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Force pairs between 5 objects (Newton's 2nd & 3rd laws)

  • Thread starter emr13
  • Start date
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1. The problem statement, all variables and given/known data

A table of mass 20kg is placed on the ground. A box of mass 3kg is placed on the table. Another box of mass 2kg is placed on the table, not in contact with the first box. Another box of mass 1 kg is placed on top of the 3kg box. The earth, table, and boxes are all at rest. Set up a table for the force that each object exerts on each other object. (1) Fill in the table with the correct values of the forces, treating this as a one-dimensional problem with the +y direction up out of the ground, and the -y direction pointing down toward the center of the earth. Use the rough approximate value g=10m/s2 for the acceleration due to gravity.

(There are two more sections, but they are wordy and long, and I think if I get this first part down, they will be much easier, so I'll leave them out for now.)

2. Relevant equations

F=ma

3. The attempt at a solution

I know there are five objects total, giving me a 5x5 table. An object doesn't exert force on itself (right?), which knocks out some boxes. If g=10m/s2, then I think the table exerts a force of -200N on the table (using F=ma). By Newton's 3rd law, the earth exerts 200N on the table, so my table looks like this so far:

View attachment Force exerted.doc

When I calculated the force pair of the earth and the table, should I have used the added mass of the table and the objects (meaning the force of the table on the earth would be -260N)? And I'm not sure how to calculate the force by the earth on the objects - does the table's mass interfere?

Should I have used the y-component force equation (Fy=F*sin([tex]\theta[/tex])) with an angle of 90 degrees (meaning my other numbers are entirely incorrect), or am I headed in the right direction?

I guess I am just wondering if I my calculations are correct so far and where exactly to go next. My professor is very, very vague in his lectures, and writes his own problems, so I find myself a little lost between his lectures and the assignments and my textbooks, because nothing seems to match up, and I'll admit, I don't seem to be that great at physics in general; a push in the right direction would be immensely helpful.

Thank you very much (especially for the time you took to read this massive post haha).
 

tiny-tim

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Hi emr13! :smile:

(have a theta: θ :wink:)
IIf g=10m/s2, then I think the table exerts a force of -200N on the table (using F=ma). By Newton's 3rd law, the earth exerts 200N on the table, …
When I calculated the force pair of the earth and the table, should I have used the added mass of the table and the objects (meaning the force of the table on the earth would be -260N)? And I'm not sure how to calculate the force by the earth on the objects - does the table's mass interfere?
To find the force of the earth on the table, use Ftotal = ma for the table …

since the table isn't going through the floor, and the force from the earth is the only force directed upward

how much does it have to be? :smile:

(and things that are not in touch don't exert contact forces on each other :wink:)
Should I have used the y-component force equation (Fy=F*sin([tex]\theta[/tex])) with an angle of 90 degrees (meaning my other numbers are entirely incorrect), or am I headed in the right direction?
hmm … always use Fcosθ, where θ is the angle between the direction and the force (in case, zero), or you'll get confused.
 
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So is -200N is incorrect for the force of the table on the earth, then? I mean, I know the table isn't moving...so it's acceleration would be zero...but that's for the net force, and I think my professor is only looking for the individual forces of each object on each other object.

And I don't know why I said 90 degrees for the y-component equation...it's been a long day. But I am wondering why you say I would use cosine rather than sine because I figured the forces were all vertical.

(And since I probably sound rather silly and challenged at the moment, I'll add that's it's 3:35am where I am, and sleep deprivation could be hindering my understanding, so maybe what you said will be clearer after I get some sleep...)
 
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I've continued working on the table, and I've come up with this so far:

table.jpg


Also, I found a hint my professor gave. He says, "the earth exerts a gravitational force on all the other objects in the problem. Additionally, the earth also exerts a "normal force" on the table. The total or subtotal force exerted by the earth on the table will be the sum of these two forces. You might want to show these two forces individually in the table."

The normal force in the y-direction is n=mgcosθ, which would be (20)(10)(cos(90)), which is 0N. Why would my professor give the hint if the normal force didn't matter?

Any advice would be appreciated...
 
Last edited:

tiny-tim

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Also, I found a hint my professor gave. He says, "the earth exerts a gravitational force on all the other objects in the problem. Additionally, the earth also exerts a "normal force" on the table. The total or subtotal force exerted by the earth on the table will be the sum of these two forces. You might want to show these two forces individually in the table."
I honestly think your professor is being ridiculously confusing. :frown:

If the gravitational force from the Earth on the table is included, then so should be the gravitational force from the table on the Earth.

Assuming we're taking the standard procedure (in virtually all exam questions and in practice) of using the Earth as our frame of reference (ie the Earth is stationary), that makes no sense.

Anyway, you need an extra row and column in your table, so that you can have one row and column marked "earth (gravitational)" and one marked "earth (contact)".

And (if we include gravitational forces), in each row or column for the table or a block, the entries should add to zero.
The normal force in the y-direction is n=mgcosθ, which would be (20)(10)(cos(90)), which is 0N. Why would my professor give the hint if the normal force didn't matter?
mgcosθ is not the normal force, neither in the y- nor in any other direction.

It is the component of the gravitational force (the weight) on a mass m at an angle θ to the vertical.

If your professor is telling you that mgcosθ is the normal force, then again he's being unhelpful and confusing. :frown:

Are you sure that you're not taking formulas from other parts of his lecture, and trying to apply them to this part?
 
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I honestly think your professor is being ridiculously confusing.
Agreed.

Anyway, you need an extra row and column in your table, so that you can have one row and column marked "earth (gravitational)" and one marked "earth (contact)".
But only the earth and table have contact forces, correct? So the earth(contact) row would be zero for everything but the table...right?

And (if we include gravitational forces), in each row or column for the table or a block, the entries should add to zero.
Do you mean for each object, all the forces totaled should be zero?
mgcosθ is not the normal force, neither in the y- nor in any other direction.

It is the component of the gravitational force (the weight) on a mass m at an angle θ to the vertical.

If your professor is telling you that mgcosθ is the normal force, then again he's being unhelpful and confusing.

Are you sure that you're not taking formulas from other parts of his lecture, and trying to apply them to this part?
He never really gave us any formulas, so I've been using the textbook. In one of the chapter summaries, it says that ny+Fgy=ma is equal to n-mgcosθ=0. So I was manipulating it, trying to find the value of the normal force, and so I was saying n=mgcosθ.
 

tiny-tim

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But only the earth and table have contact forces, correct? So the earth(contact) row would be zero for everything but the table...right?
Yes.
Do you mean for each object, all the forces totaled should be zero?
Yes … isn't that obvious from Newton's second law? If an object is in equilibrium, then all the forces on it (including its own weight) must add to zero.
He never really gave us any formulas, so I've been using the textbook. In one of the chapter summaries, it says that ny+Fgy=ma is equal to n-mgcosθ=0. So I was manipulating it, trying to find the value of the normal force, and so I was saying n=mgcosθ.
This really isn't a sensible way of doing it.

This is a one-dimensional situation … just add the forces.
 
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Yes … isn't that obvious from Newton's second law? If an object is in equilibrium, then all the forces on it (including its own weight) must add to zero.
I was just making sure...so do I need to change the numbers in the table?

This really isn't a sensible way of doing it.

This is a one-dimensional situation … just add the forces.
I'm not sure how to determine the contact force by the earth on the table then...Do I need to find the mass of the earth?

I'm sorry I'm so confused. Thank you for all of your help so far.
 

tiny-tim

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I was just making sure...so do I need to change the numbers in the table?
And add an extra row and column, yes.
I'm not sure how to determine the contact force by the earth on the table then...Do I need to find the mass of the earth?
No, that's part of g (g = GMearth/REarth2).

Use Newton's second law.

I think you've better forget the table for the moment, and write out all the equations first …

give the normal reaction forces names, Ret, Rt3, Rt2, R31

what are the equations (from Newton's second law) relating those forces?
 
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For 1kg: Fg+F3=m1a
For 2kg: Fg+Ftable=m2a
For 3kg: Fg+F(table)+F1=m(3)a
For table: F3+F2+Fg=m(table)a

Are those good? And then all of the reaction forces should add up to zero because a is zero. (sorry about the lack of formatting - I am using
my phone).
 

tiny-tim

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For 1kg: Fg+F3=m1a
For 2kg: Fg+Ftable=m2a
For 3kg: Fg+F(table)+F1=m(3)a
For table: F3+F2+Fg=m(table)a
i] Why do you keep writing Fg for four different things?

ii] You've left out F(earth).

iii] you'll risk getting your signs wrong later if you, for example, call the reaction from m1 to m3 F1 and the opposite reaction F3. :redface:
 

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