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Unformatted text preview: Summary Lecture 9
Newton’s Law of Gravity
Newton’s Law of Gravity
Force of Gravity
Directed along the line connecting the 2
masses. Always attractive.
Always attractive m1m2
FG = G
r2 Universal Gravitational Constant
G = 6.67 x 1011 Nm2/kg2 Apparent Weight Fnet = FNmg
2nd Law:
FN –mg = ma
Or
(FN = apparent weight) FN = m(g+a) Drawings Courtesy of L. Gladding RECIPE:
For each body in the problem
– Draw the free body diagram including all forces
acting ON this body
– Define your coordinate system (positive x, positive y)
– If necessary, split up forces in horizontal and vertical
components (x and y)
→ → – Determine the net force Fnet = ΣF
separately for x and y, if necessary: FNet x = ∑ Fx FNet y = ∑ Fy – Apply Newton’s second law
→ → Fnet = m a separately for x and y, if necessary: FNet x = ma x FNet y = ma y – m is the mass of this body
a is the acceleration of this body Solve equations for unknowns ...
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This note was uploaded on 12/06/2011 for the course PHYSICS 111&112 taught by Professor Unknown during the Spring '11 term at Ohio State.
 Spring '11
 UNKNOWN
 Force, Gravity, Mass

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