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F=ma (Euler form)

  1. Feb 4, 2012 #1
    1. The problem statement, all variables and given/known data
    The 200-lb block is at rest on the floor (u=0.1) before the 50-lb force is applied as shown in Figure 1. What is the acceleration of the block immediately after application of force? Assume the block is wide enough that it cannot tip over.


    2. Relevant equations

    F=ma

    3. The attempt at a solution
    See my free body diagram (FBD attachment).
    I setup the equations:

    -fi+Fcos30i+Nj-mgj-Fsin30j=m(x''i+y''j+z''k) (i,j,k are unit vectors, x'',y'',z'' are double dot or acceleration)

    Thus I get two equations:
    Fcos30-f=mx''
    mg+Fsin30=N

    I plug in for the second equation and get N=6425. Coefficient of static friction is 0.1, so my max static friction is 0.1*6425 = 642.5. So if I go back to the first equation, it looks like the block doesn't move despite the 50lb force. But the book gives the answer as 3.55ft/s^2 in the positive x direction, so I screwed up somewhere. But I can't see it. Any help would be appreciated.
     

    Attached Files:

  2. jcsd
  3. Feb 4, 2012 #2

    cepheid

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    Imperial units. Yuck. :yuck: Is the 200 pounds the weight of the block, or is it the mass of the block? In other words, is that unit pounds-force, or pounds-mass? I would have assumed the former, for consistency, since other forces in the problem are given in lbs.

    If so, then I should point out that the force balance in the y-direction says that:

    200 lbs + (50 lbs)sin(30 deg) = 225 lbs.

    Certainly not 6425 lbs.
     
    Last edited: Feb 4, 2012
  4. Feb 4, 2012 #3
    You're right. So it is 225 lbs for N, and thus friction is 0.1*225 = 22.5 pounds. But we know the F in the x direction is Fcos30 = 43.3 pounds, so the box moves and hence there is acceleration. So I plug in 22.5 for f in my first equation and get x''=0.104 which isn't right. Where am I going wrong here? Thanks.
     
  5. Feb 4, 2012 #4
    If one takes a quantity in force-pounds and divides by mass-pounds, the result is not acceleration in ft/s2.

    You could first convert the mass to slugs.

    Or you could convert the problem quantities to metric, and then convert the answer back to ft/s2.
     
    Last edited: Feb 4, 2012
  6. Feb 4, 2012 #5
    No, for this problem just assume 200 lb block = mg. I looked over and I think that's right. No need to over complicate it in the context of this problem.
     
  7. Feb 4, 2012 #6

    cepheid

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    You're right: it's not. If you do that, the result is acceleration in units of g. The reason for this is that pounds-force and pounds-mass are defined in such a way that an object will have the same mass in lbm as its weight in lbf on EARTH. So if you multiply your answer by g = 32.174 ft/s2, you'll get it in units of ft/s2, and the result is about 3.346 ft/s2.

    Yes, equivalently, you could first divide the 200 lbf by g to get mass in slugs, and then compute the result using

    a = F/m,

    where m is in slugs now (instead of using 200 lbm). You'll still get the same result though: 3.346 ft/s2.

    The two methods are arithmetically equivalent: in the second method, you divide the mass by 32.174 in order to convert it from lbm to slugs. However, since mass appears in the denominator, this is equivalent to multiplying the final answer for acceleration by 32.174, which is what you do in the first method to convert from units of g to units of ft/s2.

    Try as I might, I haven't found any way to reproduce an answer of 3.55 ft/s^2. It's close but not quite. Maybe the book is just wrong.
     
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