# How to: Force, work & dot product

1. Dec 3, 2005

### Lisa...

A force in the xy plane is given by: F= - (b/r3) (x i^ + y j^) where b is a positive constant and r= sqrt(x2 + y2)

a) Show that the maginitude of the force varies as the inverse of the square of the distance to the origin and that its direction is antiparallel (opposite) to the radius vector r= x i^ + y j^).

Could anybody please verify my work?:

The following needs to be true:
F ~ 1/r2 => F= c /r2, where c is a constant.

F * F= F2 = (b2 x2/r6) + (b2 y2/r6)

F= sqrt((b2 x2/r6) + (b2 y2/r6) )= sqrt((b2/r6) (x2 + y2)) = b/r3 sqrt(x2 + y2)

Because r= sqrt(x2 + y2) substitution gives:

b/(sqrt(x2 + y2))3 sqrt(x2 + y2)= b/ (x2 + y2).

r2= (x2 + y2) so F= c /r2 is true where the constant c= the constant b in this case.
----------------------------------------------------------------------
F= - (b/r3) (x i^ + y j^) needs to be antiparallel to r= x i^ + y j^).

If two vectors are parallel the following needs to be true: AB (dotproduct)= AB (length vectors).

The dotproduct AB gives: -bx2/r3 + -by2/r3
Substitution of r= sqrt(x2 + y2) and adding the two terms gives:

(-bx2 -by2)/(sqrt(x2 + y2))3= (-b (x2 + y2))/(sqrt(x2 + y2))3= -b/(x2 + y2)

The length of F= b/(x2 + y2), calculated in the first part of the question.

The length of r= r * r= r2= x2+ y 2
r= sqrt(x2+ y 2)

Fr (length of the vectors) gives: b (sqrt(x2+ y 2)/(x2 + y2) = b/ (sqrt(x2 + y2))

Therefore it is true that Fr (dotproduct)= Fr (length vectors):
-b/(x2 + y2)= b/ (sqrt(x2 + y2)) with the only restriction that Fr (dotproduct) is negative and therefore F and r are antiparallel.

Ok till now I get the problem, but I'm not sure how to solve b, c and d... so please help me ...

b) If b= 3 Nm2 find the work done by this force on a particle moving along a straight line path between an initial position x= 2 m, y= 0 m and a final position x= 5 m, y= 0 m.

I know W= Fs and s= 3 m but I don't know how to calculate the force with - (3/r3) (x i^ + y j^)...

c) Find the work done by this force on a particle moving once around a circle of radius r= 7 mcentered around the origin.

d) If this force is the only force acting on the particle what's the particle's speed as it moves along this circular path? Assume that the particle's mass is m= 2 kg.

2. Dec 3, 2005

### Galileo

Correct so far (didn't check the rest yet). At this stage I recommend looking back at the solution and try to see the 'big picture' of what you have done. Perhaps you will be able to see the solution in a glance or see how you could have gotten there much quicker.
What you did was subsitute r^2=x^2+y^2 back and forth again and again. When you reached the point F= b/r^3 sqrt(x2 + y2) you used $r=\sqrt{x^2+y^2}$, but subsituted r in favor of x and y (!) instead of the other way around and later you 'undid' that. Writing $\sqrt{x^2+y^2}=r$ in $F= \frac{b}{r^3}\sqrt{x^2 + y^2}$ gives $F=b/r^2$ right away.
The way to know what way to go is partly experience, but in this case you want the answer to be in terms of r, so try to eliminate x and y right away.

If you know vector notation the question becomes much easier and the solution more insightful, since $\vec r=x \hat i+y\hat j$ we have:

$$\vec F= - \frac{b}{r^3}(x \hat i + y \hat j) =-b \frac{\vec r}{r^3}=-b \frac{\hat r}{r^2}$$

Where $\hat r=\vec r/r$ is the unit vector in the direction of $\vec r$. In this case the answers are immediate.

Last edited: Dec 3, 2005
3. Dec 6, 2005

### Lisa...

Using your formula for F I've found:

b) The work is found by integrating your formula for F with respect to r between 2 and 5. W25= 3/5 - 3/2 = -0.9 J

c) Because the force is perpendicular to the path the dotproduct will equal zero. Therefore the work is zero when the particle is doing circular motion.

d) F= -b r^/r2 with r= 7 m and b= 3 N/m.
F= -3/49= -0.0612244898 N= Fres

Fres= m a= mac= m v2/r
v2= Fres*r/m= -0.06122* 7/2= -0.2142857
v= sqrt(0.2142857)= -0.46291 m/s

Is this correct?

4. Dec 6, 2005

### Galileo

For d), remember that the F you should consider in the equation F=mv^2/r is the magnitude of $\vec F$, which is always positive. You should get a warning from the fact that the v^2 you calculated was negative.

The units of b are Nm^2, not N/m. But apart from these things it looks good. Well done.

5. Dec 7, 2005

### Lisa...

Ah of course :) But then it's simple: I only need to make the negative signs positive and I'll get the right answer, correct? So the answer would be 0.46291 m/s and not - 0.46291 m/s?

6. Dec 7, 2005

### Galileo

Yes, that's right