Ace school

with brainly

  • Get help from millions of students
  • Learn from experts with step-by-step explanations
  • Level-up by helping others

A community for students.

Forces: You discover that it takes 258.0 N to set a 40.0 kg crate in motion on a concrete surface. What is the coefficient of static friction for the crate?

Physics
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Join Brainly to access

this expert answer

SIGN UP FOR FREE
@ swin u need to know what plane u are working on is it an incline plane or an horizontal plane/
coefficient of static friction=static frictional force/ normal rection
I assume the plane is horizontal: the normal reaction=weight=mg ur

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

got it buddy?
@hubertH it doesn't say. so i assumed it was horrizontal
so u knw how to get to the answer right?
but is static friction only involved in inclined plane?
No.. then I wouldn't be asking lol
No! static friction is the force required to set the body in motion in our case it 's given as 258.0N
Well they're asking from the coefficient of static friction?
coefficient of static friction=static friction force/normal reaction
for horizontal plane normal reaction=mg
|dw:1351982204492:dw|
\[mumg?\]
|dw:1351982369737:dw|
do u now have an idea of what u're doing?
so i say that Fnet = \[\mu _{s} - Fg = ma?\] a little do you?
remember u're given the Fs and the mass only
there is no Fa, no Fnet
well if it's starting to move wouldn't it be Ff = Mu(s) Fn?
\[mus =\frac{ ?Fmus }{Nr ? }\]
I have never used Nr in my FBD before?
Nr is ur normal reaction=mg
Ok so it's just normal force (Fn)
so it's Ff/Fn
exactly!
and since it's horizontal Fnet = 0N
what is Fnet standing fo?
The vector pointing upward
Ok i got it, Mu(s) = .66
that's it bro congrats and keep on going forwad
If they ask for acceleration and give kinetic friction is it still the same equation except it's Mu(k) - Fn = ma?
no! kinetic friction is the smallest force that keeps the system moving @ constant speed
Ok...
if u have more questions on science don't hesitate to massage me.....
alright because i have a big physics test and I don't really understand forces lol
we'll try to make it easier together bye for now

Not the answer you are looking for?

Search for more explanations.

Ask your own question