# Blocks/Forces - 1

1. Jan 28, 2017

### Alex126

1. The problem statement, all variables and given/known data
Two blocks of known mass m1 (20 kg) and m2 (32 kg) are at direct contact with each other, on a non-incline surface, and they are pushed by a Force F applied on the bigger block (m2). Knowing that the resulting acceleration of the Force is a (1.2 m/s2), determine:
- The contact forces between the two blocks
- How much they move after a known time t (2.5s)

2. Relevant equations
F = m*a

Motion equation:
x (space) = v0 (starting velocity) * t + 1/2 * a * (t2)
(Note: I know there is a difference in speed and velocity; let me just call it velocity here)

3. The attempt at a solution
The second point of the problem seems the easiest. The way I solved it is: knowing that the blocks are in contact with each other, they move the same distance. Using the generic equation for motion:
x = 0 + 1/2 * 1.2 * (2.52) = 3.75 meters

Correct?

The first part of the problem is the one where I have issues with.

First of all, our teacher wants us to write the forces that act on each block (or rather, the Newton law applied to each block individually). This part right here already drives me crazy. My attempt:

1. For the first body: [Weight] + [Normal force] + [Contact force between m1 and m2] = m1 * a
2. For the second body: [Weight] + [Normal force] + [Contact force between m2 and m1] + [Force F] = m2 * a

The first thing that short-circuits my brain is: when writing the forces that act on each individual body, when we consider the "contact force", which one(s) do we need to actually consider?

I'll post a picture and three possible answers (only one is correct, but I don't know which one):

Q1 (question 1): is the scheme correct, at least? Is there anything missing? (Proportionality/vectors starting in the barycenter of the bodies don't matter)

I called "Action" force the force that m2 exerts on m1, as a result of force F pushing on m2.
I called "Reaction" force the force that m1 exerts back on m1, because of the third law of Newton that says action-reaction and stuff.

Q2: assuming it is correct, what's the proper way of writing the forces for each body individually? The three options (A, B, C) that come to mind are:

A. For each body we only consider the forces exerted ON the body, but not the ones the body exerts on other objects. That's how I wrote it earlier.
A1. [Weight 1] + [Normal 1] + [Action force] = m1 * a
A2. [Weight 2] + [Normal 2] + [Reaction force] + [Force F] = m2 * a

B. Opposite of A, so for each body we only consider the forces exerted BY the body, but not the ones exerted ON the body.
B1. [Weight 1] + [Normal 1] + [Reaction force] = m1 * a
B2. [Weight 2] + [Normal 2] + [Action force] + [Force F] = m2 * a

C. Combined solution, we write all of them for each body.
C1. [Weight 1] + [Normal 1] + [Reaction force] + [Action force] = m1 * a
C2. [Weight 2] + [Normal 2] + [Action force] + [Reaction force] + [Force F] = m2 * a

D. One additional possibility/doubt that floats in my brain is that the Force F should be considered when writing the equation for the first body (m1) too, since it's at direct contact with m2. I really don't think that's the case, but it would then look like this:
D1. [Weight 1] + [Normal 1] + [Reaction force] + [Action force] + [Force F] = m1 * a
D2. [Weight 2] + [Normal 2] + [Action force] + [Reaction force] + [Force F] = m2 * a

Q3: solving the exercise. If I had to choose, I would probably pick option A or C of the above, and if I had to pick one, I would go with C. The way I see it, just like we write Weight + Normal force (and that much I know for sure) for each body, we should likewise also write Action + Reaction force for each body (after all, Weight and Normal are just another action-reaction couple).

However, if that's the case, then the first body should have no motion at all, since (C1) Weight 1 and Normal 1 null each other, and Reaction + Action also null each other. So I would exclude C.

That's why I thought that option D might be valid. However, since Action = Reaction (for Newton's third), then option D would then become Force F = m1*a and Force F = m2*a, which shouldn't be possible since m1 and m2 are different and a is the same. So I exclude D as well.

So, going by exclusion, it could either be A or B (unless there is a fifth option E which I'm missing).

I really can't figure which one is the correct one, and solutions for the exercise were not given. The more I think about it, the less I think logically, so I figured I'd ask here before having a nervous breakdown.

Anyway, with "option A" the solution would become:

Action = m1*a
Reaction + F = m2*a

What bugs me here is that we don't even need a system (mathematical system, I mean) to solve the problem, because the first equation is already sufficient: knowing m1, and knowing a, Action = 20*1.2 = 24.

Knowing the Action = Reaction (Newton's third), Action should also be 24 (+24 or -24 depending on the chosen positive direction).

With "option B" the solution would become:

Reaction = m1*a
Action + F = m2*a

I am aware that the numbers would be the same in the end, but I need to know which one is technically and formally correct.

If I had to guess, I would say that option A is the correct one. The reason is that m1 (scalar) * a (vector) should give a vector (Action or Reaction) with the same direction as a, and since Action is the vector that has that direction, while Reaction has the opposite direction, then Action = m1*a should be the correct one.

Q4: this is an additional question/"interpretation" of the problem that I have. Since the blocks are at direct contact, I thought we could also consider them as a single mass m = m1+m2.

If that were the case, then this unified block "m" would have an equation of forces like this:
[Weight] + [Normal] + [Force F] = m*a

Where Weight = Weight 1 + Weight 2; Normal = Normal 1 + Normal 2; m = m1+m2

Is this correct?

If that were the case, we could proceed by saying:
Force F = m*a
Force F = 62.4

At this point we could go back to the previous "options", option A and option B, and verify the numbers. With option A:

Action = m1*a
Reaction + F = m2*a

I'll consider positive the vectors pointing to the left (Force F, Action, and a), negative the others (Reaction).

Action = m1*a = 20*1.2 = 24
Reaction = -Action = -24
F = 62.4
m2*a = 32*1.2 = 38.4

Therefore, Reaction + F = m2*a becomes: -24 + 62.4 = 32*1.2, which would actually be correct.

Problem 1 Questions Recap:
- Q0: is the answer to the second question correct?
The answer was: x = 0 + 1/2 * 1.2 * (2.52) = 3.75 meters

- Q1: is the scheme of forces correct? (see image)

- Q2: is option A the correct representation of the forces for each individual body (block) ? Option A was:
A1. [Weight 1] + [Normal 1] + [Action force] = m1 * a
A2. [Weight 2] + [Normal 2] + [Reaction force] + [Force F] = m2 * a

- Q3: if option A is correct, why is it that for the forces on the Y axis we write both "active" (Weight) and "passive" (Normal) forces, whereas for the X axis we write only the "passive" forces?
My interpretation, "if I had to guess", is that they are all actually "passive" forces, and we always only write "passive forces". In other words, the Weight force is "passive" because it's a force external to the body, that acts on the body by pushing it down (hence "passive", from the body's perspective); likewise, the Normal force is "passive" because it's a force that "comes from the surface" and acts on the body by pushing it up. Likewise, the force F and force Reaction are obviously external, so they are also "passive" from the body's point of view.

So, am I to understand that when we write the forces for a body, in a general situation, we only write the forces that act ON the body, but never the forces that the body exerts on other objects? Is this interpretation correct?

2. Jan 28, 2017

### CWatters

Wow! I haven't read all of that but you/teacher seem to be making it way too complicated.

Lets define left as +ve.

Part 1: The problem statement just asks for the contact forces between the two blocks. You can ignore all the vertical forces and just look at the horizontal forces..

Block m1 is accelerating to the left at "a" so there must be a net horizontal force acting on it = m1*a to the left. There is no friction so the force that m2 applies to m1 is just F21 = m1a.

Newton's law says that m1 must apply an equal and opposite reaction force on m2 which is F12 = -m1a.

Those are the only "contact forces between the two blocks".

Part 2: You appear to have this correct.

The problem statement doesn't require that but you could write something like...

Vertically: The vertical forces acting on m1 sum to zero. So applying Newton's law in the vertical direction you get m1a1v=0 so a1v=0. The same applies to block m2 and a2v=0. Neither block is accelerating vertically. No surprise there.

Horizontally:
For block m1 you just have the force f21 = m1a.
For block m2 there are two forces, the applied force F and the reaction force F12 = -m1a. So applying Newtons law....

Net force = mass * acceleration
F + F12 = m2a
F + (-m1a) = m2a
rearrange..
F = m1a + m2a
F = (m1 + m2)a

which is what you expect.

3. Jan 28, 2017

### Vector1962

May I suggest a free body diagram that labels all the forces on the blocks? see attached. For clarity, I did not show the contact force between the floor and the blocks, but as a general practice you should. Almost ALL problems require a nice clean sketch showing the forces and which way is positive. Perhaps showing the blocks individually would help?

#### Attached Files:

• ###### Blocks.png
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4. Jan 29, 2017

### Alex126

Thanks for the answers guys, much appreciated.

5. Jan 29, 2017

### CWatters

Correct. The FBD normally shows just one body and the forces acting on it. Not the forces it puts on other objects.

If there are two objects then draw two separate FBDs.

6. Jan 29, 2017

### Alex126

Ok, thanks again.

7. Jan 30, 2017

### Vector1962

I'm sure others will have other opinions, But I believe before you start "pushing buttons" on the calculator or writing a bunch of equations, draw a really nice sketch and clearly label all the forces. If there are multiple bodies or parts draw a separate sketch for each body (generally speaking something like the sketch above). Also, be extra careful to neatly write each step of your solution process/algebra. Working neatly will reduce the frustration and perceived complexity of the problem. You don't want to hinder learning and understand by with carelessness and sloppiness. Just a thought. I believe you will find useful.

Not sure how to respond the "last bit". Because there are 2 bodies in contact there is a contact force between them called the "normal" force. Depending on how you say it, the normal force is the force of body "A" on body "B". On the other hand, it is also the force of Body "B" on Body "A". (every action has equal and opposite reaction). Hence in the sketch, it is shown on both bodies. Just to be clear, the normal force is perpendicular to the plane of contact between the bodies.

Last edited: Jan 30, 2017