All Mechanics DOUBTS HERE!!!

[Zishi]

Hello, anyone please solve this whole question(attached) and explain its answer fully!

Thanks~

[astarmathsandphysics]

here

[**RoRo**]

Mechanics is on it's way to drive me insane! -.-"

Guys, I need help in Dynamics, section 5.3, which is about particles inclined on a plane where you need to resolve the forces parallel and perpendicular to the plane.

I don't understand how the forces are resolved, how do they know which angle is it and when to use sin[theta] and cos[theta].

Can someone please explain this to me?

Thanks a lot!

& here's an example of a question, in case anyone would need it.

A particle of mass 5 kg is at limiting equilibrium on a rough plane which is inclined at an angle of 30 degrees to the horizontal. Find the coefficient of friction between the particle and the plane.

[The Golden Girl =D]

Mechanics is on it's way to drive me insane! -.-"

Guys, I need help in Dynamics, section 5.3, which is about particles inclined on a plane where you need to resolve the forces parallel and perpendicular to the plane.

I don't understand how the forces are resolved, how do they know which angle is it and when to use sin[theta] and cos[theta].

Can someone please explain this to me?

Thanks a lot!

& here's an example of a question, in case anyone would need it.

A particle of mass 5 kg is at limiting equilibrium on a rough plane which is inclined at an angle of 30 degrees to the horizontal. Find the coefficient of friction between the particle and the plane.

I have two pics that I hope will explain the general idea to you.


look at the worked solution for the Question on this page if you need a better idea of how to do it:

http://www.examsolutions.co.uk/A-Level-maths-tutorials/Edexcel/M1/particles-in-equilibrium/revision-guide.php

=]

[The Golden Girl =D]

& here's an example of a question, in case anyone would need it.

A particle of mass 5 kg is at limiting equilibrium on a rough plane which is inclined at an angle of 30 degrees to the horizontal. Find the coefficient of friction between the particle and the plane.

[**RoRo**]

I have two pics that I hope will explain the general idea to you.


look at the worked solution for the Question on this page if you need a better idea of how to do it:

http://www.examsolutions.co.uk/A-Level-maths-tutorials/Edexcel/M1/particles-in-equilibrium/revision-guide.php

=]

I UNDERSTOOD IT!! =D

Thanks ALOTT!!

Now a question, check the photo attached for the question, the diagram and the answer from the CD that came with the textbook.

How is the acceleration = 5/7, when in part a) it was 1.4?

[The Golden Girl =D]

I UNDERSTOOD IT!! =D

Thanks ALOTT!!

Now a question, check the photo attached for the question, the diagram and the answer from the CD that came with the textbook.

How is the acceleration = 5/7, when in part a) it was 1.4?

I'm sorry but can you tell me the exact exercise and the number of the Question please.

[**RoRo**]

I'm sorry but can you tell me the exact exercise and the number of the Question please.

I'm sorry, I forgot to attach the file! :/

There you go with it, it's Exercise 3F Q4b

[elemis]

I'm sorry, I forgot to attach the file! :/

There you go with it, it's Exercise 3F Q4b

After A hits the ground the string goes slack and there is no tension. Hence, B continues to move upwards but under the force of gravity.

That is, the acceleration of B will be = -9.8 ms^-2

[The Golden Girl =D]

I'm sorry, I forgot to attach the file! :/

There you go with it, it's Exercise 3F Q4b

my teacher taught us a different way of doing this :

we want the  GREATEST height and so we have to do TWO steps.

Step 1;
the Speed with which the Particle A strikes the ground

u = o , v = ? ,a = 1.4 ms^-1 , s = 2

v² = u² + 2as
v² = 0 +( 2 * 1.4 * 2 )
v² = 5.6
v = 2.37 ms-1

Step 2
the distance traveled by B after A has reached the ground

u = 2.37 , v = 0 , a = - 9.8 (going up hence against gravity ) , s = ?

v² = u² + 2as
0 = (2.37)2 + (2*-9.8* s )
{19.6 s = (2.37)² } / 19.6

s = 0.2865765

MAX height = 2 + 0.2865765
               = 16/7 m

=]

[elemis]

my teacher taught us a different way of doing this :

we want the  GREATEST height and so we have to do TWO steps.

Step 1;
the Speed with which the Particle A strikes the ground

u = o , v = ? ,a = 1.4 ms^-1 , s = 2

v² = u² + 2as
v² = 0 +( 2 * 1.4 * 2 )
v² = 5.6
v = 2.37 ms-1

Step 2
the distance traveled by B after A has reached the ground

u = 2.37 , v = 0 , a = - 9.8 (going up hence against gravity ) , s = ?

v² = u² + 2as
0 = (2.37)2 + (2*-9.8* s )
{19.6 s = (2.37)² } / 19.6

s = 0.2865765

MAX height = 2 + 0.2865765
               = 16/7 m

=]

LOL. You didnt answer her question

[Zishi]

here

Thanks for that, but why the two tensions are same? I just can't understand that; which word or thing in the question makes us understand that the tensions will be same. I've solved a lot of questions of like this before, but the tensions in the two strings were not same. Plus, please answer the third part too now!(Attached)

[**RoRo**]

After A hits the ground the string goes slack and there is no tension. Hence, B continues to move upwards but under the force of gravity.

That is, the acceleration of B will be = -9.8 ms^-2

mmm, alright, that makes sense too!

My question was, when calculating the acceleration for A, it was taken as 5/7 and not 1.4 like we had found in part a) of the question, why is that?

[**RoRo**]

my teacher taught us a different way of doing this :

we want the  GREATEST height and so we have to do TWO steps.

Step 1;
the Speed with which the Particle A strikes the ground

u = o , v = ? ,a = 1.4 ms^-1 , s = 2

v² = u² + 2as
v² = 0 +( 2 * 1.4 * 2 )
v² = 5.6
v = 2.37 ms-1

Step 2
the distance traveled by B after A has reached the ground

u = 2.37 , v = 0 , a = - 9.8 (going up hence against gravity ) , s = ?

v² = u² + 2as
0 = (2.37)2 + (2*-9.8* s )
{19.6 s = (2.37)² } / 19.6

s = 0.2865765

MAX height = 2 + 0.2865765
               = 16/7 m

=]

This makes more sense to me than the working on the CD, thanks a lot and sorry for all the trouble!

[**RoRo**]

Okay, another question:

I don't understand how part c) is solved, the working is given below!

Check the attachment, it has the question and the answer!

Note:
Acceleration of the system = 0.613
Tension in the string = 27.6