Mind-fu*ked (1 Viewer)

[lost1]

For some reason i cant seem to get my head around this simple circular motion/ banked road question....
2001 Q6 a:\

A road contains a bend that is part of a circle of radius r. At the bend, the road
is banked at an angle α to the horizontal. A car travels around the bend at
constant speed v. Assume that the car is represented by a point of mass m, and
that the forces acting on the car are the gravitational force mg, a sideways
friction force F (acting down the road as drawn) and a normal reaction N to the
road.



Basically i get everything in the answers except resolving the forces horizontally which i thought should be like this(let A=theta):

FCosA + NSinA = -(mv^2)/r

since the net force is directed inwards away from FcosA and NSinA

but the answers say:

FCosA + NSinA = (mv^2)/r , without the negative sign

....please explain

thanks

 

[vds700]

For some reason i cant seem to get my head around this simple circular motion/ banked road question....
2001 Q6 a:\

A road contains a bend that is part of a circle of radius r. At the bend, the road
is banked at an angle α to the horizontal. A car travels around the bend at
constant speed v. Assume that the car is represented by a point of mass m, and
that the forces acting on the car are the gravitational force mg, a sideways
friction force F (acting down the road as drawn) and a normal reaction N to the
road.



Basically i get everything in the answers except resolving the forces horizontally which i thought should be like this(let A=theta):

FCosA + NSinA = -(mv^2)/r

since the net force is directed inwards away from FcosA and NSinA

but the answers say:

FCosA + NSinA = (mv^2)/r , without the negative sign

....please explain

thanks

solutions are correct, u dont need a negative sign, make the inwards direction positive

 

[Zeber]

The centripetal force it against the road and the 'car' (meaning towards the backed track).

considering the vectors of the F and N forces, we can see that Fsin@ + Ncos@ because they both go outwards (--->) whereas the centripetal force goes inwards (<-----)

basically, net force is 0, so -----> = <------
Thus Fsin@ + Ncos@ = mv^2/r, no negative.

 

[lost1]

but why is net force zero? shouldn't it be like circular motion where the netforce is radially inwards of (mrv^2)/r ?? So the F and N forces should add up to = (mv^2)/r ??

 

[Zeber]

Ah yeah, I meant zero net force as it nothing else other than the centripetal force is acting on the mass (in terms of horiztonal forces)

 

[lost1]

yes but isn't this centripetal force(mv^2/2) the sum of all the forces acting on the mass horizontally?
ie. F<------- + N<----- = <---------mv^2/2
thus -FCosA + -NSinA = mv^2/2 ?

 

[lost1]

oh shit!

stupid me should have done more practice....I thought the car moves around anticlockwise turning left when its clearly meant to move clockwise turning right(like the banked tracks in cycling)
now it makes sooo much sense...god im stupid

i guess the mods can close this thread now