# 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.

Fgravm=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

Thanks

Homework Helper
The result is correct. But in case of static friction, Ffric≤μFMm.
At the same time Ffric=mg.

ehild

kptsilva
correct

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.