Transcript Applications & Examples of Newton’s Laws

Applications & Examples of Newton’s Laws
• Forces are VECTORS!!
• Newton’s 2nd Law: ∑F = ma
∑F = VECTOR SUM of all forces on mass m
 Need VECTOR addition to add forces in
the 2nd Law!
– Forces add according to rules of VECTOR
ADDITION! (Ch. 3)
• Newton’s 2nd Law problems:
• STEP 1: Sketch the situation!!
– Draw a “Free Body” diagram for EACH body in
problem & draw ALL forces acting on it.
• Part of your grade on exam & quiz problems!
• STEP 2: Resolve the forces on each body into
components
– Use a convenient choice of x,y axes
• Use the rules for finding vector components from Ch. 3.
• STEP 3: Apply Newton’s 2nd Law to
EACH BODY SEPARATELY:
∑F = ma
Notice that this is the LAST step, NOT the first!
– A SEPARATE equation like this for each body!
– Resolved into components:
∑Fx = max
∑Fy = may
Conceptual Example
Moving at constant v, with NO friction,
which free body diagram is correct?
Example
Particle in Equilibrium
“Equilibrium” ≡ The total force is zero.
∑F = 0 or ∑Fx = 0 & ∑Fy = 0
Example
(a) Hanging lamp (massless chain).
(b) Free body diagram for lamp.
∑Fy = 0  T – Fg = 0; T = Fg = mg
(c) Free body diagram for chain.
∑Fy = 0  T – T´ = 0; T´ = T = mg
Example
Particle Under a Net Force
Example
(a) Crate being pulled to right
across a floor.
(b) Free body diagram for crate.
∑Fx = T = max  ax = (T/m)
ay = 0, because of no vertical motion.
∑Fy = 0  n – Fg = 0; n = Fg = mg
Example
Normal Force Again
“Normal Force” ≡ When a mass is in contact with a surface,
the Normal Force n = force perpendicular to (normal to)
the surface acting on the mass.
Example
Book on a table. Hand pushing down.
Book free body diagram.

ay = 0, because of no vertical motion
(equilibrium).
∑Fy = 0  n – Fg - F = 0
 n = Fg + F = mg + F
Showing again that the normal force is
not always = & opposite to the weight!!
Example 5.4:
Traffic Light at Equilibrium
(a) Traffic Light, Fg = mg = 122 N
hangs from a cable, fastened to a
support. Upper cables are weaker than
vertical one. Will break if tension exceeds
100 N. Does light fall or stay hanging?
(b) Free body diagram for light.
ay = 0, no vertical motion.
∑Fy = 0  T3 – Fg = 0
T3 = Fg = mg = 122 N
(c) Free body diagram for cable junction (zero mass). T1x = -T1cos(37°), T1y = T1sin(37°)
T2x = T2cos(53°), T2y = T2sin(53°), ax = ay = 0. Unknowns are T1 & T2.
∑Fx = 0  T1x + T2x = 0 or -T1cos(37°) + T2cos(53°) = 0
(1)
∑Fy = 0  T1y + T2y – T3 = 0 or T1sin(37°) + T2sin(53°) – 122 N = 0 (2)
(1) & (2) are 2 equations, 2 unknowns. Algebra is required to solve for
T1 & T2!
Solution: T1 = 73.4 N, T2 = 97.4 N
Example
Example 5.6: Runaway Car
Example 5.7: One Block Pushes Another
Example 5.8: Weighing a Fish in an Elevator
Example 5.9: Atwood Machine
Example 5.10
Inclined Plane, 2 Connected Objects