# Thread: 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:

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

You can then solve for t.

For (2) if you draw the free body diagram you have forces in the vertical direction: $\displaystyle T \sin \theta + T = W$ where $\displaystyle \theta$ is the angle of the string to the first support. In the horizontal direction you have $\displaystyle T \cos \theta = P$. And from the length of the string being A you have $\displaystyle \frac B {\cos \theta} + B \tan \theta = A$. 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 $\displaystyle \cos \theta$.

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: $\displaystyle T \sin \theta + T = W$ where $\displaystyle \theta$ is the angle of the string to the first support. In the horizontal direction you have $\displaystyle T \cos \theta = P$. And from the length of the string being A you have $\displaystyle \frac B {\cos \theta} + B \tan \theta = A$. 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 $\displaystyle \cos \theta$.
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:

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

You can then solve for t.

For (2) if you draw the free body diagram you have forces in the vertical direction: $\displaystyle T \sin \theta + T = W$ where $\displaystyle \theta$ is the angle of the string to the first support. In the horizontal direction you have $\displaystyle T \cos \theta = P$. And from the length of the string being A you have $\displaystyle \frac B {\cos \theta} + B \tan \theta = A$. 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 $\displaystyle \cos \theta$.
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.
The text passage contains two distinct mechanics questions, and answers or hints to these questions are provided in the subsequent replies.
multiple_questions_with_answers
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
- NO_QUESTION
- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
system
# Thread: 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:

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

You can then solve for t.

For (2) if you draw the free body diagram you have forces in the vertical direction: $\displaystyle T \sin \theta + T = W$ where $\displaystyle \theta$ is the angle of the string to the first support. In the horizontal direction you have $\displaystyle T \cos \theta = P$. And from the length of the string being A you have $\displaystyle \frac B {\cos \theta} + B \tan \theta = A$. 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 $\displaystyle \cos \theta$.

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: $\displaystyle T \sin \theta + T = W$ where $\displaystyle \theta$ is the angle of the string to the first support. In the horizontal direction you have $\displaystyle T \cos \theta = P$. And from the length of the string being A you have $\displaystyle \frac B {\cos \theta} + B \tan \theta = A$. 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 $\displaystyle \cos \theta$.
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:

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

You can then solve for t.

For (2) if you draw the free body diagram you have forces in the vertical direction: $\displaystyle T \sin \theta + T = W$ where $\displaystyle \theta$ is the angle of the string to the first support. In the horizontal direction you have $\displaystyle T \cos \theta = P$. And from the length of the string being A you have $\displaystyle \frac B {\cos \theta} + B \tan \theta = A$. 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 $\displaystyle \cos \theta$.
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.
user
The text passage contains two distinct mechanics questions, and answers or hints to these questions are provided in the subsequent replies. 

TYPE: MULTIPLE_QUESTIONS_WITH_ANSWERS