Example Problem (difficult!)Two particles are located on the x-axis. Particle 1 has a mass of m and is at the origin. Particle 2 has a mass of 2m and is at x=+L. A third particle is placed between particles 1 and 2. Where on the x-axis should the third particle be located so that the magnitude of the gravitational force on both particles 1 and 2 doubles? Express your answer in terms of L.

Solution:

Principle – universal gravitation (no Earth), F12=Gm1m2/r2

Strategy – compute forces with particles 1 and 2, then compute forces with three particles

xm1

m2

L

Situation 1

Situation 2

Given: m1 = m, m2 = 2m, r12 = L

Don’t know: m3=?

Find: x = r13 when force on 1 and 2 equals 2F12

Situation 1:

2

2

2221

12

2)2(

L

Gm

L

mGm

r

mGmFFx

FBD

m1 F12

xm1

m2m3

L

x x1

2

2

1221

2

L

GmFFFx

FBD

F21 m2

Situation 2:

2

2

23

2

2

1312

42

L

Gm

x

Gmm

L

GmFFFx

Since in situation 2 the total force must be 2F12.

Solve for x.

F13

F12m1

FBD

223

223

2

2or

42

L

m

x

m

L

m

x

m

L

m

Lm

mx

m

L

m

x

2or

23

2

3

2

2

2

21

32

2

2321

4)2(2

L

Gm

x

mmG

L

GmFFFx

Now consider m2:

221

322

1

32

22or

422

L

m

x

m

L

m

x

m

L

m

Lm

mx

m

L

m

x 31

2

3

21 or

22

F23

F21

m2

FBD

m

mLL

m

mLxLx

Lxx

331

1

1

or

Substitute for m3

L

x

m

mL

m

mx 2

233

xLL

xLx 221

LL

Lx

Lxxx

414.2or 414.021

212

xm1

m2m3

L

x x1

m3

mmm 7.11or 343.03 Since

The Normal Force Consider the textbook on the table

F?

mg

Consider Newton’s 2nd law in y-direction:

F?

mg

? yy mamgFF

but book is at rest. So, ay=0, gives

0 ?? mgFmgFFyNew force has same magnitude as the weight, but opposite direction

New force is a result of the contact between the book and the tableNew force is called the Normal Force, N or FN

In general it is not equal to mg - we must usually solve for N``Normal’’ means ``perpendicular’’ (to the surface of contact)Now, apply an additional force, FA to the book

N

mg

FA

N

mg

FA

A

Ay

FmgNFmgNF

0

The normal force is not mg!

Normal Force (Revisited) Put textbook on a scale in an elevator

FN

mg

a

If elevator is at rest or moving with a constant velocity up or down, a=0. Then Newton’s 2nd law gives:

WmgFmgFF NNy or 0

FN

mg

If elevator is accelerating?

)(

or

agmF

mamgFmamgFF

N

NNy

If a > 0, FN > mg

If a < 0, FN < mg

If a = -g, FN = 0 (“weightless”)

Book on an Incline (Frictionless)

mg

FN

x

y

FN

mg

y

x

Using Newton’s 2nd Law, find the normal force and the acceleration of the book

As we did for 2D kinematics, break problem into x- and y-components

yyxx maFmaF

FBD

cos sin

0cos sin

mgFga

mamgFmamg

Nx

yNx

If 0°, ax = 0 and FN = mg

If 90°, ax = g, FN = 0