Force required to keep block from sliding down (check)

[Fizic]

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 $\mu$^stat=0.38, but the surface beneath the larger block is frictionless. what is the minimum magnitude of the horizontal force $\vec{F}$ required to keep the smaller block from slipping down the larger block?

Fig. 6-56 looks approximately like this.

...______ ________
...|...||...|
$\vec{F}$.|..m...||...|
...|_____||...M...|
.....|...|
.....|_______|
^{_________________________________}
FRICTIONLESS GROUND

Homework Equations

$\sum$$\vec{F}$=m$\vec{a}$

$\vec{F}$_fric=$\mu$^stat$\vec{N}$

The Attempt at a Solution

I denote $\vec{F}$ from the problem text as $\vec{F}$_A

I denote the force of block m and block M, $\vec{F}$_mM

Likewise, the force of block M and block m is denoted $\vec{F}$_Mm. I treat this as the "normal" force for the friction equation, but I'm not sure if this is correct.

F_gravm=mg=156.8 N

$\vec{a}$=$\vec{F}$_A/104


$\vec{F}$_mM=$\vec{F}$_Mm=$\frac{88}{104}$$\vec{F}$_A

$\vec{F}$_fric=$\mu$$\vec{F}$_Mm=.38*$\frac{88}{104}$$\vec{F}$_A=.322$\vec{F}$_A

To prevent downward motion $\vec{F}$_fric must cancel out (i.e. be at least) $\vec{F}$gravm

$\vec{F}$_fric=$\vec{F}$gravm
.322$\vec{F}$_A=156.8N
$\vec{F}$_A=487.7 N

Final answer 487.7 N

Thanks

 

[ehild]

The result is correct. But in case of static friction, F_fric≤μF_Mm.
At the same time F_fric=mg.

ehild

 

[kptsilva]

correct

 

[PhanthomJay]

Looks good! Might want to round your answer to 490 N. I am not particularly thrilled about using $F_mM = m/(m+M)F_A$, although it is correct. Free body diagrams and the use of Newton's laws avoid having to use this equation.