# Integrating the rational fractions

1. Sep 9, 2008

please........... i need a help in integrating the partial fractions
i cant proceed to the integration part if i don't understand the patter in finding the constant....
that is....

if the given is:

ʃ ( (x^5+1) / ((x^3)(x+1)) )dx

then;

ʃ ( x-2 + ( 4x^3+1 ) / ( x^4 + 2x^3) )
ʃ ( x-2 + ( 4x^3 + 1 ) / ( x^3 (x + 2 ) )
ʃ ( x-2 + A/x + B/x^2 + C/x^3 + D/x+2)

Ax^2(x+2) + Bx(x+2) + C(x+2) + D(x^3) = 4x^3 + 1 ----- i dont know what going on
in this part

2. Sep 9, 2008

### tiny-tim

(I assume you mean ʃ ( (x^5+1) / ((x^3)(x+2)) )dx)

Ax^2(x+2) + Bx(x+2) + C(x+2) + D(x^3) = 4x^3 + 1

is simply the previous two lines (except for both the x - 2 s), multiplied by x^3(x + 2)

A B C and D are unknowns, and you find them by solving that equation

(though I think you can find D using the usual substitution method )

3. Sep 9, 2008

Prove that F=mgtanθ

Can everyone help me to solve this problem....

A small sphere of mass hangs from a string of L as in below. A variable horizontal force F is
applied to the sphere in such a way that it moves slowly from the vertical position until the string makes angle θ with the vertical. Assuming the sphere is always in equilibrium

1. show that F=mgtanθ
2.show that work done by F is mgL(1-cosθ)

__________________________
|\............................
|..\..........................
|.θ.\.L......................
|......\.......................
|........\.....................
|.........0------F..........
0.........|....................
...........|....................
...........|....................
..........W...................

4. Sep 9, 2008

Prove that F=mgtanθ

Can everyone help me to solve this problem....

A small sphere of mass hangs from a string of L as in below. A variable horizontal force F is
applied to the sphere in such a way that it moves slowly from the vertical position until the string makes angle θ with the vertical. Assuming the sphere is always in equilibrium

1. show that F=mgtanθ
2.show that work done by F is mgL(1-cosθ)

__________________________
|\
| \
| θ \ L
| \
| \
| 0------F
0 |
|
|
W

5. Sep 9, 2008

### tiny-tim

("Can anyone help me to solve this problem...." would be more realistic! )

"moves slowly" is examiners' code for saying that, at any particular time, the acceleration is so small that you can treat it as zero.

Does that help?

6. Sep 10, 2008

so, let d be the displacement W(work)=F*dcosθ
∑F=ma=0 then,
∑F=W+T+F
W+T+F=0
-W*dcosθ + T*dcosθ + F*dcos(θ+90)

[T*dcosθ=0 because they are perpendicular]

W*dcosθ=F*dcos(θ+90)

but cos(θ+90)=sinθ

w*dcosθ=F*dsinθ
F=Wtanθ
F=mgtanθ

is my solution correct?

7. Sep 10, 2008

thnks tim....

plz... i need a hint for the problem number 2.....

8. Sep 10, 2008

### tiny-tim

Let's do part 1 first:
You should say which direction you are taking components in, and why.

Anyway, if w*dcosθ=F*dsinθ, then F = w cotθ.
Whatever is this supposed to be??

Hint: work done = energy gained.

(this is the only reason we bother about work done)

9. Sep 10, 2008

okay....

There three forces acted on the sphere:
tension(T), applied force F and gravitational force(mg)

so..
W=T*Scosθ=0
the displacement of the sphere along the arc length is perpendicular to the tension
While..
W=-mg*d cos(θ+90)=-mg*dsinθ (trigonometric identity of cos(θ+90)=sinθ)
because the angle between the displacement of the sphere is more than 90 degrees..
thus...work done by weight is negative
however..the displacement of the sphere started from the vertical position to the slanting position until the two position made an angle θ
and,
W=F*dcosθ (angle between F and d is θ)
so...
equate W to W:
F*dcosθ= -mg*dsinθ (cancel d)
F*cosθ=-mg*sinθ
F=-mgtanθ--------this is my problem...i just end up with negative sign...

10. Sep 10, 2008

work done by gravitational force(W=-mgcos(θ+90)) is negative because it opposed the direction of W=F*dcosθ

11. Sep 10, 2008

### tiny-tim

I'm sorry, I can't follow any of this …

you seem to be using work done to find the value of F.

1. to find F, just take components perpendicular to the unknown force, whch is … ?

2. to find the work done, either:

i] integrate the force "dot" the displacement (which works, but is really complicated ), or:

ii] just use the work-energy theorem: work done equals energy gained …

so what is the energy gained?