# Calculating Forces Question

1. Nov 26, 2009

### DeathByVirus

1. The problem statement, all variables and given/known data

Two boxes stay beside each other touching on a flat ground.
The ground is frictionless.
________ _______
|~~~~~||~~~~~|
|~1.5KG`||~1.0KG|
|~BoxA~||~BoxB~|
|_______||______|

A) If 5.0N
of force is applied to BoxA what is the acceleration of BoxB

B) During this, what is their contact force?

M1 = 1.5KG
M2 = 1.0KG
Fapp = 5.0N

3. The attempt at a solution

a) so what i did was add the boxes together so total mass is 2.5KG
then the Fnet will equal applied force since no other forces are relevant
so by second law

Fnet = ma
5.0 = 2.5a
a = 2.0 m/s^2

the second box will accelerate at 2.0 m/s^2.

b) since only applied force of 5N is on BoxA
the third law states that for every action there is an equal and opposite reaction
so if the boxA is pushed by 5N that means boxB will back on boxA with 5N
therefore the contact force = applied force = 5.0 N

did i do it right?​

2. Nov 26, 2009

### rl.bhat

Hi DeathByVirus, welcome to PF.
Both the blocks are moving with the same acceleration.
The applied force pushes block A. Block A acts on block B. Block reacts on Block A.
Net contact force pushes the block B towards right. You know the mass of the block B and its acceleration. Find the contact force.

3. Nov 26, 2009

### DeathByVirus

the contact force isn't 5?
its ...
fnet = ma
= 1.5 x 2
= 3 N

or is it..

fnet = ma
5 = 1.5 a
a = 3.33 for box A alone

then the net force box A pushes box B with
fnet = 1.5x3.33
which will round back to

fnet = 5N

4. Nov 26, 2009

### wisvuze

Okay, we know that Fnet is equal to 5, since it is the only force significantly acting on the system. But we also know that if
Fnet = m1a + m2a = (m1+m2)a, then the force effecting m1 cannot be 5, unless m2 has no mass. By Newton's third, we know that m1 is 'held back' by the reaction force from m2. Meaning, the force on m1 should be 5-(reaction force) = m1a; and again, by Newton's third, this force is equal and opposite to the force from m1 to m2.

Last edited: Nov 26, 2009
5. Nov 26, 2009

### DeathByVirus

if a = 2
then fnet(boxA) = m(boxA)a

fnet(boxA) = 1.5x 2
= 3N

shouldnt it be 3?

6. Nov 26, 2009

### wisvuze

I edited my post to include some explanation, read above

7. Nov 26, 2009

### DeathByVirus

so you're saying the force on m1 should be 5-m2a = m1a and
the force on m2 is 5 - m1a = m2a?

but m1a represents the net force on mass 1...it allows mass 1 to accelerate and also is the force that it pushes mass 2 with?

sorry ..im not very smart

8. Nov 26, 2009

### wisvuze

oops, I meant that m1a = 5 - some reaction force/contact force not m2a, sorry

9. Nov 26, 2009

### DeathByVirus

huh? now i'm just confused all over again

m1a=5- what reaction/contact force?

then what happens to m2a?

10. Nov 26, 2009

### wisvuze

the reaction force is the force equal to the contact force pushing back onto m1, so it would have had a force of 5 N if it weren't for that block:

5N ---> [Block 1] <--- ---> [Block 2]

You see that Block 1 pushes onto Block 2, but by Newton's third law, Block 2 exerts an equal but opposite force back onto Block 1, giving the equation m1a = 5 - reaction force

11. Nov 26, 2009

### DeathByVirus

ah.. i see =p thank you very much...i just lost 3 marks on my test (:

12. Nov 26, 2009

### wisvuze

it's okay :), so then since the reaction force is equal to the contact force, m2 is pushed by the contact force

Forces from the applied force to Block 2

5N ---> [Block 1] ---> [Block 2]

we know that m1a = 5 - reaction force

m1a = 3 , 3 = 5 - ReactionForce, ReactionForce = 2, but Reaction Force = Contact Force
so m2a = Contact force, since there are no other forces involved.

So m2a = 2

Now we can see that F net = 5 = m1a + m2a = 3 + 2

13. Nov 26, 2009

### DeathByVirus

lol i think im getting it (:
great...i now i have a 95 ish in phys >.>