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Friction problemby tony873004
Tags: friction 
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#1
Nov104, 02:59 PM

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PF Gold
P: 1,542

Figure 636: http://www.webassign.net/walker/0636alt.gif
5. [Walker2 6.P.071.] Two blocks, stacked one on top of the other, slide on a frictionless, horizontal surface (Figure 636), where M = 5.0 kg. The surface between the two blocks is rough, however, with a coefficient of static friction equal to 0.47. If a horizontal force F is applied to the bottom block, what is the maximum value F can have before the 2.0 kg top block begins to slip? The answer in the back of the book is 32 N. But that's not what I get. I get the same wrong answer 2 different ways: friction = mu * mg friction = 0.47 * 2 * 9.81 friction = 9.2214 N at this point, I'm stuck. I'd like to say that if pushing the 2 kg block with a force of 9.2214 will move it, then how much harder would I have to push a 5 kg block to make its push equivalent to the 2 kg push. 9.2214 * (5/2) = 23.0535 which is the wrong answer. Or I could do it like this: f=ma 9.2214 = 2 * a a = 4.6107 now use this for a for the 5 kg block F = ma F = 5 * 4.6107 F = 23.0535 which is exactly what I got before and it is wrong Any thoughts...?? 


#2
Nov104, 03:15 PM

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P: 41,315

You are very close. What is the maximum acceleration that the top block can have? (You already figured that outthat's when the frictional force on it is at maximum.) Now treat the two blocks as a single object (why not?). What force F is required to accelerate both blocks to that value?
(One of your mistakes was using F = ma, but not using the net force on the object. If you treat the bottom block by itself  nothing wrong with that!  don't forget that there are two horizontal forces on it. Recall Newton's 3rd law.) 


#3
Nov104, 03:31 PM

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PF Gold
P: 1,542

lol... I feel so stupid for not realizing that I'm pushing 7kg, and not 5 !!! I stared at this for an hour. 


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