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.