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Two Blocks, Springs, and Friction

  1. Jan 7, 2007 #1
    1. The problem statement, all variables and given/known data
    This problem has actually been on here twice (with varying numbers) but not yet solved entirely

    A 30.0-kg block is resting on a flat horizontal table. On top of this block is resting a 15.0-kg block, to which a horizontal spring is attached, as the drawing illustrates. The spring constant of the spring is 345 N/m. The coefficient of kinetic friction between the lower block and the table is 0.640, while the coefficient of static friction between the two blocks is 0.940. A horizontal force F is applied to the lower block as shown. This force is increasing in such a way as to keep the blocks moving at a constant speed.


    At the point where the upper block begins to slip on the lower block determine the following.
    (a) the amount by which the spring is compressed
    (b) the magnitude of the force F

    2. Relevant equations
    [tex] F = kx [/tex]
    [tex]f = \mu F_N [/tex]
    [tex]F = ma [/tex]

    3. The attempt at a solution

    a) correctly solved by me (with no assistance)
    [tex] F_{spring} \geq f_{UL} [/tex]

    [tex]k x \geq \mu_{UL} m_u g [/tex]

    [tex]x \geq \frac{\mu_{UL} m_u g}{k} , k \gt 0 [/tex]

    plugging in my values, 0.40m

    b) most logically I need to figure out how [tex] F \propto f_{UL} [/tex] .. I can't quite figure out the interaction between the blocks.. which is quite frustrating

    as an alternative effort I focused on the concept of constant speed.. so concentrating on forces applied to the the lower block
    [tex] \Sigma F_x = 0 = F - f_{LT} = F - \mu_{LT} F_N [/tex]
    [tex]\Sigma F_y = - W_u - W_L + F_N = 0 [/tex]
    [tex]F_N = (m_u + m_l)g [/tex]
    [tex]F = \mu_{LT}(m_u + m_l)g [/tex]

    entering my numbers.. I get 28.2 N .. which .. according to WebAssign .. is wrong .. so any hints?

    when deciphering my subscripts .. U is upper box, L is lower box, T is table

    Thanks in advance for any help,
    Last edited: Jan 7, 2007
  2. jcsd
  3. Jan 7, 2007 #2


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    Welcome to the Forums,

    Your section (a) is correct, although you should be a little more accurate with your final answer (i.e. 3sf). Since, the applied force is horizontal we can ignore the vertical components of the forces. However, you are missing one vital force in your expression for the sum of the forces in the x direction.
  4. Jan 7, 2007 #3
    yea .. it should probably be written with more significant figures, if it were .. it would be 0.401 m

    As for the missing force, the only horizontal force I see that I have not included is the restoring force of the spring, however, this is applied to the upper block, not the lower. I'd imagine it would be relevant, if I were to consider the two blocks as a system, but then I am unsure how to handle the static friction between the blocks; is it omitted because it is internal to the system...

    or (eureka.. maybe) .. is it the friction applied to the top of the lower block by the upper block?!

    in that case it would be
    (for the purposes of getting a positive answer, as expected by the problem, towards the left is considered the positive direction)
    [tex] \left\begin{array}{rcl}\Sigma F_y & = & F + f_{UL} - f_{LT} = 0 \\
    F & = & \mu_{LT}(m_u + m_l)g - \mu_{UL} m_u g \\
    F & = & g [ \mu_{LT}(m_u + m_l) - \mu_{UL} m_u ] \\
    F & = & (9.8 \frac{\mbox{m}}{\mbox{s}^2})[ (0.640)(15.0\mbox{kg} + 20.0\mbox{kg}) - (0.940)(15.0\mbox{kg}) ] \approx 81.4\mbox{N} \end{array}\right

    actually.. on second thought considering those two things would be equivalent other than direction .. but I think the way I've calculated it is correct.

    Can anyone verify my calculation?

    Thanks for help so far and thanks for any forthcoming help,
    Last edited: Jan 7, 2007
  5. Jan 7, 2007 #4


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    Don't you just love eureka moments :biggrin: Your almost there, but there is one mistake. In which direction is the frictional force fUL acting on the lower block?
  6. Jan 7, 2007 #5
    hmm .. yea .. I've been debating that

    .. its opposite the intended motion of the lower block .. so .. in effect to the right .. or as I set it before, negative.. which of course .. for the moment considered, makes it precisely the same as the restoring force of the spring

    therfore making it .. F = fUL + fLT

    which would be ... 358 N ... that seems large-ish (although, maybe not cause I really have no clear concept of exactly how much force 1 N is )

    While it would be nice if I could a clandestine "yes thats right," I understand that could be seen as cheating ... so .. before I go and enter that for my final answer (of my last submission) .. one last concern.. am I correct in thinking that fLT is a function of the weight of both blocks?

    anyway, thanks for all the help, Hootenanny:biggrin:
    Last edited: Jan 7, 2007
  7. Jan 7, 2007 #6


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    Correct! In fact as an aside, an alternate method would be to use the compression of the spring you calculated in section (a). As this is the point at which the blocks begin to slip we can say that;

    [tex]F - F_{spring}-f_{LT} = 0[/tex]
    Well, I ain't actually punching the numbers into the calculator :yuck:, but your method seems about right. Yes, fLT is a function of the weight of both blocks. And I think you've done enough work yourself (in fact very nearly all of it) for it not to be considered cheating :approve:

    As for the thanks, thank you and it was my pleasure :smile:
  8. Jan 7, 2007 #7
    :cry: apparently not..

    it got marked wrong (and we only have two submissions cause our teacher is mean like that)

    I wonder what we/I did wrong? oh well, it probably doesn't matter grade-wise cause its the only one I've missed so far this grading period and he curves it twice (he gives us a few free misses .. last quarter five .. but that was out of more so probably less, then he does a square root * 10 curve) (and yet most people still get a C or D).

    Anyway either way thanks for trying .. I'll ask what I did wrong and post the correction here on Tuesday night (if I remember), in case you want to know.

    Done with physics for now, off to do Calculus, US History, English, O-chem, and an application for a ECE-type research mentorship at NASA Langley,

  9. Jan 8, 2007 #8
    No .. apparently the method is correct (verified with friends who have gotten it correct) and I reran numbers and got 421 ... so apparently I just fail at math

    Specifically .. I should have noted that [tex] 20 \mbox{kg} \neq 30 \mbox{kg} [/tex]

    and yes its a double post, but its new info
    Last edited: Jan 8, 2007
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