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Find force box exerts on other box up incline

  1. Oct 31, 2015 #1
    1. The problem statement, all variables and given/known data
    A force F pushes two boxes, box 1 and box 2 up an inclined plane of 30 degrees. If F=25 N and box 1 is 2 kg and box 2 is 3 kg, what is the force exerted on the 2 kg box by the 3 kg box. Assume no friction.

    2. Relevant equations
    F=ma.

    3. The attempt at a solution
    Since the boxes' accelerations are the same, I decided to resolve the 2 kg box's weight into mgsin(theta) and mgcos(theta). Since mgcos(theta) is equal and opposite to the normal force on the 2 kg box, I thought that the magnitude of the force with which the 3 kg box exerts on the 2 kg box equals mgsin(theta) for the 2 kg box. So the answer to the problem I thought would be mgsin(theta) which would be (2 kg)(9.8 m/s^2)(30 degrees) which is 9.8 N.
     
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  3. Oct 31, 2015 #2

    PhanthomJay

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    Ouch, what happened to Newton's laws? The weight of an object is the force of the earth (gravity) on the object. An objects weight never acts on another object. Try Newton's 2nd law on the system of the 2 boxes by first finding the net force acting on the system to solve for the acceleration. Then draw free body diagrams of each block separately to find the contact force between blocks, again using his laws.
     
  4. Oct 31, 2015 #3
    ok my bad. Yeah you're right: I guess Newton's laws don't exist in my world. I think I got confused when I resolved the weight into its components. That's what happens when you do a problem too quickly. Yikes indeed
     
  5. Oct 31, 2015 #4
    I found that the acceleration of the system is 5 m/s^2. The forces acting on the 2 kg box are its weight, normal force from incline, and the reaction force from the 3 kg box. So would the force that 3 kg box exerts on 2 kg box be (3 kg)(5 m/s^2) which would be 15 N?
     
  6. Oct 31, 2015 #5
    Depends on which way the force vector points.
     
  7. Oct 31, 2015 #6
    It points in the left direction, in direction of 2 kg box.
     
  8. Oct 31, 2015 #7
    Is it parallel to the ground or the 30 degree incline?
     
  9. Oct 31, 2015 #8
    At a 30 degree angle
     
  10. Oct 31, 2015 #9
    I just don't see how I could resolve that force vector if it is parallel to incline
     
  11. Oct 31, 2015 #10

    Student100

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    Why would the acceleration be ##5\frac{m}{s^2}##? Ask yourself if this is the same as if it were a horizontal case. Why would the incline case be the same?

    Did you draw your force diagrams?

    Not sure what this means.
     
  12. Oct 31, 2015 #11
    You need to find the net force of the system. After you do that you can determine the acceleration. In this particular problem there is normal, weight due to gravity, and the 25N.
     
  13. Oct 31, 2015 #12

    Student100

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    LOL my bad, got who was asking the question confused.
     
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