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anonymous
 4 years ago
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?
anonymous
 4 years ago
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?

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@ swin u need to know what plane u are working on is it an incline plane or an horizontal plane/

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0coefficient of static friction=static frictional force/ normal rection

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I assume the plane is horizontal: the normal reaction=weight=mg ur

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@hubertH it doesn't say. so i assumed it was horrizontal

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so u knw how to get to the answer right?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0but is static friction only involved in inclined plane?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0No.. then I wouldn't be asking lol

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0No! static friction is the force required to set the body in motion in our case it 's given as 258.0N

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Well they're asking from the coefficient of static friction?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0coefficient of static friction=static friction force/normal reaction

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0for horizontal plane normal reaction=mg

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1351982204492:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1351982369737:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0do u now have an idea of what u're doing?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so i say that Fnet = \[\mu _{s}  Fg = ma?\] a little do you?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0remember u're given the Fs and the mass only

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0there is no Fa, no Fnet

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0well if it's starting to move wouldn't it be Ff = Mu(s) Fn?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[mus =\frac{ ?Fmus }{Nr ? }\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I have never used Nr in my FBD before?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Nr is ur normal reaction=mg

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Ok so it's just normal force (Fn)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0and since it's horizontal Fnet = 0N

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0what is Fnet standing fo?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0The vector pointing upward

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Ok i got it, Mu(s) = .66

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0that's it bro congrats and keep on going forwad

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0If they ask for acceleration and give kinetic friction is it still the same equation except it's Mu(k)  Fn = ma?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0no! kinetic friction is the smallest force that keeps the system moving @ constant speed

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0if u have more questions on science don't hesitate to massage me.....

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0alright because i have a big physics test and I don't really understand forces lol

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0we'll try to make it easier together bye for now
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