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Dynamics problem using Newton's third law.

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  • #1
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Homework Statement


Two packages at UPS star sliding down a 20° ramp. Package A has mass of 5.0 kg and coefficient of friction of 0.20. Package B has a mass of 10 kg and a coefficient of friction of 0.15. How long does it take package A to reach the bottom?

Homework Equations



N/A

The Attempt at a Solution


Here's my drawing of the diagram in the textbook:
http://imageshack.us/photo/my-images/802/dynamicsdiagram1.jpg/
Here are my free-body diagrams of the forces acting on both boxes:
Free-body diagram for box A: http://imageshack.us/photo/my-images/15/freebodydiagramforblock.jpg/

where n is the normal force, FG is the force due to gravity, F_f is the frictional force, F_BonA is the force box b exerts on A, and F_g_x is the x component of the force due to gravity, as the y component of gravity balances out the normal force.

Free-body diagram for block B:
http://imageshack.us/photo/my-images/256/freebodydiagramforblock.jpg/

Where n is the normal force, FG is the force due to gravity, F_f is the frictional force, F_AonB is the force box A exerts on B, and F_g_x is the x component of the force due to gravity. The friction force and the force block A exerts on B oppose its motion.

Now I begin to work, first starting with newton's second laws:
[itex]\sum f_{A_{y}} = n - F_{G_{y}} = n - mg\cos\theta[/itex]
[itex]\sum f_{A_{y}} = 0 [/itex] so, [itex]n = mg\cos\theta[/itex]
That was for the y component, now for the x component:

[itex]\sum f_{A_{x}} = F_{BonA} + F_{G_{x}} - F_{k_{A}} = m_{A}a_{A}[/itex]
where F_{k_{A} is the kinetic friction on block A, and m_{A}a_{A}
is the net force acting on block A.

Now I further simplify this:
[itex]\sum f_{A_{x}} = m_{B}g\sin\theta + m_{A}g\sin\theta - \mu_{k}m_{A}g[/itex]
Now I substitute in the numbers and solve for the force:
[itex]\sum f_{A_{x}} = (10kg)(-9.8m/s^{2})(\sin20) + (5.0 kg)(-9.8 m/s^{2})(\sin20) - (0.2)(5.0 kg)(-9.8m/s^{2}) = -40.5 N.[/itex]

Okay, this is where I start to doubt my answer, but I continue to solve for the acceleration of box A:
[itex]a_{A} = \frac{\sum f_{A_{x}}}{m_{A}} = \frac{-40.5 N}{5.0 kg} = -8.1m/s^{2}[/itex]

Now I am even more doubtful, as the acceleration does not seem reasonable for the box given the circumstances.

So I use one of the kinematics equations for constant acceleration to find the time it takes for the box to reach the bottom of the ramp:

[itex]S_{1} = S_{0} + V_{0}\Delta t + \frac{1}{2}a_{A}\Delta t^{2}[/itex]
[itex] 0 = 2.0 m + 0 + \frac{1}{2}a_{A}\Delta t^{2}[/itex]
Rearranging and solving for t:
[itex] t = \sqrt{\frac{-4}{-8.1}} = 0.7[/itex] seconds.

The book's answer was 1.48 seconds. I know the book's answer is correct, but they did their solution in a completely different way. I want to know just exactly what I did wrong, what assumptions or inferences I incorrectly made, and where I can improve next time. Thanks in advance.
 

Answers and Replies

  • #2
ehild
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[itex]\sum f_{A_{x}} = F_{BonA} + F_{G_{x}} - F_{k_{A}} = m_{A}a_{A}[/itex]
where F_{k_{A} is the kinetic friction on block A, and m_{A}a_{A}
is the net force acting on block A.

Now I further simplify this:
[itex]\sum f_{A_{x}} = m_{B}g\sin\theta + m_{A}g\sin\theta - \mu_{k}m_{A}g[/itex]
The force on A from B is not equal to mBg sin (θ ). What would be the force on B from A then? There is a force of interaction FAB between A and B, the force from B to A is equal in magnitude with the force on B from A.
The force of friction is μ n = μ mAg cos(θ ).
Write out the equations both for A and B. The accelerations are equal. You have two equations for FAB and a. Eliminite the force of interaction and solve for a.


ehild
 
  • #3
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The force on A from B is not equal to mBg sin (θ ). What would be the force on B from A then? There is a force of interaction FAB between A and B, the force from B to A is equal in magnitude with the force on B from A.
The force of friction is μ n = μ mAg cos(θ ).
Write out the equations both for A and B. The accelerations are equal. You have two equations for FAB and a. Eliminite the force of interaction and solve for a.


ehild
So, using Newton's second law for both blocks:
[itex]\sum f_{A_{x}} = F_{BonA} + m_{A}g\sin\theta - \mu_{k}m_{A}g\cos\theta = m_{B}a_{B} + m_{A}g\sin\theta - \mu_{k}m_{A}g\cos\theta = m_{A}a_{A}[/itex]
[itex]\sum f_{B_{x}} = F_{AonB} + F_{G_{x}} - f_{k_{B}} = -F_{BonA} + m_{B}g\sin\theta - \mu_{k}m_{B}g\cos\theta = -m_{B}a_{B} + m_{B}g\sin\theta - \mu_{k}m_{B}g\cos\theta = m_{A}a_{A}[/itex]

Correct? If yes, so now I would just eliminate the force interactions?
 
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  • #4
ehild
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No, its is wrong. Think of Newton's second law: The resultant of the forces acting on B equal to mass of B multiplied by the acceleration of B. Why is mBaB = FBonA? And why is the sum of forces acting on B equal to mAaA?

ehild
 
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  • #5
309
7
No, its is wrong. Think of Newton's second law: The resultant of the forces acting on B equal to mass of B multiplied by the acceleration of B. Why is mBaB = FBonA? And why is the sum of forces acting on B equal to mAaA?

ehild
Because that is the force B exerts on A, and since that is that is the direction the net force points towards, that is the direction B will accelerate towards. As for your second question, how did you get that? Unless FAonB + mBaB = -FBonA + mBaB = mAaA ?
 
  • #6
ehild
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Read your equations again and try to understand why are they wrong.

ehild
 
  • #7
309
7
I just realized the mistake I made in my previous post. -FBonA + mBaB = 0, since -FBonA = -mBaB.

Anyways, the sum of the forces acting on B should be equal to mBaB and the sum of the forces acting on A should be equal to mAaA, by my interpretations of course. I can understand due to the acceleration constraint, both boxes will have the same acceleration, hence aA = aB. But since mA ≠ mB, the net force that acts on both boxes is not the same. Correct?
 
  • #8
ehild
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I just realized the mistake I made in my previous post. -FBonA + mBaB = 0, since -FBonA = -mBaB.
Wrong.

Anyways, the sum of the forces acting on B should be equal to mBaB and the sum of the forces acting on A should be equal to mAaA, by my interpretations of course. I can understand due to the acceleration constraint, both boxes will have the same acceleration, hence aA = aB. But since mA ≠ mB, the net force that acts on both boxes is not the same. Correct?
Right.

ehild
 
  • #9
309
7
So, if the sum of all forces on box B is mBaB and the sum of all forces on box A is mAaA? But, if that is the case, wouldn't the equations become:
[itex]\sum f_{A_{x}} = m_{B}a_{B} + m_{A}g\sin\theta - \mu_{k}m_{A}g\cos\theta = m_{A}a_{A}[/itex]
[itex]\sum f_{B_{x}} = -m_{B}a_{B} + m_{B}g\sin\theta - \mu_{k}m_{B}g\cos\theta = m_{B}a_{B}[/itex]

Since FBonA = mBaB and FAonB = -FBonA = -mBaB?

If that is also the case, wouldn't the equation for box B give me a constant as the unknowns will cancel with each other?
 
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  • #10
ehild
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Since FBonA = mBaB and FAonB = -FBonA = -mBaB?
As I said nearly one hundred times, it is wrong.

ehild
 
  • #11
309
7
As I said nearly one hundred times, it is wrong.

ehild
Why is it wrong? All you've been doing is stating "wrong, wrong, WRONG" without ever saying justifying WHY it is wrong.

How is it wrong? You seem to be implying that there isn't an action/reaction pair, that FBonA ≠ -FAonB. Then how is Newton's third law even relevant to this problem?
 
  • #12
ehild
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FBonA = -FAonB, but it is not mBaB what you keep on stating. The force B exerts on an other body does not accelerate B. Why should be mBaB=FAonB right? I asked the same in post 4 and explained what are the correct forces in post 2.

ehild
 
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  • #13
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FBonA = -FAonB, but it is not mBaB what you keep on stating. The force B exerts on an other body does not accelerate B. Why should be mBaB=FAonB right?

ehild
My mistake. So,
[itex]F_{BonA} = m_{B}a_{B} + m_{B}g\sin\theta - \mu_{k}m_{B}g\cos\theta = m_{A}a_{A}[/itex]
Since this is the sum of all forces for B which would imply that it is the same total force B exerts on A to cause it to accelerate. If this is not correct, please explain why rather than just stating "wrong, please try again". You have said I was right in saying that the forces acting on both objects are not the same, yet have also posted mAaA is equal to [itex]\sum f_{B_{x}}[/itex]
 
  • #14
ehild
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Let's start again.

Forces acting on A are: the component of gravity along the slope mAgsin(θ),
friction: μmAgcos(θ),
B with a force F, what we do not know.
The sum of all three forces equals to maaA.

The same holds for B, with the exception that it has a force from A which is equal to -F, and the sum of all forces equals to mBaB.

[tex]m_Agsin(\theta)-\mu_A m_A g cos(\theta) +F=m_A a_A[/tex]
[tex]m_Bgsin(\theta)-\mu_B m_B g cos(\theta) -F=m_B a_B[/tex]

aB=aA=a

There are only forces on the left-hand side and accelerations are only on the right-hand side of the equations. Do not mix them.


ehild
 
  • #15
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Thanks for that, ehild. That was really helpful.

I just wanted to know if the same problem can be solved without using Newton's third or second law and just using energy and kinematics?
 
  • #16
ehild
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The energy method can be used if you are sure that the bodies really move together. The method based on FBD and Newton's laws will give the force of interaction, too, and you can check if the bodies really touch each other. As B is behind A, B can only push A, so F has to be positive. Now A has greater coefficient of friction than B, so it needs an extra push to have the same acceleration as B. If the body A with greater coefficient of friction were behind B, it would lag after B and they would move with different accelerations.

If you calculate the acceleration, than substitute it back into one of the original equations, you will find that B really pushes A.

Try to apply the energy method. Take care, there are two different terms for the work of friction, one for A and one for B. At the end you get the final speed and use the equation 0.5(vi+vf)t=s to find the time.

ehild
 

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