Explanations & Calculations
- By resolving the pulling force into orthogonal components we can find the verticle & horizontal applications of it.
- The vertical component of the pulling force that acts downwards on the ring is
Pv=T×cos60=2T
- And it's horizontal component which applies horizontally on the ring (pulling away) is
Ph=T×sin60=23T
- Now, the ring's weight is 4g downwards and the frictional force is holding the ring sliding downwards the rod hence it applies on the rod upwards.
- Since the ring is in equilibrium, all the vertical forces should be balanced or the net vertical force on the ring should be zero. Then,
↑ΣFF=F−4g−2T=0=4g+2T
- Frictional force is found then.
- Treating the ring's horizontal equilibrium the same way, we can find the normal force that acts on the ring.
→23T−RR=0=23T
- The contact force on the ring is obtained by R2+F2
Comments
Leave a comment