Maximum force on 1.0 kg block before 0.5 kg block slips

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Fusilli_Jerry89
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A 0.5 kg mass is on top of a 1.0 kg mass on a table. The static coefficient of friction between the blocks is 0.35, and the kinetic coefficient of friction between the table and1.0 kg block is 0.20. What's the maximum force that can be applied horizontally to the 1.0 kg block without letting the 0.5 kg block slip?

the acceleration of the small block has to be zero, so the friction between the two blocks cannot exceed 1.715 N.I got the equation for the first blockdown to F-3.675=1.0a but what do I do after that? Does the acceleration of the large block have to be zero as well, because if it does then the answer is 3.675 N.
 
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Fusilli_Jerry89 said:
the acceleration of the small block has to be zero, . . . .
No it does not. Think about the motion that will occur if there is no slipping.
 
well won'tthe accelration of the small block havet to match that of the big one? Wouldn't that mean it is 0 in relation to the big block?
 
Fusilli_Jerry89 said:
well won'tthe accelration of the small block havet to match that of the big one? Wouldn't that mean it is 0 in relation to the big block?
Yes they have to be the same. Yes that means it is zero relative to the big block, but the big block is accelerating. You don't know how to do problems in accelerating reference frames. Do the problem in the laboratory frame. Both blocks will be accelerating.
 
Fusilli_Jerry89 said:
So is the answer 7.105 Newtons then?
I don't think so. How did you get it?
 
well i found theat the equation for the large block was:
F-Ff-Ff=ma
F-1.96-1.715=1.0a
F-3.675=a

and for the small block:
Ff=ma
1.715=0.5a
a=3.43

then I substituted a into the first equation because they are accelerating at the same rate
F-3.675=3.43
F=7.105 N
 
Why 1.96? What is the normal force acting on the bottom of the 1kg block?
 
woops i forgot to add the 0.5 kg. So it would be 2.94 N instead?
 
Fusilli_Jerry89 said:
woops i forgot to add the 0.5 kg. So it would be 2.94 N instead?
That looks better.