1. 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

2. x m1 m2 m3 L x1 m2 m1 x 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:
FBD m1
F12

3. FBD
F21 m2
Situation 2:
FBD
Since in situation 2 the total force must be 2F12.
Solve for x.
F13 m1
F12

4. FBD
F23 m2
Now consider m2:
F21

5. or
Substitute for m3

6. m3 x m1 m2 m3 L x1
Since

7. The Normal Force Consider the textbook on the table F? F? mg mg
Consider Newton’s 2nd law in y-direction:
but book is at rest. So, ay=0, gives
New force has same magnitude as the weight, but opposite direction

8. New force is a result of the contact between the book and the table
New 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 FA N FA N mg mg
The normal force is not mg!

9. Normal Force (Revisited)
Put textbook on a scale in an elevator FN FN a mg mg
If elevator is at rest or moving with a constant velocity up or down, a=0. Then Newton’s 2nd law gives:

10. If a = -g, FN = 0 (“weightless”)
If elevator is accelerating?
If a > 0, FN > mg
If a < 0, FN < mg
If a = -g, FN = 0 (“weightless”)

11. Book on an Incline (Frictionless)y y
FBD FN FN mg x    x mg
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

12. If   0°, ax = 0 and FN = mg
If   90°, ax = g, FN = 0