Thanks for your interest, Simon, I appreciate it.
As I've said in my original post I solved the problem using a non-inertial frame.
But I was surprised to see the trigonometry of the forces and accelerations holding up in the intertial frame also, involving components of components, etc. For...
That's exactly the point Gracy...it doesn't need to be resolved.
(My answer to haruspex was purely trigonometric, to show it added up to g)
Any comments on my original post? Am I making some elementary mistake in physics or trigonometry?
Thanks Haruspex.
In my original post I was asking whether the answer I got from the inertial frame was correct because I resolved a component twice.
If you would kindly look at my original post my difficulty may become clearer. I obtained a weight of 0.25mg using both non-inertial and...
Indeed it is, but if you resolve the other original component gcos(60) also you get gcos(60)sin(30).
So adding both vertical components (sorry, of the original components!) you end up with Sin(60+30) which is 1, so you get the original g back.
Haruspex, thanks for your response. Since gsin(60) is the vector (hypotenuse) that is being resolved I was wondering whether it would be gsin(60)cos(30).
Hope I'm not making some very elementary mistake.
Thanks for your reply, Simon. Your description is spot on.
I managed to attach a picture to my original post but forgot to correct my post. If you would care to see it please look at the attachment to my post.
As you say, you can resolve a vector ad infinitum but in this particular case I...
Homework Statement
A man is standing on a small wedge that is sliding down a frictionless large wedge (base on the ground - horizontal) with angle 60. I don't have a picture so I'll just say that the surface of the smaller wedge is parallel to the ground. It has a weighing machine on it and the...