What is the coefficient of kinetic friction and tension in this physics problem?

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In summary, the system has an acceleration of 1.20 m/s2 and the coefficient of kinetic friction between the block and incline is the same. The tension in the string is 60.6 N.
  • #1
kulai
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Homework Statement


The system shown in Figure 4.54 has an acceleration of magnitude 1.20 m/s2, where m1 = 3.70 kg and m2 = 7.60 kg. Assume that the coefficient of kinetic friction between block and incline is the same for both inclines.

http://www.webassign.net/serpop/p4-54alt.gif

(a) Find the coefficient of kinetic friction.
(b) Find the tension in the string.


The Attempt at a Solution



ive tried to compile all the forces into one equation and it turned out like this:

Fk on m1 + Fk on m2 + Fg m1 - Fg m2 = Fnet
n1*muk + n2*muk + m1g - m2g = m2a

-please i need help as fast as possible
 
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  • #2
The angle matters. You shouldn't have m1g and m2g, but rather the components of those forces that are along the slope. The n1 and n2 will also be components of the mg forces, so they'll have sin 35 or cos 35 in their expressions, too.

Welcome to PF!
 
  • #3
ya i know. sorry i didnt type it in more detail cause i was late for another class. if i added the numbers to it, it came out

7.6*9.8*cos35*muk +3.7*9.8*sin35 + 3.7*9.8*cos35*muk - 7.6*9.8*sin35 = 9.12

-please tell me if I'm missing anything or if my numbers are off. thank you for the fast responses
 
  • #4
For the ma on the right, your number doesn't seem right.
 
  • #5
Delphi51 said:
For the ma on the right, your number doesn't seem right.

i multiplied the acceleration given and the mass of m2 because its the block pulling the the other. i tried adding m1 and m2 together and using that for the net force but i still go the answer wrong
 
  • #6
ma = (3.7 + 7.6)*1.2 = 13.56
I'd say the signs on the left side are all wrong. Friction should be negative.
 
  • #7
ok so I've got

7.6*9.8*sin35 - 7.6*9.8*cos35*muk - 3.7*9.8*sin35 - 3.7*9.8*cos35*muk) = 13.56

when i factor out the muk n balance out the equation i get muk = 0.0919. is that what you got? sorry I'm being really picky about this but i have to make sure because this is my last attempt on this problem. I'm really grateful for all the help
 
  • #8
I actually got .0924, but I use g=9.81 so we probably agree.
 
  • #9
ok it was right. thank you so much. now i have to do work on getting tension on the string. I am not too sure how to start it. i kind of tried mixing the components around to make it

(m2g - n2muk) + (m1g + n1muk) = T

but my answer ends up being 60.6N and I'm very sure that's not right.
 
  • #10
Recommend you concentrate on just m1. The sum of the forces parallel to the ramp equals ma.
 

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