Acceleration & tension in a cable

[Robb]

Homework Statement

Homework Equations

F=ma

The Attempt at a Solution

Does the total Force F=50N divide evenly between m1+m2 and m3? Meaning the tension between m1/m2 & m3 is 25N given m1+m2=m3. Also, is the acceleration between m1/m2 & m3 the same as the acceleration at m3?

a=50/.098=510.2

a=25/.049=510.2

 

[Chestermiller]

Have you tried drawing free body diagrams, or do you feel like you have advanced beyond the point where you need to draw free body diagrams?

Chet

 

[Robb]

Not quite yet. And yes I have. Thought the answer might be pretty straight forward though? No?

 

[Chestermiller]

Not quite yet. And yes I have. Thought the answer might be pretty straight forward though? No?

Apparently not. Please show us your force balances on the three masses.

If the length of the cable doesn't change, how do the displacements, velocities, and accelerations of the three masses compare?

chet

 

[Robb]

Guess I'm not sure about displacement and velocity.

Robb

 

[Chestermiller]

View attachment 90064

Guess I'm not sure about displacement and velocity.

Robb

I don't see any force balance equations on any of the masses. Let T be the tension on the cable, and let f be the contact force between masses 1 and 2. Let's see some force balance equations (in the horizontal direction). One equation for each mass.

If the cable doesn't get any longer, how would it be possible for the velocities and accelerations of the three masses to be anything but the same? They're "joined at the hip."

Chet

 

[Mister T]

a=50/.098=510.2

a=25/.049=510.2

What do the 0.098 and 0.049 represent?

 

[Mister T]

View attachment 90064

You say that 3 kg = 0.029 N,
2 kg = 0.020 N,
5 kg = 0.049 N.

This doesn't make sense to me. What are you doing to come up with this?

 

[Robb]

Sorry for the bad conversion! Anyway, we have learned zero about contact force (engineering physics 1) so not sure about that. .098 is m1+m2+m3 and .049 is m1 + m2.

 

[Chestermiller]

I want a separate free body diagram for each of the masses, and a separate force balance equation for each of the masses. Didn't they teach you to do this in your course?

Chet

 

[Robb]

Is this more like it?

 

[Chestermiller]

No. That's only slightly like what I had in mind. Here is what I really had in mind:

Please tell me if it makes sense. If not, please pose questions. Also, now please solve these equations for a, T, and f.

Chet

 

[Mister T]

Anyway, we have learned zero about contact force (engineering physics 1) so not sure about that.

You've learned about the normal force? That's a contact force. The table pushes up on the block, the block pushes down on the table.

Block 1 pushes rightward on Block 2, Block 2 pushes leftward on Block 1. Those are contact forces.

.098 is m1+m2+m3 and .049 is m1 + m2.

Huhh? ##3+2+5=10##.

By the way, multiplying a mass in kilograms by 9.8 m/s² to get a force in Newtons is not a conversion. Also, it's not good to get in the habit of writing things like 3 kg = 29.4 N.

 

[Robb]

Thanks, Chet. Yes it makes sense. I'm not sure how to find T, though. Does this make sense for m3? I tried a couple of different ideas for solving for a and T. Obviously very different results.

I'm not believing T is .02N. This is like the example we have from class except we weren't finding T.

 

[Chestermiller]

None of this makes any sense to me. We are not dealing with the force balances in the y direction at all. We are only looking at the force balances in the x direction. So I have no idea why you are using angles and y-components.

The solution to this problem involves taking the three linear algebraic equations in post #12 and solving for the three unknowns, a, T, and f. This is a 9th grade algebra problem. The easiest thing to do is to add the three equations together; this eliminates T and f immediately, so that you are left with an equation for a. What is the algebraic solution to this equation for a (in terms of m1, m2, m3, and F)?

Chet

 

[Robb]

Sorry, Chet. Believe it or not I'm a pretty good math student I've just never had physics before and I feel like I have to figure a lot of this out on my own. Anyway, given that T is eliminated does that suggest that there is no tension on the connecting cable?

 

[Mister T]

Does the total Force F=50N divide evenly between m1+m2 and m3? Meaning the tension between m1/m2 & m3 is 25N given m1+m2=m3.

Yes, but you will to need understand things in more depth if you want to be able to predict what will happen when you have a less simple relationship between m₁, m₂, and m₃.

Many times in math classes the focus is on answer-making. But in a physics class that's just one layer. The other layer is sense-making.

In this case, Blocks 1 and 2 are being treated like a single object, making the contact forces they exert on each other internal forces.

 

[Chestermiller]

Sorry, Chet. Believe it or not I'm a pretty good math student I've just never had physics before and I feel like I have to figure a lot of this out on my own. Anyway, given that T is eliminated does that suggest that there is no tension on the connecting cable?

If you are good at math, you should be able to add my 3 equation together. What do you get? No, the.tension is not zero

 

[Robb]

I think I understand that. Basically they null each other mathematically. So if the acceleration is constant then the acceleration at m1/m2= acceleration at m3?

 

[Robb]

adding them and solving for a I get a=F/(m1+m2+m3)= 50/98= .51m/s

m1a=T-f
m2a=f

T=a(m1+m2)=(.51)(49)=25N

 

[Chestermiller]

adding them and solving for a I get a=F/(m1+m2+m3)

The part above is correct.

= 50/98= .51m/s

This part is incorrect. m1= 3, m2 = 2, m3= 5; (3+2+5)=10, not 98. Where the heck did the 98 come from? 50/10 = 5 m/s²

m1a=T-f
m2a=f

T=a(m1+m2)

The part above is correct.

=(.51)(49)=25N

This part is incorrect.

m1=3, m2 = 2
=5(3+2)=25

 

[Robb]

I converted the masses to Newtons. Obviously I didn't need to do that for acceleration. Thanks for the help. Much appreciated. And GO BLUE! Beat state!