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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: dino on November 01, 2009, 05:18:18 pm

Title: Label all forces acting...
Post by: dino on November 01, 2009, 05:18:18 pm: [img]http://img.skitch.com/20091101-f186igq8ugpbe3bewwxec4qe84.preview.jpg[/img]

If I have this, do I only need to label the orange forces, or can I resolve in those types of questions. I find it easier to just resolve - minimises mistakes.

Also, will they accept μN, or just leave as Fr?
Title: Re: Label all forces acting...
Post by: NE2000 on November 01, 2009, 05:19:39 pm: Don't resolve.

And best to leave as Fr as sometimes Fr does not equal μN (that is the limiting friction)
Title: Re: Label all forces acting...
Post by: dino on November 01, 2009, 05:26:14 pm: Thanks mate.

Didn't think of the Fr≠μN situations.
Title: Re: Label all forces acting...
Post by: monokekie on November 01, 2009, 05:32:51 pm: I LOVE YOUR DIAGRAM
Title: Re: Label all forces acting...
Post by: dino on November 01, 2009, 05:34:54 pm: =]
Title: Re: Label all forces acting...
Post by: QuantumJG on November 01, 2009, 06:04:24 pm: Quote from: NE2000 on November 01, 2009, 05:19:39 pm
Don't resolve.

And best to leave as Fr as sometimes Fr does not equal μN (that is the limiting friction)

With a case like this you have your box situated on an inclined plane. Static friction on a microscopic level is basically bonds that the box forms with the surface its on, once the exerted force is greater than μN, these bonds break and your object slides across the surface and you get a new friction called static friction with is much weaker (μ_s > μ_k) but is a constant force.

But in diagrams like this it is physically more correct to use Fr to represent friction than μN.
Title: Re: Label all forces acting...
Post by: timmay12 on November 01, 2009, 08:34:15 pm: Would mg and W both be fine... and can you sub in a value for m?
Title: Re: Label all forces acting...
Post by: arthurk on November 01, 2009, 08:53:47 pm: Fr does not equal uN when does this occur?
Title: Re: Label all forces acting...
Post by: timmay12 on November 01, 2009, 09:41:07 pm: uN is the maximum friction force. Fr will only equal the opposing force. So say if uN = 20N, but the opposing frictional force is 10N. Then Fr would would = 10N, not the uN of 20N

If that makes any sense.
Title: Re: Label all forces acting...
Post by: /0 on November 01, 2009, 09:43:38 pm: Yeah, if the pushing force is less than or equal to $\mu N$ then $Fr = pushing \ force$ .

I think this would be an excellent MC question
Title: Re: Label all forces acting...
Post by: TrueTears on November 02, 2009, 01:38:42 am: Quote from: /0 on November 01, 2009, 09:43:38 pm
Yeah, if the pushing force is less than or equal to $\mu N$ then $Fr = pushing \ force$ .

I think this would be an excellent MC question
Yeah MAV and Kilbaha had heaps of these.
Title: Re: Label all forces acting...
Post by: kamil9876 on November 02, 2009, 02:15:28 am: yeah, if pushing force is less than $\mu N$ (say 0), and you mistakenly thought that $friction=\mu N$ then you would have friction causing things to move :P.

Also, as quantumjg said, $\mu_s>\mu_k$ , because if otherwise; say pushing force, P satisfied: $N\mu_s<P<N\mu_k$ then the object would move since $N\mu_s<P$ but yet $P<N\mu_k$ would imply that the net force is backwards, so friction is pushing! I thought of this once and felt a bit scared!, but luckily it's not true :P surprisingly, it can be reasoned out unlike certain laws of physics :P *hides from physics nerds*
Title: Re: Label all forces acting...
Post by: NE2000 on November 02, 2009, 09:49:55 am: Quote from: kamil9876 on November 02, 2009, 02:15:28 am
yeah, if pushing force is less than $\mu N$ (say 0), and you mistakenly thought that $friction=\mu N$ then you would have friction causing things to move :P.

Although as I think the 2007 exam shows. Conveyer belts are an exception to the rule that friction doesn't cause things to move.
Title: Re: Label all forces acting...
Post by: kamil9876 on November 02, 2009, 05:35:53 pm: Hrmm.. but relative to the conveyer belt's frame, the friction is causing the object to be stationary. Conversely dragging something across a surface; from your reference frame it may look like the friction is causing the thing's motion from your reference frame.