# Math Help - Couple of general/abstracted mechanics questions

1. ## Couple of general/abstracted mechanics questions

Hi I have two questions that I'm stuck on. I don't want solutions but any hints/pointers would be much appreciated.

1) An object is projected vertically upwards with a velocity of u m/s, and after t s another object is projected upwards from the same point with the same velocity. Find, in terms of u, t and g the time which elapses between the second object's projection and the collision between the objects.

At first I tried finding the times when the displacements of equal, using (T+t) and t for A and B respectively. Then got nowhere...

2) A light string of length a is attached to two points A and B on the same level and a distance b apart, where b<a. A smooth ring of weight W is threaded on the string and is pulled by a horizontal force, P, so that it rests in equilibrium vertically below B. Show that the tension in the string is

[W(a^2 + b^2)]/(2a^2)

No idea on this one...any hints?

2. ## Re: Couple of general/abstracted mechanics questions

For (1) you are on the right track. Where did it fail? You should have equations:

$u(t+T)-\frac 1 2 g(t+T)^2 = ut - \frac 1 2 g t^2$

You can then solve for t.

For (2) if you draw the free body diagram you have forces in the vertical direction: $T \sin \theta + T = W$ where $\theta$ is the angle of the string to the first support. In the horizontal direction you have $T \cos \theta = P$. And from the length of the string being A you have $\frac B {\cos \theta} + B \tan \theta = A$. From this you have enough to get the final expression, but it takes some playing around. I suggest you get this last expression into terms of $\cos \theta$.

3. ## Re: Couple of general/abstracted mechanics questions

Originally Posted by ebaines

For (2) if you draw the free body diagram you have forces in the vertical direction: $T \sin \theta + T = W$ where $\theta$ is the angle of the string to the first support. In the horizontal direction you have $T \cos \theta = P$. And from the length of the string being A you have $\frac B {\cos \theta} + B \tan \theta = A$. From this you have enough to get the final expression, but it takes some playing around. I suggest you get this last expression into terms of $\cos \theta$.
Ok all that is clear to me, but I can't seem to get an equation for T without P or any trig functions...What is the next step?

thanks

4. ## Re: Couple of general/abstracted mechanics questions

Originally Posted by ebaines
For (1) you are on the right track. Where did it fail? You should have equations:

$u(t+T)-\frac 1 2 g(t+T)^2 = ut - \frac 1 2 g t^2$

You can then solve for t.

For (2) if you draw the free body diagram you have forces in the vertical direction: $T \sin \theta + T = W$ where $\theta$ is the angle of the string to the first support. In the horizontal direction you have $T \cos \theta = P$. And from the length of the string being A you have $\frac B {\cos \theta} + B \tan \theta = A$. From this you have enough to get the final expression, but it takes some playing around. I suggest you get this last expression into terms of $\cos \theta$.
WRONG APPROACH!!!!
Solving the above suggested equation yields t=(5T^2)/(u-10T) which is wrong!!!

Kinhew...

The first object needs (u/10 ) seconds to reach the max height and then it stops for an instant before it starts to fall free down.
At this moment the second object is at a height H travelling (u/10 -t) seconds because the delay is t seconds.
The height H can be calculated easily from the formula of the Uniform Decelarated Motion with g =10 and it is H = (u^2/20)-5t^2
the gap between the two object at this moment A at the max height and B at a height H is H = 5t^2.
The two objects will continue the motion but now the object A is falling down with U A. Motion while the onbect B continues the U.D M upwards.
They will meet at a point P x seconds after.
Get the UAM and UDM formulas (put x as time)and add them together and equate them with 5t^2. solve the equation to get x = t/2 !!! this is the corect solution
therefore the total time needed for the coallision is u/10+t/2
EXample: If u=40 m/s and t =1 second then they will meet at a height of 78.75 m after 4.5 seconds the first object was launched.

5. ## Re: Couple of general/abstracted mechanics questions

Originally Posted by MINOANMAN
WRONG APPROACH!!!!
Solving the above suggested equation yields t=(5T^2)/(u-10T) which is wrong!!!

The approach I suggested is correct. Check your math, my approach it yields t = u/g - T/2. I think the difference is that I define t=0 when the second projectile is fired, not the first.