Mechanics question! (1 Viewer)

[Aysce]

Yo.

So I'm having trouble with resolving the forces for these 2 questions (will add the 2nd one later) and although it may seem easy, I just don't understand how you arrive at some of these expressions. One instance:



I understand how to get T + N but not T - N?

To be specific, why is it " -N " if that makes sense?

Thanks in advance!

 

[Carrotsticks]

When we resolve forces, it is "Everything vs. mg" when resolving vertically.

So if Ncosa + Fsina (just an example) is going up and mg is going down, then Ncosa + Fsina = mg (everything vs. mg)

But when we do it horizontally, we want to see what MAKES UP centripetal force instead of 'Everything vs etc etc"

So if centripetal force is going left, and Nsina + Fcosa (again, an example) is also going left, then Nsina + Fcosa = mrw^2 or mv^2/r.

 

[Aysce]

When we resolve forces, it is "Everything vs. mg" when resolving vertically.

So if Ncosa + Fsina (just an example) is going up and mg is going down, then Ncosa + Fsina = mg (everything vs. mg)

But when we do it horizontally, we want to see what MAKES UP centripetal force instead of 'Everything vs etc etc"

So if centripetal force is going left, and Nsina + Fcosa (again, an example) is also going left, then Nsina + Fcosa = mrw^2 or mv^2/r.

Okay, okay I understand the horizontal part now. But is it T-N because N is going the opposite direction....or something...? I'm just clueless tbh.

 

[Carrotsticks]

Okay, okay I understand the horizontal part now. But is it T-N because N is going the opposite direction....or something...? I'm just clueless tbh.

Think of component vectors. For a vector to be like y=x pointing upwards (just as an example), I can 'break it up' into 2 smaller vectors that make up this diagonal vector.

One is going to the right, and one is going to the left.

Suppose the vector is now y=x pointing downwards. Then I can break it up into 2 smaller vectors, 1 going left and another going down:

 

[D94]

(-N) is merely a direction. Usually, the question will ask you to explain by resolving the forces to find T+N and T-N, e.g. look at the 2011 MX2 paper Q5.

 

[Aysce]

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Alright, to make it more clear to you guys I'll show my working.

So resolving forces on P vertically: (forgot to put P in the diagram )

Tsin45 + Ncos45 = mg

T + N = (2)^1/2 mg

Resolving horizontally:

Tcos45 + Nsin45 = mrw^2 - So this here is the mistake. So I'm proposing that since this force (Nsin45) is "away" or "opposing" the other horizontal forces, it is taken as a negative rather than positive NSin45?

 

[Carrotsticks]

When drawing your component vectors (the dotted lines), it helps if you draw little arrows for them too.

And yep should be Nsin45 since it's going opposite to centripetal force (going towards centre)