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Coefficient of kinetic friction 
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#1
Oct109, 04:59 PM

P: 128

1. The problem statement, all variables and given/known data
When the three blocks in Fig. 633 are released from rest, they accelerate with a magnitude of 0.800 m/s^2. Block 1 has mass M, block 2 has 2M, and block 3 has 2M. What is the coefficient of kinetic friction between block 2 and the table? 2. Relevant equations F_f = mu_k*F_N 3. The attempt at a solution So far I have: The xcomponent of the force on the second block is equal to the sum of the ycomponent of 1, the ycomponent of 3, and the friction between the table and 2 : F_2,x = 2*M(0.8) = F_1,y + F_3,y + F_f For friction, we have: F_f = mu_k*F_N = mu_k*2*M*g Since the normal force is just gravity in this case. Substituting, we have F_2,x = 2*M*g  (M*g + mu_k*2*M*g) = (1  2*mu_k)*M*g Since the acceleration is 0.8 then we solve for mu_k F_2,x = (0.8)*(2.0)*M = 1.6*M = (1  2*mu_k)*M*g 1.6 = (9.8)*(1  2*mu_k) (1/2)*((1.6/9.8)  1) = mu_k (1/2)*(8.2/9.8) = 0.418 But this answer isn't right. I have a hunch I messed up the beginning by assuming that F_2,x = 2*M(0.8) = F_1,y + F_3,y + F_f But I'm not sure exactly. 


#2
Oct109, 09:38 PM

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P: 3,440

You need to draw three separate free body diagrams for each block. An important item that you forgot to include is the tension in the ropes. Note that the tension in the right rope is not the same as the tension in the left rope. In fact the difference in the tensions minus the force of friction is the net force on the middle block. Sorry, you have to start all over again.



#3
Oct109, 09:58 PM

P: 128

I should probably draw up another diagram, but I was just doing another, much simpler problem, and got it wrong because I don't fully understand the way these two forces interact. Here's the problem I got wrong: (a) An 11.0 kg salami is supported by a cord that runs to a spring scale, which is supported by a cord hung from the ceiling (Fig. 535a). What is the reading on the scale, which is marked in weight units? (b) In Fig. 535b the salami is supported by a cord that runs around a pulley and to a scale. The opposite end of the scale is attached by a cord to a wall. What is the reading on the scale? (c) In Fig. 535c the wall has been replaced with a second 11.0 kg salami, and the assembly is stationary. What is the reading on the scale? Now the first two are fairly obvious, which gives 108 N for both the freely hanging body and the body attached to a cord, by a pulley. However, for the third question, I assumed that the two forces would combine to produce a total of 216 N on the scale. However, the book says it's also 108 N. This is kind of lost on me because there should be a force in both directions, away from the scale, causing it to read double. I've looked through the book and I don't quite see an example that explains this phenomenon. 


#4
Oct109, 10:10 PM

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Coefficient of kinetic friction



#5
Oct109, 10:13 PM

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#6
Oct109, 10:15 PM

P: 128

What would happen if we put two scales there, one facing each direction.. Would each read 108 or would each read 54? 


#7
Oct109, 10:18 PM

P: 128




#8
Oct109, 10:20 PM

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#9
Oct109, 10:51 PM

P: 128




#10
Oct209, 12:13 AM

P: 128

This guy seems to have had the same problem I do:
http://www.physicsforums.com/showthread.php?t=17376 It seems to me that tension is only a onesided thing, despite the fact that Newton's Third Law says it ought to exist in both directions. 


#11
Oct209, 08:11 AM

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Look at it this way. Suppose you are looking at one of the salamis wondering when it will be time to eat them and I sneak unseen over to the other side of the table, hold the string with one hand and cut it with the other then tie it to the wall so that you get picture (b). I have stolen your salami but as far as you are concerned, nothing has changed. It doesn't matter whether the other side of the string is tied to the wall, is held by a person or another salami. As long as the salami does not accelerate, the net force on it is zero which means that the tension is equal to the weight. 


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