Solve Friction Problem: Max Force on 1.0 kg Block

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The discussion centers on calculating the maximum horizontal force that can be applied to a 1.0 kg block without causing a 0.5 kg block on top to slip. The static friction between the blocks is 0.35, resulting in a maximum friction force of 1.715 N. Participants clarify that both blocks will accelerate together, and the acceleration of the smaller block must match that of the larger block. The correct normal force acting on the 1.0 kg block is 2.94 N after accounting for the weight of the 0.5 kg block. Ultimately, the calculations lead to a reevaluation of the forces involved to find the accurate maximum force.
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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?
 
So is the answer 7.105 Newtons then?
 
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?
 
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Fusilli_Jerry89 said:
woops i forgot to add the 0.5 kg. So it would be 2.94 N instead?
That looks better.
 
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