A Simple Acceleration Problem - two different answers

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In summary, the main point of confusion in the given problem is the inconsistency in the signs of the forces in equations 1-3. In order for F to be the net force on block 1, the force of block 2 on block 1 should be negative, which is not reflected in the given equations. This can be resolved by consistently using either positive or negative signs for all forces. Additionally, the use of scalars instead of vectors can influence the signs in equations and should be kept in mind when solving physics problems.
  • #1
PeterPumpkin
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Homework Statement



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?

Homework 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

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).

Thanks in advance.


 

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  • #2
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
Ah! I guess that's the key point: F(block 2 on block 1) is negative.
 
  • #4
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
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.
 

FAQ: A Simple Acceleration Problem - two different answers

What is an acceleration problem?

An acceleration problem is a type of physics problem that involves calculating the acceleration of an object. This is usually done by using the equation a = Δv/Δt, where a is acceleration, Δv is the change in velocity, and Δt is the change in time.

Why are there two different answers in a simple acceleration problem?

In a simple acceleration problem, there can be two different answers because there are two different types of acceleration: average acceleration and instantaneous acceleration. Average acceleration is calculated over a period of time, while instantaneous acceleration is the acceleration at a specific moment in time. Therefore, depending on which type of acceleration is being asked for in the problem, there can be two different answers.

How do you know which answer to use in a simple acceleration problem?

To know which answer to use in a simple acceleration problem, you need to carefully read the problem and identify which type of acceleration is being asked for. If the problem specifies a certain time frame, then you should use the average acceleration. If the problem asks for the acceleration at a specific moment in time, then you should use the instantaneous acceleration.

What are some common mistakes to avoid in a simple acceleration problem?

Some common mistakes to avoid in a simple acceleration problem include using the wrong equation, confusing average and instantaneous acceleration, and not paying attention to units. It is important to carefully read the problem and identify which type of acceleration is being asked for, and to use the correct equation and units throughout the problem.

How can I check my answer in a simple acceleration problem?

To check your answer in a simple acceleration problem, you can use the equation a = Δv/Δt to calculate both the average and instantaneous acceleration. If your answer matches one of these calculations, then it is likely correct. You can also double check your units to make sure they are correct and consistent throughout the problem.

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