• Best of luck to the class of 2025 for their HSC exams. You got this!
    Let us know your thoughts on the HSC exams here

Mechanics (1 Viewer)

Joined
Mar 13, 2015
Messages
100
Gender
Male
HSC
2015
A particle P is undergoing uniform circular motion, while being pulled by two strings, one connected at A with the other string connected at B, vertically under A. At A, there is a hollow ring that is suspended.

I drew the tensions in string AP, one tension going away from B towards P and one going away from P towards B, both labelled as T1. There is also another tension (T2) going from P towards A.
The angle between AB and BP, AP are both theta.

My question is, what are the F_net components for the hollow ring?
I did:
F_net vertical = mg = T1costheta
However, my teacher said there is also a horizontal vector (N)
She wrote F_net horizontal = N = T1sintheta
But why is there a force of N?
THe ring is stationary in the horizontal plane, so F_net horizontal = 0? But if that's the case then T1sintheta = 0 ==> theta = 0??
 

porcupinetree

not actually a porcupine
Joined
Dec 12, 2014
Messages
661
Gender
Male
HSC
2015
Unless you misread your teacher's writing, she should have written:

F_net horizontal = 0
--> T1sintheta - N = 0 (where N is the horizontal normal force of the rod AB on the ring [assuming that rod AB exists; you didn't specify it, but the fact that there's a ring involved strongly suggests it)
--> N = T1sintheta
 
Joined
Mar 13, 2015
Messages
100
Gender
Male
HSC
2015
why is there a horizontal force of N? Is it because the ring touches and pulls against the rod so the rod has to have an opposing force?
 

leehuan

Well-Known Member
Joined
May 31, 2014
Messages
5,768
Gender
Male
HSC
2015
If one surface touches another, consequently exerting a force on it, there is always a normal reaction force against it from the surface. That is essentially one application of Newton's Third Law of Motion.
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top