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What's the acceleration of the blocks and the tension?

  • #1

Homework Statement

:

[/B]
2731039e19740e7b86019e274367b851.png




Homework Equations

: F = ma[/B]


The Attempt at a Solution

:
[/B]
The sum of the vertical forces for the second block is (mg) - T (tension). The tension of the second block is the parallel component of the first block which is (mgsin40). Therefore, the sum of the forces is (m2g - m1gsin40) = m(net)a. With numbers inserted, it's the following: [ (4 kg * 9.8 m/s^2) - (2 kg * 9.8 m/s^2 * sin40) = (4 kg + 2 kg) * a ] = a = 4.43 m/s^2. The answer in the back of the book is 0.933 m/s^2. What was my mistake?
 

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  • #2
Orodruin
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The tension of the second block is the parallel component of the first block which is (mgsin40)
This is not correct. You are assuming that the first block is in equilibrium, which it cannot be if the second is accelerating.
 
  • #3
The two main principles you should have in mind for this problem are $$\sum{F}=0$$ $$\sum{F=ma}.$$ These two principles, provided by Sir Isaac Newton, can be applied to free body diagrams (individual graphical vector layouts of all of the forces and their components acting on a particular object). In your problem, you will have two free body diagrams.. one for block 1 and one for block 2. You must define a coordinate system for each FBD (free body diagram). If you define your coordinate system for the FBD of block 2 to have a positive upward vertical axis and a positive rightward horizontal axis (as a familiar conventional two-dimensional coordinate system would), then you can use Sir Newton's principles to deduce that $$\sum F_y = T-m_2g=-m_2a_y $$ $$\sum F_x =0. $$ As for the free body diagram of block 1, it turns out to be a much easier analysis if you tilt the coordinate system such that the x-axis is parallel to the incline. If you construct such a FBD for block 1, you will find that in fact the tension in the FBD for block 2 is not parallel to the y-component of tension in the FBD of block 1. Moreover, the quantity $$mgsin(40)$$ represents the x-component of the gravitational force vector in the coordinate system of the FBD for block 1, which points in the -x direction.
 
  • #4
ehild
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The Attempt at a Solution

:
[/B]
The sum of the vertical forces for the second block is (mg) - T (tension). The tension of the second block is the parallel component of the first block which is (mgsin40). Therefore, the sum of the forces is (m2g - m1gsin40) = m(net)a. With numbers inserted, it's the following: [ (4 kg * 9.8 m/s^2) - (2 kg * 9.8 m/s^2 * sin40) = (4 kg + 2 kg) * a ] = a = 4.43 m/s^2. The answer in the back of the book is 0.933 m/s^2. What was my mistake?
No mistake, your derivation and result are correct. But you should not write [ (4 kg * 9.8 m/s^2) - (2 kg * 9.8 m/s^2 * sin40) = (4 kg + 2 kg) * a ] = a = 4.43 m/s^2, use an arrow instead the second "=" : [ (4 kg * 9.8 m/s^2) - (2 kg * 9.8 m/s^2 * sin40) = (4 kg + 2 kg) * a ] → a = 4.43 m/s^2.
 
  • #5
Orodruin
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No mistake, your derivation and result are correct.
The reasoning is not correct, at least not the one he is presenting:
The tension of the second block is the parallel component of the first block which is (mgsin40).
The tension in the string is not mg sin(40). However, he is using this:
Therefore, the sum of the forces is (m2g - m1gsin40) = m(net)a
The sum of forces on what? There is an equivalent situation that will lead to this, but it is not how his argument goes.
 
  • #6
ehild
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The reasoning is not correct, at least not the one he is presenting:

The tension in the string is not mg sin(40). However, he is using this:

The sum of forces on what? There is an equivalent situation that will lead to this, but it is not how his argument goes.
I understood the OP that he used the sum of forces on the whole system consisting of both masses, although the sentence "The tension of the second block is the parallel component of the first block which is (mgsin40)." is wrong.
The sum of external forces is equal to the acceleration of the CM of the system, multiplied by the whole mass. The components of forces and acceleration can be taken along the path, the string. I saw this argument many times.
 
  • #7
Orodruin
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I saw this argument many times.
Yes, but it is not the argument he is presenting in the OP, which means that he either did not intend to do it or did not understand the argument when demonstrated to him.
 
  • #8
ehild
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Yes, but it is not the argument he is presenting in the OP, which means that he either did not intend to do it or did not understand the argument when demonstrated to him.
The OP calculated the common acceleration with the equation (m2g - m1gsin40) = m(net)a, which is correct.
It is true that the concept of "acceleration along the string " is difficult to understand and most students use formulas without fully understanding them. That is why I prefer using FBD-s for every mass and pointing out that the magnitude of accelerations of both masses are equal because of the constant length of the string.
 
  • #9
Orodruin
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The OP calculated the common acceleration with the equation (m2g - m1gsin40) = m(net)a, which is correct.
Yes, I never said it was not. I said that the argument he is presenting does not justify it. Making it very likely that he has a conceptual misunderstanding and I do not think it is good to say that "everything is ok" in this situation.
 
  • #10
ehild
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Yes, I never said it was not. I said that the argument he is presenting does not justify it. Making it very likely that he has a conceptual misunderstanding and I do not think it is good to say that "everything is ok" in this situation.
You are right, I did not notice/ did not understand his first sentence.
 
  • #11
No mistake, your derivation and result are correct. But you should not write [ (4 kg * 9.8 m/s^2) - (2 kg * 9.8 m/s^2 * sin40) = (4 kg + 2 kg) * a ] = a = 4.43 m/s^2, use an arrow instead the second "=" : [ (4 kg * 9.8 m/s^2) - (2 kg * 9.8 m/s^2 * sin40) = (4 kg + 2 kg) * a ] → a = 4.43 m/s^2.
Thank you for correcting my organization. I'll improve that.
 

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