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Homework Help: Static Friction and Frictional Force Ranking Task of Crate

  1. Oct 14, 2011 #1
    Below are six crates at rest on level surfaces. The crates have different masses and the frictional coefficients [the first value is the static friction and the second is the kinetic friction] between the crates and the surfaces differ. The same external force is applied to each crate, but none of the crates move.
    Box 1= 600kg (Static friction-0.8)(Kinectic friction-0.5)
    Box 2=750kg (Static friction-0.6)(Kinetic Friction-0.5
    Box 3=1500kg(Static friction-0.3)(Kinetic friction-0.1)
    Box 4=500kg(Static Friction-0,6)(Kinectic friction-0,3)
    Box 5=750kg(Static Friction-0.4)(Kinectic Friction-0.3)
    Box 6=250kg(Static Friction-0.2)(Kinetic Friction-0.1)

    I need to rank the crates on the basis of the frictional force acting on them

    The equation I would use here is the static friction equation (Static Friction is less than or equal to the miu times the normal force)

    Since the boxes arent moving I would just multiply the mass of the box times the coefficient of the static friction (For example the first box would be 600x0.8 and then rank those numbers i get from greatest to least correct? However I tried this and it doesnt like my answer. Am i doing something incorrect?
  2. jcsd
  3. Oct 14, 2011 #2


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    If the crates don't move, and the external force applied to each is the same, and the static friction is less than or equal to uN, use Newton 1 to calculate the friction force acting on each one. Whch one, if any, has the greater friction force acting on it???
  4. Oct 14, 2011 #3
    I calculated for example the first box to be 600kg x the static coeffiecient which is 0.8. I have done this for all the boxes and my order from least to greatest was the 600 kg box being the greatest frictional force acting on it, then stacking box 2 and 3 because they had the same frictional force, and then stacking boxes 4 and 5 becuase they also had the same number and finally the least box was box 6 . To get the frictional force on these boxes that do not move i should use the static coefficient number given rather than the kinetic friction. So i did this for all these boxes and it did not like my order of answers I guess i am just still confused
  5. Oct 14, 2011 #4
    that order is from greatest to least, my apologies
  6. Oct 14, 2011 #5


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    Although you clearly stated in your first post that
    , you since have stated that
    Do you see the difference between these 2 statements? Apply Newton's first Law to each box. I don't understand how you can 'order' the choices, unless I've misunderstood the problem statement or you have written it incorrectly.
  7. Oct 15, 2011 #6


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    The force of static friction is given by the following relation:
    [itex]F_\text{static friction }\le\mu_sN\,.[/itex]​

    Why is there an inequality sign, ≤, rather than an equal sign, = ?

    Can the force of static friction be greater than the external force being applied.
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