Circular Motion, why is it R = MG - F and not R = MG + F?

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#1
Hi, Doing this circular motion exam question:

and I'm on part a)ii).

The mark scheme uses the fact that R = mg - F but I can't work out why. The reaction force is acting upwards, the weight is acting downwards, and the centripetal force, F, is acting downwards too.

The car isn't in the air or going through the floor, so in my mind, R must be equal to the sum of the downwards forces, aka R = MG + F.

Can someone explain this to me please? Thanks
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3 years ago
#2
The reaction force is up, the weight is down, and the centripetal force is the resultant of these

F= mg - R (because F is downwards towards the centre of the circle)

Then rearrange.

Whenever you look at a circular motion question, look at the forces which are acting and make the resultant of these the centripetal force.
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3 years ago
#3
(Original post by stuart4)
The reaction force is acting upwards, the weight is acting downwards, and the centripetal force, F, is acting downwards too.
There is no such thing as centripetal force. It is simply a name given to the result of other forces that act on a body. It is not in itself a separate force.
Using Newton's 2nd Law, a force must act if there is an acceleration. If you do Mechanics 3, you will derive that this acceleration is towards the centre for circular motion (or you are just told this and never taught why).
Hence you know that the resultant of weight and normal reaction acts towards the centre, and ergo mg > R when in contact with a surface. The resultant (named centripetal for the sole purpose of tagging a name on things to help understand them) force is simply mg - R (resolving forces perpendicular to tangent of circular motion).

(Original post by stuart4)
The car isn't in the air or going through the floor
No, however without the resultant force acting downwards, the car would simply fly over the bridge at a tangent as if it were a ramp.
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