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 inthe 2nd Law!
–Forces add according to rules of VECTORADDITION! (Ch. 3)
•Newton’s 2nd Law problems:
•STEP 1: Sketch the situation!!
–Draw a “Free Body” diagram for EACH body inproblem & draw ALL forces acting on it.
•Part of your grade on exam & quiz problems!
•STEP 2: Resolve the forces on each body intocomponents
–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
–A SEPARATE equation like this for each body!
–Resolved into components:
∑Fx = max ∑Fy = may
Notice that this is the LAST step, NOT the first!
Conceptual Example
Moving at constant v, with NO friction,
which free body diagram is correct?
ExampleParticle 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
ExampleParticle 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
ExampleNormal 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
A box of mass m = 10 kg is pulled by an attached cord along ahorizontal smooth (frictionless!) surface of a table. The force exertedis FP = 40.0 N at a 30.0° angle as shown. Calculate:
a. The acceleration of the box.
b. The magnitude of the upward normal force FN exerted by thetable on the box.
Free Body
Diagram
The normal force, FN is NOT
always equal & opposite to the weight!!
Two boxes are connected by a lightweight (massless!) cord & areresting on a smooth (frictionless!) table. The masses are mA = 10 kg &mB = 12 kg. A horizontal force FP = 40 N is applied to mA. Calculate:a. The acceleration of the boxes. b. The tension in the cord connecting theboxes.
Example
Free Body
Diagrams
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 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 4-13 (“Atwood’s Machine”)
Two masses suspended over a (massless frictionless) pulley by a flexible(massless) cable is an “Atwood’s machine”. Example: elevator &counterweight. Figure: Counterweight mC = 1000 kg. Elevator mE = 1150kg. Calculate a. The elevator’s acceleration. b. The tension in thecable.
aE = - a
aC = a
a
a
Free Body
Diagrams
Conceptual Example
mg = 2000 N
Advantage of a Pulley
A mover is trying to lift apiano (slowly) up to a second-story apartment. He uses arope looped over 2 pulleys.
What force must he exert onthe rope to slowly lift thepiano’s mg = 2000 N weight?
Free Body Diagram
Example: Accelerometer
A small mass m hangs from a thinstring & can swing like apendulum. You attach it above thewindow of your car as shown.What angle does the string make
a. When the car accelerates at aconstant a = 1.20 m/s2?
b. When the car moves atconstant velocity, v = 90 km/h?
Free Body Diagram
Example
= 300 N
FT2x = FTcosθ
FT2y = -FTsinθ
FT1x = -FTcosθ
FT1y = -FTsinθ
Free Body
Diagram
Inclined Plane Problems
Understand ∑F = ma & how to resolve it into x,y components in the tilted coordinate system!!
Engineers & scientists
MUST understand these!
a
The tilted coordinate
System is convenient,
but not necessary.
A box of mass m is placed on a smooth (frictionless!) incline thatmakes an angle θ with the horizontal. Calculate:a. The normal force on the box. b. The box’s acceleration.c. Evaluate both for m = 10 kg & θ = 30º
Example: Sliding Down An Incline
Free Body
Diagram
Example 5.10Inclined Plane, 2 Connected Objects
