TBR Translational Motion

[Chocolatebear89]

I have a question on Passage 1 of Translational Motion of Berkeley Review.

Its a pully system where one block with mass M1 is on a surface tied to another block with mass M2 that is hanging.

#5 If the force of kinetic friction acting on M1 goes to zero, the resulting NET forces on M1 and M2 will be:

A. less than M1g and less than m2g, respectively
B. less than M1g and more than M2g, respectively
C. more than M1g and less than M2g,respectively
D. more than M1g and more than M2g, respectively

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[collegestud2013]

if there's no kinetic friction then the acceleration of the system is:

a = F/m = M2g/(M1 + M2)

but the net force on M2 should be F2 = [M2/(M1 + M2)](M2g) < M2g
and the net force on M1 should be F1 = [M2/(M1 + M2)](M1g) < M1g

so my guess would be choice A

 

[Chocolatebear89]

How did you come to the conclusion that net force on M2 should be F2 = [M2/(M1 + M2)](M2g)?

 

[fizzgig]

Net force on 2(my axes point left and down)
m2g-T=m2a
Net force on 1
T=m1a

combine (acceleration of 1 and 2 are the same - they move together)
m2g-m1a = m2a
m2g = m2a+m1a
g*m2/(m2+m1) = a

now, F=ma for both 1 and 2
Fon1 = m1*(g*m2/(m2+m1)) = m1g * fraction less than 1
Fon2 = m2*(g*m2/(m2+m1)) = m2g * fraction less than 1