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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
m2 m1 m3 x x x1 L 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
FBD F21 m2 Situation 2: FBD Since in situation 2 the total force must be 2F12. Solve for x. F13 m1 F12
FBD F23 m2 Now consider m2: F21
or Substitute for m3
m2 m1 m3 x x x1 L m3 Since
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
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!
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:
If elevator is accelerating? • If a > 0, FN > mg • If a < 0, FN < mg • If a = -g, FN = 0 (“weightless”)
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
If 0°, ax = 0 and FN = mg • If 90°, ax = g, FN = 0
