1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Law of Motion: Horizontal Force

  1. Sep 18, 2011 #1
    Physics: Newton Law of Motion Question?
    I am a bit confused about this elementary question. If anyone can guide me in the right direction, I would appreciate it.

    Picture: http://tinypic.com/r/14w84g1/7"

    2 boxes are placed next to 1 another on a horizontal surface. The box on the left has a
    mass of 5 kg while the box on the right has a mass of 2 kg. Assume that there is no friction
    is negligible. By pulling on a horizontal string attached to the 2 kg box, Stephane is exerting a force T to the right, while you are exerting a force F to the right by pushing on the 5 kg box.

    Using F = 12 N and T = 2.0 N,

    (a) Determine the acceleration of the 2 kg box.

    [ There is an acceleration exerted on the 2 kg box by the 5 kg. Is this 12N/ 5 kg or 2.4 m/s^2. The tension applies solely to the 2kg box and is 2 N/2 kg or 1 m/s^2. The net acceleration should be 3.4 m/s^2. If that is the case, then the 2 kg box would be pulled by itself. Thus is the total net acceleration 2.4 m/s^2 because that acceleration outweighs the tension on the 2 kg box?]


    (b) Which box experiences the largest net force in this situation?

    [Should be the same based on part a?]

    (c) Determine the magnitude of the force exerted by the 2 kg box on the 5 kg box
    in this situation.

    If F = 10.0 N and T = 10.0 N, (d) Determine the acceleration of each of the boxes now.

    Thank you in advance.
     
    Last edited by a moderator: Apr 26, 2017
  2. jcsd
  3. Sep 18, 2011 #2

    PeterO

    User Avatar
    Homework Helper

    Firstly I am worried by this statement of yours:
    "There is an acceleration exerted on the 2 kg box by the 5 kg."

    A force may be exerted , but not an acceleration!

    1). If the sole force acting was Stephane's, the 2kg box would accelerate away from the 5 kg box - you can easily calculate that rate of acceleration.
    2). If the sole force acting was the Force you applied, then you would be accelerating the 7 kg combination. You could calculate that acceleration.
    But both forces are applied:
    If Stephane's force had really accelerated the 2 kg box away from the 5kg box, then Your force would only have to accelerate the 5kg box - giving a higher acceleration than you will have calculated in 2)., above.
    If in that scenario, the 5kg box would be accelerating at a higher rate than the 2kg box [an impossibility] then effectively the 7kg combination is being accelerated by the combined force of you and Stephane.
     
    Last edited by a moderator: Apr 26, 2017
  4. Sep 18, 2011 #3
    I am not sure of the last step you mentioned.

    a). The acceleration of 2.0 kg box. I tried to use a few household objects to understand this. If the T= 2 N on the 2 kg box and the F=12 N, in the picture, then the tension applied on box 2kg would be canceled, would it not? The whole system is pushed by a 12 N force. The acceleration of the system is 12 N/ (2kg +5kg) = 1.71 m/s^2. The tension on box 2kg is 2 N and causes the box to accelerate at 1 m/s^2. Should these forces be added for the net acceleration on box 2kg, 2.71 m/s^2, or is it simply the acceleration of the whole system, 1.71 m/s^2.

    b.) Which experiences the largest net force. The the whole system experiences the same acceleration, don't both boxes have the same net force. However, this doesn't seem correct to me. Is the Force T not taken into account here at all. That 1m/s^2 acceleration or 2 N force of and on the box 2 kg does not matter? I am rather confused.

    c.) Is this simply 10 N, F-T=10 N. Because the T is a smaller magnitude, while F is a larger magnitude, box 2 kg stands in the way of the 5 kg box.

    If anyone can explain this a bit more clearly, I would be thankful and less frustrated with myself.
     
  5. Sep 19, 2011 #4
    bump 10char
     
  6. Sep 19, 2011 #5

    PeterO

    User Avatar
    Homework Helper

    The Tension would not be cancelled out by a force in the same direction! Augmented perhaps.

    You push the 5kg box with a force of 12 N to the right.
    The 2 kg box pushes the 5 kg box with a force I will call F2 to the left.
    Net force is 12 - F2 to the right.

    Steph pulls the 2kg box to the right with a force of 2N
    The 5 kg box pushes the 2 kg box with a force F2 to the right. [Newtons 3rd law couple]
    Net force on 2kg box is 2 + F2

    Those forces result in an acceleration of the blocks, [12-F2 accelerates the 5kg box, 2+F2 accelerates the 2kg box], and the acceleration is the same for each box.

    Simultaneous equations, solve for a and the contact force F2
     
  7. Sep 19, 2011 #6
    Thank you very much for your help, PeterO. I think the trouble I had was grasping on the idea that the contact forces from A to B is the same as B to A. Thanks again.
     
  8. Sep 19, 2011 #7

    PeterO

    User Avatar
    Homework Helper

    That last bit - the A to b, B to A, equal forces - is Newton's Third law. From experience, approx 3% of Senior High students have a confident grasp of Newton's Third law.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Law of Motion: Horizontal Force
Loading...