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A 5.0-kg block rests on a 30 degree incline. The coefficient of static friction between the block and the incline is 0.20. How large a horizontal force must push on the block if the block is to be on the verge of sliding (a) up the incline and (b) down the incline?

Ans. (a) 43 N (b) 16.6 N

Now how does one get these answers...

By Fairyofwind (Fairyofwind) on Sunday, April 25, 2004 - 10:18 am: Edit

bump

By Patmanj20 (Patmanj20) on Sunday, April 25, 2004 - 03:48 pm: Edit

do u have an e-mail address, i can send u the work

By Jason817 (Jason817) on Sunday, April 25, 2004 - 03:54 pm: Edit

For part b, its:

force= mgsin(theta) - umgcos(theta) where u=coefficient of friction

By Jason817 (Jason817) on Sunday, April 25, 2004 - 03:55 pm: Edit

and for part a, dont you just switch sin and cos in the above equation?

By Fairyofwind (Fairyofwind) on Sunday, April 25, 2004 - 04:52 pm: Edit

that doesn't work...
force= mgsin(theta) - umgcos(theta) = 16.03

Furthermore, I think the "horizontal force" has to be parallel to the ground or something like that. but when you divide 16.03 by cos 30 you get 18.something, which isn't 16.6....

my email is in my profile

By Jason817 (Jason817) on Sunday, April 25, 2004 - 05:16 pm: Edit

well then, I want to know how to do it. Can you plz post here if you ever get the answer.

By Fairyofwind (Fairyofwind) on Sunday, April 25, 2004 - 08:20 pm: Edit

help!

By Beero1000 (Beero1000) on Sunday, April 25, 2004 - 09:43 pm: Edit

1.
its Fhcos(theta)=mgsin(theta) + umgcos(theta) + uFhsin(theta)
solve for Fh = 43

By Beero1000 (Beero1000) on Sunday, April 25, 2004 - 09:47 pm: Edit

2
Fhcos(theta) + mgsin(theta) = umgcos(theta) - uFhsin(theta)
Fh = 16.6

By Jason817 (Jason817) on Sunday, April 25, 2004 - 10:08 pm: Edit

can you explain the concept behind that please? Why the hell do you add 3 things

By Beero1000 (Beero1000) on Sunday, April 25, 2004 - 10:32 pm: Edit

for part a - where you are pushing the block up the incline the forces are Fn, mg, F, Ff, and Fh
F and Ff both point in the same direction - down the incline so you add them. Ff = uFncos(theta) and F = mgsin(theta)
the part of fh that is parallel to the incline is Fhcos(theta)
that has to equal F + Ff for it to be on the verge of sliding up
so its Fhcos(theta) = F + Ff = mgsin(theta) +uFncos(theta)
Fn = mg but its increased by Fhsin(theta) because the Fh pushes down on the block increasing Fn
so Fhcos(theta) = mgsin(theta) + umgcos(theta) + uFhsin(theta) Fh = 43

By Beero1000 (Beero1000) on Sunday, April 25, 2004 - 10:35 pm: Edit

part b
now the parallel component of Fh is working with F against Ff
so its
Fhcos(theta) + mgsin(theta) = uFn = umgcos(theta) - uFhsin(theta)
because its lifting up and decreasing fn