written as you have it, conventional expression evaluation rules would multiply by $m_0$ rather than dividing]

Indeed. What is wrong with that answer?

 

[Feodalherren]

It would be good if you would have filled in your reasoning for that. Something along the lines of:

Point 3 is located at the center of mass of the structure. Accordingly, the velocity, position and acceleration of the center of mass are the same as the velocity, position and acceleration of point 3. The two are interchangeable.

By definition, acceleration is the first derivative of velocity. So the first derivative of the velocity of the center of mass is equal to the acceleration of point 3. That is to say that it is equal to "a3".​

That is all true and correct. But is not the answer I was hoping for. What is the acceleration of the center of mass in terms of the given's of the problem? What does F=ma tell you?

That's what I'm trying to find out :).
Force equals mass times acceleration...? Not really sure what you're going for here.

Yes, it does. Can you tell us by pure force of reason whether that detail in the drawing is correct or incorrect?

Well my first line of reasoning is that there can't be any acceleration in Y since there are no forces acting on the system from the Y direction. However, I assumed that I was missing something since the vector is clearly drawn with an y-component.

[Repairing formatting, that's $a_3=\frac{F_1}{5m_0}$. Written as you have it, conventional expression evaluation rules would multiply by $m_0$ rather than dividing]

Indeed. What is wrong with that answer?

I have no idea. I'm literately out of ideas now. This problem looks trivial but I'm missing something. There is no friction and rotation is irrelevant so... It should be easy but I can't figure out why the acceleration of the COM isn't $a_3=\frac{F_1}{5m_0}$.

 

[jbriggs444]

That's what I'm trying to find out :).
Force equals mass times acceleration...? Not really sure what you're going for here.

You are trying to find acceleration. You know force. You know the mass. You have the formula F=ma. It is just that simple.

Well my first line of reasoning is that there can't be any acceleration in Y since there are no forces acting on the system from the Y direction. However, I assumed that I was missing something since the vector is clearly drawn with an y-component.

Yes, it is drawn that way.

I have no idea. I'm literately out of ideas now. This problem looks trivial but I'm missing something. There is no friction and rotation is irrelevant so... It should be easy but I can't figure out why the acceleration of the COM isn't $a_3=\frac{F_1}{5m_0}$.

Sometimes drawings are wrong. Sometimes drawings are wrong on purpose. In this case, the drawing is wrong. Your reasoning that there cannot be a Y acceleration without a Y force was spot on.

 

[Orodruin]

For every action there's an equal but opposite reaction.

Mistyped. Meant Newton 2 ...

It should be easy but I can't figure out why the acceleration of the COM isn't $a_3=\frac{F_1}{5m_0}$.

Isn't it?

 

[Feodalherren]

So wait, now I'm confused, the answer is or isn't $a_3=\frac{F_1}{5m_0}$ ..? Can it really be that trivial? I mean that was my initial answer but it just felt way too easy.

 

[Orodruin]

So wait, now I'm confused, the answer is or isn't $a_3=\frac{F_1}{5m_0}$ ..? Can it really be that trivial? I mean that was my initial answer but it just felt way too easy.

Is. Yes.

You can also see it by considering the force-torque couple acting on the structure. Translating it to the CoM, the force does not change, but the torque does. The torque is irrelevant for the acceleration of the CoM.

 

[Feodalherren]

Is. Yes.

You can also see it by considering the force-torque couple acting on the structure. Translating it to the CoM, the force does not change, but the torque does. The torque is irrelevant for the acceleration of the CoM.

Man... I really need to stop overcomplicating things. I always end up over analyzing everything and then I crash and burn. Nothing can ever be "that simple" in my mind.

Are you picturing the force as an invisible hand moving rightward and pushing on the structure? And then picturing the structure rolling up and off as the hand continues in a purely horizontal path? That would be an incorrect mental picture. Better to picture a smooth wall pushing at the indicated point without resisting any resulting upward or downward deflection. [Or take the formalist approach, close your eyes and see where the math takes you].

The first case is exactly what I was picturing. I even tried it on my desk but I wasn't sure what to think of the result as there's friction and I'm sure I applied some force in the Y direction too but the structure did move in the Y direction too.

 

[Orodruin]

The first case is exactly what I was picturing. I even tried it on my desk but I wasn't sure what to think of the result as there's friction and I'm sure I applied some force in the Y direction too but the structure did move in the Y direction too.

In the case with friction, things will be much more complicated and generally you will end up with the friction giving you a significant y-component. The best place to try this out IRL would be to go to an arcade with some air-hockey games ... (put a square piece of carton on the puck if you are worried about applying the force along a line not going through the CoM)

 

[Feodalherren]

In the case with friction, things will be much more complicated and generally you will end up with the friction giving you a significant y-component. The best place to try this out IRL would be to go to an arcade with some air-hockey games ... (put a square piece of carton on the puck if you are worried about applying the force along a line not going through the CoM)

I'd love to try that but I haven't seen an arcade hall for over a decade :). It's interesting that friction creates a reaction in the y-direction though, any quick and dirty way of explaining why that is?

 

[Orodruin]

I'd love to try that but I haven't seen an arcade hall for over a decade :). It's interesting that friction creates a reaction in the y-direction though, any quick and dirty way of explaining why that is?

It would depend a lot on how you actually hit the object and its shape and construction. Consider the simplest case where you have an object consisting of two equal masses joined with a rigid massless rod and you hit one of the masses in a direction perpendicular to the rod. The other mass will then start its acceleration in the rod direction and the friction at that point will tend to oppose that motion, leading to a net force on the object in the y-direction.

You can do a rather simple experiment at home if you have a fairly slippery and flat floor and using a normal pen. Put the pen down and hit it (hard) with one finger in a direction perpendicular to the pen at one of its ends. Watch the direction the pen goes and enjoy physics. There might be some small corrections due to friction, but you will get the general direction right.

 

[jbriggs444]

You can do a rather simple experiment at home if you have a fairly slippery and flat floor and using a normal pen. Put the pen down and hit it (hard) with one finger in a direction perpendicular to the pen at one of its ends. Watch the direction the pen goes and enjoy physics. There might be some small corrections due to friction, but you will get the general direction right.

I just attempted this with a small, glossy cardboard box (about 3 inches by 4) that happened to be sitting on my desk. Whacking it on the right-hand corner with the end of handy pen, perpendicular to the facing side, I watched the direction it travelled. Measurement was not accurate, but there was no observable leftward deflection. A second trial was consistent and appeared fairly accurate.

Leftward deflection, had it occurred, would have been due to the leftward friction from pen on box as the contacted box surface rotated rightward during the brief collision.

One can also do this on a pool table, applying "english" to a cue ball. Properly done, the ball will travel parallel to the (well-chalked) cue, rather than skittering off to one side.

 

[Orodruin]

Whacking it on the right-hand corner with the end of handy pen,

Physics is fun! Not only do we get to understand the universe. We also get to whack things around!

 

[Feodalherren]

I took a Qtip and attached two small rollers on it, worked pretty nicely. Thanks for the help gentlemen!