# Homework Help: Forces question. please take me through this

1. Oct 31, 2009

### holezch

forces question....... please take me through this

1. The problem statement, all variables and given/known data
two blocks are in contact on a frictionless table. A horizontal force is applied to one block. If m1 = 2.0 kg and m2 = 1.0 kg, and F = 3.0 N, find the force of contact between the blocks

2. Relevant equations

f = ma?

3. The attempt at a solution

I thought that when the force pushes one block against another,t he other block will push back with a force of 3 newtons, then I don't know what to say

thank you!.

2. Oct 31, 2009

### rl.bhat

Re: forces question....... please take me through this

Both the blocks move with same acceleration.
The acceleration a = Applied force/ Total mass
To have the same acceleration in m2, what should be the contact force?

3. Oct 31, 2009

### holezch

Re: forces question....... please take me through this

hi, thank you so much for the reply.. why do we suppose that the accelerations are the same? I thought only the force would be the same? doesn't acceleration vary between objects of different masses given the same force for all of them? like m2/m1 = a1/a2? thank you again

4. Nov 1, 2009

### traffic_boy

Re: forces question....... please take me through this

hi ,, about the acceleration it should be the same because both the two masses can be
considered as one system or it can be said as one mass..but the question that came to my mind is that is it correct to say that the forces exerted by the two masses are the same..
i have the same problem and am still trying to find the explanation....any help is ver appreciated...

5. Nov 1, 2009

### holezch

Re: forces question....... please take me through this

if block#1, 2.0kg, pushes the block#2, 1.0 k.g. with a force of 3.0 N, block#2 will accelerate more when pushed by block#1, and block#1 will accelerate less when pushed by block#1 with the same amount of force...?

6. Nov 1, 2009

### holezch

Re: forces question....... please take me through this

how do I know when I'm supposed to factor in the force exerted by gravity? thank you

7. Nov 1, 2009

### rl.bhat

Re: forces question....... please take me through this

The forces appear in pairs, action and reaction.
If you apply a force F on the masses, you will experience an equal and opposite force.
Since masses are moving in the forward direction, m1 pushes the mass m2 in the forward direction and m2 pushes m1 in the reverse direction.. Net force on m1 is ( F- R), where R is the reaction force. If a is the acceleration of the system, (F - R) = m1*a.
Find R. R is equal to force on m2 by m1, or contact force.

8. Nov 1, 2009

### traffic_boy

Re: forces question....... please take me through this

well , i agree with you and i have used this way before but with a small change..i found (a) and then i used this formula : f2 = (m1+ m2) a - m1 (a). but the final result is not consistent with the answer of the book...then it's not my fault if this is the right way,,

9. Nov 1, 2009

### holezch

Re: forces question....... please take me through this

hello, wouldn't the reaction force just be the same magnitude but reverse direction? :S I'm confused.. or does m2 not get pushed by m1 with a force of 3.0 because its weight will change the magnitude of the applied forcE?
other than gravity, what else would make m1's force on m2's anything different from 3 Newtons? and wouldn't the acceleration of m1 be 3/2?

Last edited: Nov 1, 2009
10. Nov 1, 2009

### holezch

Re: forces question....... please take me through this

I got the answer by (F-R) = m1*a a = force/ total mass so 3-R = 2 R=1.. I dnt get it though, why would I consider the mass of m2 in force/m1+m2 ? so (F-R) is supposed to be the net force after both the applied force and reaction force is considered.. so acceleration would still be centered around m1 wouldn't it.. how come m1+m2?

11. Nov 1, 2009

### ApexOfDE

Re: forces question....... please take me through this

(F-R) is not net force but a force applied in m1.

When you apply a force F in one of two blocks, they will move. At that time, m1 will affect a force R in m2. Base on Newton's 3rd law, m2 will affect the same force but in opposite direction R in m1.

From this, you would have:
(m1): F + R1 = m1a (R1's sign is negative)
(m2): R2 = m2a

Moreover, in this case, 2 blocks can be considered one block with mass = m1 + m2 => F = ma

12. Nov 1, 2009

### holezch

Re: forces question....... please take me through this

thank you for the reply, if I am calculating what happens to m1, how come I need to have my mass m1+m2? and then if the mass is m1+m2, how come the mass in the equation is just m1? F-R = m1*a , a = F/m1+m2?

13. Nov 1, 2009

### holezch

Re: forces question....... please take me through this

I hate this question, how will I ever know if "contact force" means the force on M1 from M2 or the force on M2 from M1?? its so ambiguous.. how do I find acceleration?? ahhh

14. Nov 1, 2009

### rl.bhat

Re: forces question....... please take me through this

In tug of war who is exerting force on whom?
If you press two blocks why don't they stick together?
If you apply a force of 3.0 N on a block of mass 3 kg, what will be its acceleration?
If you apply a force of 3.0 N on two blocks of 2 kg and 1 kg kept in contact with each other, what will be the acceleration of each mass?

15. Nov 1, 2009

### holezch

Re: forces question....... please take me through this

in a tug of war, both sides are being exerted on by the rope, since they pull away and by newton's 3rd law, the rope pushes back

if I apply a force 3.0 N on a mass of 3, the acceleration is 1
?
well, I guess the acceleration of m2 in the same direction as m1 would be the same.. but aren't we supposed to find the happens when m2 pushes back on m1? I guess that's what "force of contact" means?
thank you

16. Nov 1, 2009

### holezch

Re: forces question....... please take me through this

okay , so R = m2a.. a is the same as the a from (F=3.0) = 2.0a.. so the acceleration in R is 3/2?

17. Nov 1, 2009

### rl.bhat

Re: forces question....... please take me through this

All the 3 N force is not utilized to accelerate 2 kg mass. Part of it is transferred to 1 kg mass due to contact force.
Atoms in the blocks repel each other when they try to come closer. This is the contact force. It is equal and opposite.

18. Nov 1, 2009

### holezch

Re: forces question....... please take me through this

this is what I understand up till now..

3.0 is applied to m1 which pushes m2 as a consequence.. to find the contact force, I need to find the force of m2 pushing back on m1, which is equal to the force of m1 pushing onto m2.

I have no idea why m1 pushing onto m2 is not equal to the force 3 N applied to m1, but I think that this transferring thing you are talking about will be my answer? why is it necessarily true that 3 N is "transferred" about m1 and m2? couldn't it be said that 3 N is totally applied to move m1 and m2 just moves as a side-effect? even if 3 N is "transferred" around m1 and m2, how can I calculate the force of m1 and m2?

is the contact force the part of 3N that has been distributed to m2?

Last edited: Nov 1, 2009
19. Nov 1, 2009

### holezch

Re: forces question....... please take me through this

if the contact force is the part of 3N that has been distributed to m2...

F(force on m1) + F2(force on m2) = 3.0

F = 2a
F2 = a

2a+a = 3a = 3 , a = 1 then F2 = 1 N... I'm not sure if this is right because I don't even understand the whole transferring thing..

20. Nov 1, 2009

### holezch

Re: forces question....... please take me through this

okay, I think that the "transferring" thing works, because m1 doesn't move with a force of 3 N, since there is another force from m2 that "pushes back" on m1, retarding its movement
so since the repelling force from m2 is the same magnitude as the force making m2 go forward, F + F2 should equal to 3. Right?