Why isn't the normal force mgsin60 in this rotating rod problem?

[Doc Al]

But back to the case of the inclined rod, can i say that
nsin60=mg?

Tried it. But the answer wasn't right. Could you tell me why?

Of course it's not right, as I've been saying since my first post. Why do you think it's right? (For my reasons, reread this thread.)

 

[Latios1314]

Reread it.
But i still don't really get it.

There is a resultant centripetal force pointing towards the centre and thus there is a centripetal acceleration pointing towards the centre. The forces along BC is caused by friction and a component of the slider's weight mgcos60.

But how is it different from a car on a banked road?

In the case of a car on a banked road.

ncosθ=mg. It is a component of the normal force ncosθ=mg

but why doesn't a component of the normal force=mg here?

 

[Doc Al]

But how is it different from a car on a banked road?

It's very similar.

but why doesn't a component of the normal force=mg here?

It does. You are confusing ncosθ = mg with n = mgcosθ. Big difference!

 

[Latios1314]

i mean for the case of a car on a banked road.

ncosθ = mg

But why doesn't nsinθ = mg for the case of the slider?

 

[Doc Al]

i mean for the case of a car on a banked road.

ncosθ = mg

But why doesn't nsinθ = mg for the case of the slider?

Because there is friction. To get an equation for the vertical forces you must include all vertical force components. Friction will have a vertical component.

For the car on a banked road, ncosθ = mg only if there is no friction.