Force required to keep block from sliding down (check)

  • Thread starter Fizic
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  • #1
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
 

Answers and Replies

  • #2
ehild
Homework Helper
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The result is correct. But in case of static friction, Ffric≤μFMm.
At the same time Ffric=mg.

ehild
 
  • #3
kptsilva
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correct
 
  • #4
PhanthomJay
Science Advisor
Homework Helper
Gold Member
7,179
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Looks good! Might want to round your answer to 490 N. I am not particularly thrilled about using [itex] F_mM = m/(m+M)F_A [/itex], although it is correct. Free body diagrams and the use of Newton's laws avoid having to use this equation.
 

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