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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.

F

_{grav}

_{m}=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