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Homework Help: The Vector nature of Newton's Second Law

  1. Jan 11, 2009 #1
    1. The problem statement, all variables and given/known data
    A wagon carries a child. Together mass is 28.5kg. You pull on the handle of the wagon at an angle of 40degrees from the horizontal. The wagon travels over a level horizontal sidewalk. A force of friction of 12.0N acts on the wagon.

    a) What is the net force acting on the wagon?

    2. Relevant equations

    Fnet = F1 + F2

    3. The attempt at a solution

    I missed a day of Physics class but the teacher gave me this assignment. I thought it was suppose to be solving right triangles but now, I'm not too sure. I've been looking through my notes! Please, I need a visual representation.
    Last edited: Jan 11, 2009
  2. jcsd
  3. Jan 11, 2009 #2


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    Draw the wagon, the frictional force acts opposite to the motion. Resolve the applied force into vertical and horizontal components.

    Can you do those two?
  4. Jan 11, 2009 #3
    Would it look like this?

  5. Jan 11, 2009 #4


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    Yes, it would, though on the handle, there should be a force going away from the wagon since you are pulling it.
  6. Jan 11, 2009 #5
    So there should be an arrow on the handle line. With that, which vector's magnitude am I suppose to figure out...? There are unknown sides to that triangle.
  7. Jan 11, 2009 #6


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    Put the unknown force as, F. Draw the right angled trianle now.
    Can you express the sides of right angled triangle in terms of F?
  8. Jan 11, 2009 #7


    Would it look like that? and I'm suppose to find Fx, Fy and F?

    I'm trying to figure out the net force and that includes friction I believe.
  9. Jan 11, 2009 #8


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    yes, that is how it would look.



    [tex]sin40 = \frac{F_y}{F}[/tex]

    Correct? Then Fy=?

    Now consider what cos40 should be
  10. Jan 11, 2009 #9
    I understand this but the only thing is that F, Fy and Fx have no values.
  11. Jan 11, 2009 #10


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    Forget that they have no values for the moment. (After you resolve the forces, you will be able to get the value for F)

    [tex]sin40 = \frac{F_y}{F} \Rightarrow F_y= ?[/tex]
  12. Jan 11, 2009 #11
    Fy/Sin40 = F
  13. Jan 11, 2009 #12


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    Other way around, Fy=Fsin40

    Similarly cos40= Fx/F, so Fx=Fcos40 right?

    In your Free body diagram, include the weight of the wagon.

    EDIT: Now that you have Fx and Fy found. Knowing that the weight acts downwards and Fy upwards. Is there any net motion upwards?
  14. Jan 11, 2009 #13
    Alright, I understand the equations, but how am I suppose to find the values now?
    What do you mean by include the weight of the wagon...? Is it another vector?
  15. Jan 11, 2009 #14


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    Well the question stated that the mass of the wagon and child was 28.5kg, so it has an assoiciated weight. Yes, it is another vector.
  16. Jan 11, 2009 #15
    Is it possible for yourself to draw it because I'm having trouble visualizing it... If I add a weight vector, wouldn't it throw me off with the fx fy and f vectors...
  17. Jan 11, 2009 #16


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    http://img352.imageshack.us/img352/6447/diagramii8.jpg [Broken]

    So, when you pull a wagon, it only moves horizontal right? Meaning that there is no vertical movement, therefore, the forces Fy and weight should balance.

    So Fy=Weight

    => Fsin40=28.5g where g is the acceleration due to gravity. Can you now find F?
    Last edited by a moderator: May 3, 2017
  18. Jan 11, 2009 #17
    F = 28.5g / sin40

    is that the net force? or do I have to figure out Fx and add F + Fx + Friction(with direction) to find F net
    Last edited by a moderator: May 3, 2017
  19. Jan 12, 2009 #18
    I disagree with the statement that Fy = 28.5g. If this was the case, either there would be no friction (as the wagon isn't exerting any force on the ground) or there is a moment acting on the wagon, in which case it will have a rotational acceleration - and that seems much more complicated than this problem is supposed to be. I would say that there should be a net force between the wagon and the ground, meaning you will need one more force in your diagram.

    The problem, as originally stated, is not determinate. If the wagon is not moving at a constant velocity, then it is being accelerated (or decelerated). To find the net force, you will need to be given either the acceleration or information on how the frictional force is calculated (are the wheels dragging? or is the friction in the hub of the wheels? In either case, you would need a coefficient for friction). Again, I think this is more complicated than the problem is supposed to be.

    So after all that, I believe it is safe to assume the wagon is supposed to be moving at a constant velocity. If this is indeed the case, the problem has just become amazingly easy (check Newton's First Law of Motion). However, I also think the question is actually asking for the force applied on the handle, F. To find this, consider what the sum of the forces in the horizontal direction will be. Once that is determined, you will be able to find Fx, which will then be a simple calculation to find F.
  20. Jan 13, 2009 #19


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    Resultant Vertical force=0
    => Fsin40=28.5g...F can be found

    Resultant horizontal force,Fnet=Fcos40-12....when F is found from above, the resultant horizontal force can be found.
  21. Jan 13, 2009 #20
    The sum of the vertical forces will be equal to zero, but there are three forces acting in the vertical direction:
    F(sin 40) (upward force)
    28.5g (downward force)

    and the normal force of the ground upwards on the wagon (I'll call it Fn)

    F(sin 40) - 28.5g + Fn = 0

    If there was not a force between the wagon and the ground, there would be no frictional forces involved in this problem (no force between the wheels and the ground means the wheels would neither turn nor drag).

    So we are left with a missing piece of information - either we need the friction coefficient so we can find Fn, or we need to know what the acceleration of the wagon is.
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