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Force pushes three blocks, find mag of force exerted on last two blocksby gap0063
Tags: forces friction 
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#1
Sep2410, 01:37 PM

P: 66

1. The problem statement, all variables and given/known data
Three blocks in contact with each other are pushed across a rough horizontal surface by a 75 N force as shown. The acceleration of gravity is 9.8 m/s2 . If the coefficient of kinetic friction between each of the blocks and the surface is 0.076, find the magnitude of the force exerted on the 3.5 kg block by the 4.1 kg block. Answer in units of N. 2. Relevant equations sum of Fy=Nmg=0 => N=mg sum of Fx= fk=max fk=uN 3. The attempt at a solution I have no clue how to figure out the force just between those two block... Do I add the first two blocks together to act a one force against the third block? 


#2
Sep2410, 01:48 PM

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P: 41,579

Hint: Start by treating the three blocks as a single system. What can you deduce?



#3
Sep2410, 01:59 PM

P: 66

Then the forces acting in the y direction is N1+N2+N3 and (m1+m2+m3)g forces acting in the x direction is fk and F which fk=u*(N1+N2+N3) right? 


#4
Sep2410, 02:01 PM

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P: 41,579

Force pushes three blocks, find mag of force exerted on last two blocks



#5
Sep2410, 02:17 PM

P: 66

so N1+N2+N3=Mg=> where N1+N2+N3=94.08 so the sum of Fx= fk=Ma fk=u(N1+N2+N3) fk=7.15008 so 7.15008=9.6a a=0.7448 m/s^2 this is small... is this right? 


#6
Sep2410, 02:29 PM

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P: 41,579

But ΣF ≠ fk. Friction is not the only horizontal force acting. 


#7
Sep2410, 02:46 PM

P: 66




#8
Sep2410, 03:02 PM

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P: 41,579

What other horizontal forces act? Look at your diagram. 


#9
Sep2410, 03:03 PM

P: 66




#10
Sep2410, 03:10 PM

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P: 41,579




#11
Sep2410, 03:14 PM

P: 66

7.15008+75=9.6a => a=7.0677 m/s^2 (wrong) what did I do wrong? 


#12
Sep2410, 03:18 PM

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P: 41,579




#13
Sep2410, 03:19 PM

P: 66

So can I do Sum of F= (m2+m3)*a? 


#14
Sep2410, 03:24 PM

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P: 41,579




#15
Sep2410, 03:28 PM

P: 66

in the y direction it is N and m3g So for the force in the x direction from m1 and m2 (i'll call P12) FP12=m3a? I'm not sure how to calculate P12... 


#16
Sep2410, 03:32 PM

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P: 41,579




#17
Sep2410, 03:39 PM

P: 66

and P12 is my force bwtween m2 and m3 right? 


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