Relative acceleration: Need help with this problem. Please see the attached JPG

- anonymous

Relative acceleration: Need help with this problem. Please see the attached JPG

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- anonymous

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- radar

It would appear since the acceleration is upward and the collar can only move at a path which is 45 degrees from the vertical, the acceleration force in that direction would be .707(4 meter/sec^2) Just a swag lol.

- anonymous

Thanks for replying but book answer says that a=7.49 m/s^2 at an angle of 22.8 degrees measured from neg-x axis counterclockwise.

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## More answers

- radar

It was just a guess (with some wrong thinkin)

- Mani_Jha

|dw:1332554742554:dw|
The collar C would have two accelerations - one in the upper direction, and one along the shaft AB. Can you find the resultant of these two vectors?
\[a=\sqrt{a 1^{2}+a2^{2}+2a1a2\cos 45}\]
a1=4
a2=4cos45

- anonymous

What about component of weight of collar C?, it doesnt come to the picture?

- anonymous

ans=6.324

- Mani_Jha

Oh, yes you are right. The downward acceleration net will be mg-m4 and the inclined acceleration will be mgcos45-4cos45,
Now see

- anonymous

I made a Free Body Diagram, could you review it please?

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- Mani_Jha

The component of the weight along AB is mgcos45 and perpendicular to it is mgsin45.

- anonymous

I was guessing with numbers and I got correct answer if a(c/a) =(g+4)*COS(45) at an angle of -45 deg measured from -X axis counterclock wise, but I cant set 2nd Newton's Law correctly for this.

- anonymous

a(c) = a(c/a) + a(a) (vectorial addition) gives the answer stated in the book. My problem is I cant get a(c/a) correctly.

- Mani_Jha

The acceleration along the incline would be (g-4)cos45. I am pretty sure.

- anonymous

\[(g+4)SIN(45)\left\{ 225\deg \right\}+4\left[ 90\deg \right]\]

- anonymous

I got the same result, but with that value I cant get the answer stated in the book. This is pretty weird.

- Mani_Jha

The answer is not important. gcos45 is the acceleration due to wight acting downwards, and 4cos45 is the acceleration due to A moving upwards. They are in opposite directions, see?
Maybe the answer in your book is wrong.

- anonymous

A guy showed me the solutions manual for this problem, and the result is the same stated at the book, but I didnt understand how the author sets the 2nd Newton's Law.

- anonymous

Please see.

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- anonymous

It supossed that 2nd Newtons Law can only be expressed relative to an inertial frame, but AB is moving upward, and this is not an inertial frame!!!.

- anonymous

This book is plenty of problems of this caliber!!!.

- Mani_Jha

Oh ok. then you've got to use 'fictitious' or 'pseudo' forces. We generally add a force opposite to the direction of the frame's acceleration.
|dw:1332558039957:dw|
a=4
Now if you take components you would get acceleration along the incline to be (g+4)cos45

- anonymous

Â¿such as an action-reaction pair?

- anonymous

Your pseudo force is directed downwards (-90degs)??

- anonymous

Did you mean this??See JPG

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- anonymous

The component in the graph should be COS45. Sorry

- Mani_Jha

No, it's not an action-reaction pair. From an inertial frame, we read measurements wrong because of the frame's acceleration. So,we just add a force in the other direction to make things right.
Yup, pseudo force should be taken opposite to the direction of the frame's acceleration. The forces should be both downwards, so that
mgcos45+mg=ma

- anonymous

But that's not resulting in (g+4)COS45. Â¿Am I right?

- Mani_Jha

Oh sorry, along the incline the net force would be macos45+mgcos45. ma and mg both act downwards and so their components must be in the same direction. get it now?

- anonymous

I get now your point, but still cant dig ma downwards along with mg.

- anonymous

Numerically your model makes sense. but, Â¿did you reviewed the jpg from the solutions? ma is directed upwards!!!!.

- anonymous

Â¿Are you studying at university? Â¿is there any physics teacher that you can consult to about this problem?

- Mani_Jha

I am in high school. what about you? I will think on this, and post a few hours later.

- anonymous

Haha, I live in Ecuador; if I would ask any universitary physics teacher about this, surely he would falls down and die of a brain attack. Anyway, many thanks to you for your efforts. You deserve a medal mate. Cheers.

- Mani_Jha

You study in university?

- anonymous

I was, but I retired because those administratives are a bunch of corrupts and didnt allow me to get academic documentation for a change of university. Now I'm selfstudying in order to take SAT and search an scholarship for study outside this country.

- anonymous

Where do you come from?

- Mani_Jha

I am from India, and even I am preparing for the SAT!

- anonymous

When do you expect to take the test?

- anonymous

Also I have to improve a lot my English!!!.

- Mani_Jha

in 2013. Even I have to work on my english. Join the English and Writing sections here

- anonymous

Sure I will, best wishes for you. If you have any update for this problem, please let me know. Thanks again for coming in my help. Bye.

- anonymous

|dw:1332580841706:dw|
|dw:1332580956697:dw|
the normal reaction shud balance 6cos45 in th upward direction of C
so only 6sin45 operates in the direction toward AA right?
how did u guys get 2 components of acceleration and direction of acceleration?

- anonymous

Hi Salini, Did you see this?, let me know your comments.
http://assets.openstudy.com/updates/attachments/4f6d1696e4b0772daa0812c0-juancarlosquintero_ec-1332557149257-dibujo1.jpg

- anonymous

To make sense the analysis of the previous post: Net "Real" forces at X and Y sould "equal" "Pseudoforces", let me know what do you think about this.

- anonymous

What could be an accurate concept of Pseudo forces? any force caused by relative/alien accelerations to a body being analyzed?, if so, what about gravity (g)?

- anonymous

hi juancarlosquintero,
i saw that i dint get the method...will tell u if i figure it out
abt pseudo forces,dont worry abt them while solving problems as pseudo forces arent that necessary
but when u do a perceived gravity concept,it might be more than helpful
for example if u are accelerating in your car in the left direction then u perceieve or feel a force pulling you in the opposite direction to the cars motion(that is towards the right side)
to get to know more abt this pseudo perceieved gravity concept take alookk at this video(it helps a lot!)
http://ocw.mit.edu/courses/physics/8-01-physics-i-classical-mechanics-fall-1999/video-lectures/lecture-7/
i will find the exact part and lett u know abt it

- anonymous

@salini : thanks for your reply. I will check the links right now.

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