Two identical trains are moving on rails along the equator on the earth in opposite directions with the same speed. Will they exert same pressure on the rails ?
my sister also hates instant noodles; she said she won't eat instant noodles anymore after leaving the home where she was fed on instant noodles every meal.
Problem Statement: Two particles of mass $m$ each are tied at the ends of a light string of length $2a$. The whole system is kept on a frictionless horizontal surface with the string held tight so that each mass is at a distance $a$ from the center $P$. Now, the mid-point of the string is pulled vertically upwards with a small but constant force $F$. As a result, the particles move towards reach other on the surface. The magnitude of acceleration, when the separation is $2a$ is ?
When an object is thrown in air and having a constant acceleration A due to air find out about the time of the whole motion, weather it's greater than or lesse than the time which it takes without air friction
Now, we can say that for one string the force acting is $F/2$ (sir I really want to know why I can do distribute half forces, but it seems correct to me) and hence we have
my try - F pulling up, the string pulls back with 2 forces G, G' of equal magnitude, G on the left ball say. Looking at the force diagram for the left ball, there's the force of gravity, normal force, and this G. The normal force will change so that the net upward force on the ball is 0 and the ball moves only horizontally (this is the $|F|\ll 1$ assumption), and the size of the horizontal component of $G$ is $|F|\cos^{-1}(x/a)/2$. Since $F=ma$ the acceleration is $|F|\cos^{-1}(x/a)/(2m)$
Now, $F_{\parallel}$ is the force on the mass. We have $\sin \theta = x/a$ $$ F_{\parallel} = F/2 \cos \theta \\ F_{\parallel} = \frac{F}{2a} \sqrt{a^2 -x^2}$$
We have $\cos \alpha = \frac{x}{a}$, therefore horizontal force on the mass $m$ is $$ \frac{F}{2a} \sqrt{a^2 - x^2} \times \frac{x}{a}\\ \frac{F}{2a^2} x \sqrt{a^2 -x^2}$$
since you know $\text{stuff} = F_\parallel\sin\alpha$ and you want $F_\parallel\cos\alpha$, this is $\text{stuff} \times \frac{\cos\alpha }{\sin\alpha} $
Now, $F_{\parallel}$ is the force on the mass. We have $\sin \theta = x/a$ $$ F_{\parallel} = F/2 \cos \theta \\ F_{\parallel} = \frac{F}{2a} \sqrt{a^2 -x^2}$$
whatever force diagram you drew to say that $T=F_\parallel = \frac{F}2 \cos\theta$ must have a mistake
we can't use the horizontal components of the forces on the ball because that equation is what we use to solve for the acceleration (using newton's second? law)
in other news @Knight i cant believe i managed to convince the captain to try something, and simultaneously i cant believe i found an adult who hasnt had cereal before (same person)
i dont like counting, but my guess is like this. Say $d$ is an arbitrary derangement. its naturally one of the functions $f$ that we can get. if we switch one of the $d(i)$ with 0, we get a new function, say $g$. Its impossible to get $g$ if you start with a different derangement. I think this is all possible $f$s...
so i guess its just 5 times the number of derangements on 5 things
since we can either not switch with 0, or switch one of $d(i)$ for $i=2,3,4,5$ with $0$
wait no this is wrong, there is double counting...
