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!

Two crates and a spring

  1. Mar 4, 2015 #1
    1. The problem statement, all variables and given/known data
    Two large crates, with masses 575 kg and 255 kg, are connected by a stiff, massless spring (k = 7.8 kN/m) and propelled along an essentially frictionless, level factory floor by a constant force applied horizontally to the more massive crate. If the spring compresses 4.7 cm from its equilibrium length, what is the applied force?

    2. Relevant equations
    F=kx?
    F=ma

    3. The attempt at a solution
    I started by drawing a picture and setting up some free body diagrams. The question kind of confuses me because it just says horizontally but not which way, but if the spring is being compressed then the bigger block is moving towards the smaller block?

    There is a hint in the homework that says there is a non zero acceleration but I don't understand how you can find it?
     
  2. jcsd
  3. Mar 4, 2015 #2

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    If the spring is being compressed, then it gets smaller.
    This means that the crates get closer together ... so the big crate is being pushed towards the smaller one.
    Just go through the free body diagrams ... what are the forces on the big crate?
     
  4. Mar 4, 2015 #3
    Mg, the normal force, the applied force, and the force of the spring.
     
  5. Mar 4, 2015 #4

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Cool - so can you write Newton's law relations for the forces?
     
  6. Mar 4, 2015 #5
    Fapp - Fspring = ma?
     
  7. Mar 4, 2015 #6

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    You got the directions right anyway. Lets call the +ve direction the direction of the applied force - this is also the direction of the acceleration.
    Lets just call the applied force F, avoid subscripts = good, what is the equation for the force in the spring?

    (We can also use M for the mass of the big crate and m for the mass of the little one.)
     
  8. Mar 4, 2015 #7
    F=kx I believe
     
  9. Mar 4, 2015 #8

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Well done - so what is the newtons law relation for the big crate then?
     
  10. Mar 4, 2015 #9
    F-kx=ma?
     
  11. Mar 4, 2015 #10

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Sounds bashful, with the question mark...

    F - kx = Ma is correct. (I follow Simon's suggestion to call the big one's mass M).

    You know the values for k, x, and M, so to find F you need a second equation. There is one given we haven't used yet. Any chance to link it to an unknown with a suitable equation ?
     
  12. Mar 4, 2015 #11
    F=ma of the second block?
     
  13. Mar 4, 2015 #12

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Again this question mark ..

    Now we run the risk of confusion. F was the applied force in Simon's advice in post #6 (I'm not so much against subscripts, but never mind).

    m is the mass of the small crate

    We assumed that the spring compression is a constant 4.7 cm -- meaning the distance between the crates is constant -- meaning the speed of both crates is the same at all times -- meaning the acceleration of both crates is the same at all times (*)
    This means we can use the same a as we used in F - kx = Ma

    And we have the force the spring exercises on the small crate, that is: we have an expression for that force in terms of known variables. (Think of Newton's third law)

    So force = mass times acceleration can be expressed as an equation with one unknown, namely a: ##\qquad## kx = ma

    And now we are in business to work out F.


    (*)In fact that is pretty hard to achieve on a completely frictionless floor !
     
  14. Mar 4, 2015 #13
    So you use kx=ma to find the acceleration and then plug that in to F - ka = ma? I'm afraid I don't fully understand.
     
  15. Mar 4, 2015 #14

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    a Please stick to the correct notation. M was the big crate m was the small one. You want to eliminate a, so you want to substitute knowns for it.

    And there never can be ka to subtract from a force. It doesn't have the dimension of a force.

    But basically, yes. You use kx = ma to find the acceleration a and stick that a into F - kx = Ma .

    Your notation & typo (I hope) threw me off.

    Furthermore it's good practice to keep working with symbols as long as possible, preferably until you have one single answer expression:
    • factors and/or terms may cancel
    • less chance of calculation errors
    • possibility to check dimensions
     
    Last edited: Mar 4, 2015
  16. Mar 4, 2015 #15
    Oh sorry, I did not realize you had done that. Subscripts are much easier for me to work with, I think.

    kx = m2a
    a = kx/m2

    Then I plug that in to the other equation
    F - kx = m1(kx/m2)

    Is this the right thought process?
     
  17. Mar 4, 2015 #16

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Yes, well done.
     
  18. Mar 4, 2015 #17
    I'm not getting the correct answer with this, so I'm must be messing up in the algebra somewhere.

    F - (7.8 * 0.047) = 575 * ((7.8 * 0.047)/255)
    F = 1.19N
    Which is not the right answer here.

    Edit: nvm, had to convert kN/m, to N/m. Got it now. Thanks a bunch!
     
    Last edited: Mar 4, 2015
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted