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 2008-11-18, 07:18 #1 Joshua2     Sep 2004 13·41 Posts Physics problem Please I know this is math, but I don't know any physics forum, and I've been a member here a long time and like these people. Anyway, I think it is mostly a math problem. The lower block in the figure is pulled on by a rope with a tension force of 20 N. The coefficient of kinetic friction between the lower block and the surface is 0.30. The coefficient of kinetic friction between the lower block and the upper block is also 0.30. What is the acceleration of the 2.0 kg block? http://jjoshua2.googlepages.com/a.p8.28.jpg Thats what it looks like. The answer is 1.77 m/s^2 the book says and that is what I got. Now on the test there is an almost identical problem, same picture. http://jjoshua2.googlepages.com/knight_Figure_08_28.jpg A rope pulls on the lower block in the figure with a tension force of 20 N. The coefficient of kinetic friction between the lower block and the surface is 0.37. The coefficient of kinetic friction between the lower block and the upper block is also 0.37. What is the acceleration of the 2.0 kg block? It says the answer is 2.7, which I guessed since it was the lowest answer, but I think the answer should be lower than 1.77, I calculated about 0.6. I can ask my teacher in a couple days, but she'll probably get mad at me if I bother her with what should be a non-issue. Thanks so much!
 2008-11-18, 08:24 #2 hockmeng     Oct 2008 2×7 Posts I tried the problem but I couldn't get 1.77 m s$^{-2}$ for the first question. Looks like I'm having some problem with my physics concepts. Would you mind showing me how you got that figure? Thanks.
 2008-11-18, 09:07 #3 axn     Jun 2003 132·29 Posts Code: (20 - 4*9.8*0.37)/2 = 2.748 Is this not how it should be calculated? EDIT:- Probably should be (20 - 4*9.8*0.37)/3 = 1.8 Last fiddled with by axn on 2008-11-18 at 09:17
 2008-11-18, 14:45 #4 davieddy     "Lucan" Dec 2006 England 2·3·13·83 Posts Applying Newton's 2nd law to the top block we get T - 9.8*1*0.37 = 1*a For the bottom block we get 20 - T - 9.8*1*0.37 - 9.8*3*0.37 = 2*a Adding the equations gives 20 - 9.8*5*0.37 = 3*a David Sorry to be too explicit for the homework help roolz :) Last fiddled with by davieddy on 2008-11-18 at 15:18
 2008-11-18, 18:31 #5 Joshua2     Sep 2004 13·41 Posts 20 - 2*9.8*1*mu - 9.8*3*mu = 3*a is how I thought of it, giving 0.62 for the 2nd and 1.77 for the first. So I think the 2nd answer is wrong, right?
2008-11-18, 20:41   #6
davieddy

"Lucan"
Dec 2006
England

194A16 Posts

Quote:
 Originally Posted by Joshua2 20 - 2*9.8*1*mu - 9.8*3*mu = 3*a is how I thought of it, giving 0.62 for the 2nd and 1.77 for the first. So I think the 2nd answer is wrong, right?

 2008-11-18, 21:05 #7 Joshua2     Sep 2004 13×41 Posts Right, I was agreeing with yours. I was saying that the tests answer of 2.7 isn't correct and doesn't agree with the similar problem with mu of 0.3 the answer being 1.77. :)
 2008-11-18, 21:49 #8 davieddy     "Lucan" Dec 2006 England 2×3×13×83 Posts Yes. More friction means lass acceleration. Note that we are assuming motion in the direction of the 20N force, Last fiddled with by davieddy on 2008-11-18 at 21:56
 2008-11-28, 19:25 #9 davieddy     "Lucan" Dec 2006 England 2×3×13×83 Posts ou could make this "problem" more interesting by speifying a moment of inertia, radius and frictional torque to the pulley. BTW the problem should have specified g (which we have been taking as 9,8 m/s^2) David
2009-03-02, 03:39   #10
flouran

Dec 2008

15018 Posts

Quote:
 Originally Posted by Joshua2 I know this is math, but I don't know any physics forum
Ever tried http://www.physicsforums.com?

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