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Fizic
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Homework Statement
The two blocks (with m=16kg and M=88kg) shown in Fig. 6-56 are not attached. The coefficient of static friction between the blocks is [itex]\mu[/itex]stat=0.38, but the surface beneath the larger block is frictionless. what is the minimum magnitude of the horizontal force [itex]\vec{F}[/itex] required to keep the smaller block from slipping down the larger block?
Fig. 6-56 looks approximately like this.
...______ ________
...|...||...|
[itex]\vec{F}[/itex].|..m...||...|
...|_____||...M...|
.....|...|
.....|_______|
_________________________________
FRICTIONLESS GROUND
Homework Equations
[itex]\sum[/itex][itex]\vec{F}[/itex]=m[itex]\vec{a}[/itex]
[itex]\vec{F}[/itex]fric=[itex]\mu[/itex]stat[itex]\vec{N}[/itex]
The Attempt at a Solution
I denote [itex]\vec{F}[/itex] from the problem text as [itex]\vec{F}[/itex]A
I denote the force of block m and block M, [itex]\vec{F}[/itex]mM
Likewise, the force of block M and block m is denoted [itex]\vec{F}[/itex]Mm. I treat this as the "normal" force for the friction equation, but I'm not sure if this is correct.
Fgravm=mg=156.8 N
[itex]\vec{a}[/itex]=[itex]\vec{F}[/itex]A/104
[itex]\vec{F}[/itex]mM=[itex]\vec{F}[/itex]Mm=[itex]\frac{88}{104}[/itex][itex]\vec{F}[/itex]A
[itex]\vec{F}[/itex]fric=[itex]\mu[/itex][itex]\vec{F}[/itex]Mm=.38*[itex]\frac{88}{104}[/itex][itex]\vec{F}[/itex]A=.322[itex]\vec{F}[/itex]A
To prevent downward motion [itex]\vec{F}[/itex]fric must cancel out (i.e. be at least) [itex]\vec{F}[/itex]gravm
[itex]\vec{F}[/itex]fric=[itex]\vec{F}[/itex]gravm
.322[itex]\vec{F}[/itex]A=156.8N
[itex]\vec{F}[/itex]A=487.7 N
Final answer 487.7 N
Thanks