# A Simple Acceleration Problem - two different answers

1. Jun 11, 2009

### PeterPumpkin

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

AN EMBARRASSINGLY SIMPLE QUESTION:
Suppose we have two rectangular blocks (masses m1 and m2) at rest on a frictionless flat table. They are in contact. Block 1 is on the left. We apply a force F to block m1. (Positive is to the right.) What is the equation relating the applied force, F, and the acceleration, a?

2. Relevant equations
(Resolving horizontally)
(1) F - F(block 2 on block 1) = m1 * a
(2) F(block 1 on block 2) = m2 * a
(3) F(block 1 on block 2) = - F(block 2 on block 1)
(where equations (1) and (2) follow from Newton’s second law, equation (3) follows from the third law and F(block 2 on block 1) is the force exerted by block 2 on block 1.)

(4) F = (m1 + m2) * a

3. The attempt at a solution
FIRST SOLUTION:
Considering the two blocks as one, the answer is obviously F = (m1 + m2) * a

ALTERNATIVE SOLUTION WHICH GIVES A WRONG ANSWER!
Considering each block separately ie using equations (1), (2) and (3).

Substituting (3) into (1) we get
F + F(block 1 on block 2) = m1 * a

Using (2) and rearranging we get:
F = (m1 - m2) * a which is clearly wrong!

QUESTION:
Where's the mistake? I suspect the sign is wrong in one of equations (1), (2) or (3).

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

2. Relevant equations

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

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2. Jun 11, 2009

### diazona

In equations 1-3, you're being inconsistent in the signs of the forces. For example, the initial force F is to the right. If you choose your axis such that positive forces are to the right, then the force of block 2 on block 1 is negative, which means that equation 1 should be
F + F(block 2 on block 1) = m1 * a
F is a positive number, F(block 2 on block 1) is a negative number, which means that you're going to find that the net force on block 1 is less than F, as you expect.

3. Jun 12, 2009

### PeterPumpkin

Ah! I guess that's the key point: F(block 2 on block 1) is negative.

4. Jun 12, 2009

### PeterPumpkin

I'm now not sure this is the right reason. I have just looked at least a dozen studies of Atwood's (or Attwood's) machine on edu sites on the Internet. This machine also involves opposing forces (tension acting up and gravity acting down) and in each case they write:

T - m1 * g = m1 * a

They don't write: T + m1 * g = m1 * a.

Why should we have to use + in the two blocks case and - in the Atwood case? Can anyone throw any light on this?

5. Jun 12, 2009

### cepheid

Staff Emeritus
It's just confusion between whether something is considered to be intrinsically negative, or whether the negative sign is included explicitly, and all quantities in the expression are considered to be positive.

In T - m1g = m1a,

we've assumed g means +9.81 m/s2, therefore the negative sign must be included explicitly. If we just wrote:

T + Fg = m1a

we would be NOT assuming anything specific about the sign of Fg. When its value was calculated, it would come out either negative or positive depending on what the force balance required.

Another way of looking at it is that you have to think carefully about whether Fg refers to the magnitude of the force vector, in which case it would be intrinsically positive and you would have to put the appropriate signs in manually, or whether it is in fact a scalar force value whose sign could be positive or negative depending on its direction (which is more convenient than working with vectors in 1D problems). Since, in more complicated scenarios, you may not know from before hand what the directions of the forces are, using scalars that could potentially be any real number and letting the math sort itself out is more advisable.