Force Exerted by 100g & 50g Masses on Each Other

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In this discussion, participants analyze the forces exerted by two masses (100g and 50g) when an applied force of 1N is exerted on the larger mass. Clarification is sought regarding the configuration of the masses, whether they are stacked or side by side, as this affects the calculations. The acceleration of the system is determined to be 6.7 m/s², and the forces exerted between the masses are discussed in the context of Newton's third law. Participants emphasize the importance of understanding the setup of the problem to accurately calculate the forces involved. Ultimately, the conversation highlights the complexities of force interaction in systems with multiple masses.
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If friction is negligible, and masses of 100g and 50g are pushed across with an applied force of 1N of the greater mass, how do I find the magnitude of the force exerted by each of the two masses on each other?
 
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What do you mean by "applied force of 1N of the greater mass"? That's kinda confusing. I might be able to help you.
 
oops sorry...

I meant applied 1 N of force on the greater mass, that being the 100g mass.
 
OOOHHH!

Ok, so let me see if I've got this...your 100g mass is moving and the 50 g mass is stationary...right?
 
Both masses are in contact with each other when they are being moved. Like the 50g mass is on top of the 100g one.
 
Alright, that makes this a bit easier. What do you have for work so far if anything? Because you can count both of them as one mass since there in contact. That elminates one variable.
 
I've only found that acceleration is 6.7m/s^2
 
Ok...I have notes on a similar problem. Let me see if I can point you in the right direction.
 
Couldn't you use the Fnet=ma formula?

Input the total mass as m and use the 6.7 m/s^2 as your a.
 
  • #10
That's what I thought but it didn't work. I think that 6.7 is the horizontal acceleration, that's why.
 
  • #11
Hmm, this is interesting. It would've been eaiser had they asked you for the coefficient of kinetic friction. lol.

Well, there is no vertical acceleration so that's out. Are you looking for WORK or FORCE?
 
  • #12
Do you have any other information?
 
  • #13
I'm looking for the magnitude of the force exerted by each of the masses on the other. I've mentioned all the info that's given.
 
  • #14
Grr, unfortunately I've never seen a problem like this. All I can think of is Newton's thrid law: the forces of two objects acting on one another or of one object acting on another object, the contact forces will be equal and opposite.

That's as far as my thinking takes me right now. Sorry love.
 
  • #15
Draw the freebody diagram of the top mass. Draw all the forces being exerted on the top mass. There is weight... what else is there?

Then use F=ma in both the horizontal and vertical directions to solve for the unknowns. You know a=0 in the vertical direction, and a=6.7 in the horizontal direction.
 
  • #16
harhar,

Can you please state the problem exactly as it was given to you, all in one post?

- Warren
 
  • #17
How did you come up with the acceleration of 6.7 m/s^2?
 
  • #18
Ok the EXACT question is the same concept but w/ different numbers because I just tried to use easy numbers to deal with..

2 masses in contact w/ each other are 0.113kg and 0.139kg. No friction. The applied horiz. force of magnitude 5.38x10^-2 N is exerted on the heavier mass.

Determine:
a)magnitude of accel. of the 2 mass system.
b)magnitude of the force exerted by each of the 2 masses on the other

I've already gotten a) which is 0.213m/s^2
 
  • #19
I assume there's a picture, also, showing one mass above the other. You need to tell us about those parts of the question when posting your question here, or we won't be able to make sense of it.

The answer to b) is simply the weight of the mass on top. The top mass presses its weight down on the bottom mass, and the bottom mass in turn pushes back up, supporting it.

- Warren
 
  • #20
harhar said:
Ok the EXACT question is the same concept but w/ different numbers because I just tried to use easy numbers to deal with..

2 masses in contact w/ each other are 0.113kg and 0.139kg. No friction. The applied horiz. force of magnitude 5.38x10^-2 N is exerted on the heavier mass.

Determine:
a)magnitude of accel. of the 2 mass system.
b)magnitude of the force exerted by each of the 2 masses on the other

I've already gotten a) which is 0.213m/s^2

Are the masses on top of each other, or just side by side with one pushing the other along? Is that mentioned in the question? That will affect the answer to b). If it is one on top of each other, have a look at my previous post.
 
  • #21
Also, you mentioned "no friction". There is no friction anywhere? Even between the two masses? Or just no friction between the bottom mass and the ground?
 
  • #22
lp,

I assumed he meant nowhere at all.

- Warren
 
  • #23
chroot said:
I assume there's a picture, also, showing one mass above the other. You need to tell us about those parts of the question when posting your question here, or we won't be able to make sense of it.

The answer to b) is simply the weight of the mass on top. The top mass presses its weight down on the bottom mass, and the bottom mass in turn pushes back up, supporting it.

- Warren

I'm pretty sure there can be friction between the two masses.
 
  • #24
"2 masses in contact w/ each other are 0.113kg and 0.139kg. No friction." Sounds like no friction between the bodies to me.

- Warren
 
  • #25
No friction between mass and ground. There's no picture...but anyways if the mass isn't on the top then I assume they're side by side b/c I haven't been able to get the answer. How do I find the force if they are side by side?
 
  • #26
*sigh*

If the two bodies are side-by-side, then pushing on one pushes on the other, too. The force you apply to one body gets applied directly to the other, same magnitude, same direction. Weight is not involved.

- Warren
 
  • #27
learningphysics,

That's very true. I have the feeling the two blocks are actually side-by-side. If I were harhar, I would be yelling at the teacher by now for writing such an incomprehensible problem.

- Warren
 
  • #28
chroot said:
learningphysics,

That's very true. I have the feeling the two blocks are actually side-by-side. If I were harhar, I would be yelling at the teacher by now for writing such an incomprehensible problem.

- Warren

lol. Yes, me too.
 
  • #29
chroot said:
*sigh*

If the two bodies are side-by-side, then pushing on one pushes on the other, too. The force you apply to one body gets applied directly to the other, same magnitude, same direction. Weight is not involved.

- Warren

But Warren, this isn't true. The force acting on the second mass isn't equal to the force being applied to the first.

The accelerations of the two masses are the same.
 
  • #30
Yeah sry, turns out they were side by side...I was overcomplicating it :redface:
Thanks for your help guys.
 
  • #31
harhar said:
Yeah sry, turns out they were side by side...I was overcomplicating it :redface:
Thanks for your help guys.

That's ok. What did you get for part b)?
 
  • #32
learningphysics said:
But Warren, this isn't true. The force acting on the second mass isn't equal to the force being applied to the first.

The accelerations of the two masses are the same.
LOL, yes, of course... I appear to be going nuts.

- Warren
 
  • #33
chroot said:
LOL, yes, of course... I appear to be going nuts.

- Warren

lol. That's ok. I think I went nuts years ago.
 

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