@sammygerbil Wait my explanation is not over yet. It has just begun
So in the bottommost position $F\sin \theta = R$ and $F \cos \theta = mg$
@sammygerbil Now this is how components of forces look like when the ball is in the dotted position
@sammygerbil This clearly shows that $F \cos \theta$ has decreased which means $F $ has decreased but the angle $\theta$ is same
@sammygerbil And previously $F \sin \theta=R \sin \alpha$, but now $F \sin \theta=R\sin \theta$. And $R \sin \theta$ is less than R. So R has increased
@sammygerbil Now my explanation is over
Sorry my last sentence is incorrect. let me rewrite it
@sammygerbil And previously $F \sin \theta=R$, but now $F \sin \theta=R\sin \alpha$. And $R \sin \alpha$ is less than R. So R has increased
@LoopBack Sorry it is not clear to me. I didn't understand the significance of the two positions of the ball. I shall have to study this to try to understand what you are saying.
@sammygerbil See the first figure, where the string is intersecting the hemisphere. It shows that there are two points where the string makes same angle with the vertical. You can read my explanation again
@sammygerbil And previously $F \sin \theta=R \sin \alpha$, but now $F \sin \theta=R\sin \theta$. And $R \sin \theta$ is less than R. So R has increased
Here's the proof : In your last diagram the string is tangent to the hemisphere at the position of the upper ball, because it is at right angles to the normal. But in your earlier diagrams the string is tangent at the midpoint of the two positions of the ball.
@LoopBack If your derivation is based on your diagram in which you show that $\theta+\alpha=90$ then I think it's wrong. Have you looked at my disproof?
Here's the proof : In your last diagram the string is tangent to the hemisphere at the position of the upper ball, because it is at right angles to the normal. But in your earlier diagrams the string is tangent at the midpoint of the two positions of the ball.
@sammygerbil you are misinterpreting my diagram. I said the string will be parallel to a tangent at some point of the hemisphere. But this doesn't mean that the ball will be at that point. Wait I'll explain you with a diagram
I understand that the ball will not be at the tangent point. But in your diagram showing $\alpha+\theta=90$ degrees it is at the tangent point of the string.
I guess, take 2 extreme cases. Once when the small sphere is just above the ground (near the equator). Let the angle string makes with vertical be theta. Resolving tension, you get Normal force = mgTan theta
@user64829 That is what I thought initially. But when I looked at the equations I found it is not correct. It works only if you are exactly at the top.
@user64829 You can balance the ball exactly at the top with no tension in the string. However, if you go just a tiny distance either side you need some tension in the string to keep the ball from sliding off the hemisphere.
Another point is that when the ball is at the top you can divide the weight of the ball between normal reaction and tension in the string in any ratio you choose so that $F+R=W$. So instead of the ball resting on the sphere with $F=0, R=W$ you could have the ball totally suspended by the string with $F=W$ and $R=0$.
@LoopBack Sorry I am getting too tired to think. I shall have to sleep now. I have to go somewhere in the morning but will be back around 3pm UK time tomorrow. Goodnight chhaatr.
@blue_eyed_... $\cos(\omega t + kx)$ gives you a wave moving left while $\cos(\omega t - kx)$ gives you a wave moving right. They are identical at time $t=0$ but start moving apart for $t>0$.
One sec, I think I got how to do it, give me few seconds
I got it. If the arrow moves from left to right, we gotta consider work as +ve and vice versa. Then we have to add the corresponding pieces of areas keeping proper sign convention right?
Conversely A is still connected to the battery so the voltage on A has to remain constant but the charge can change when the dielectric is inserted.
@user64829 I don't think there is a trick. You just have to play with the layout until you get something that makes sense. I like to colour the different bits of wire so I can keep track of them as I rearrange the circuit.
A rope of length 30 cm is on the horizontal table with maximum length hanging from edge A of the table. The coefficient of friction between the rope and the table is 0.5. The distance of center of mass of the rope from A is?
Firstly, 20 cm of rope will be on the table right?
@JohnRennie, yeah..., y=2a cos(kx).sin(wt) is the equation for standing wave which I've derived using the superposition of y=a sin(wt-kx) and y=a sin(wt+kx)
