Transcript Part IV

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
A box of mass m = 10 kg is pulled by an attached cord along a
horizontal smooth (frictionless!) surface of a table. The force exerted
is FP = 40.0 N at a 30.0° angle as shown. Calculate:
table
a. The acceleration of the box.
b. The magnitude of the upward normal force FN exerted by the
on the box.
Free Body
Diagram
The normal force, FN is NOT
always equal & opposite to the weight!!
Example
Two boxes are connected by a lightweight (massless!) cord & are
resting 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 the
boxes.
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 = 1150
kg. Calculate
a. The elevator’s acceleration. b. The tension in the
cable.
a


a
aE = - a
Free Body
Diagrams
aC = a
Conceptual Example
Advantage of a Pulley
A mover is trying to lift a
piano (slowly) up to a secondstory apartment. He uses a
rope looped over 2 pulleys.
What force must he exert on
the rope to slowly lift the
piano’s mg = 2000 N weight?
mg = 2000 N
Free Body Diagram
Example: Accelerometer
A small mass m hangs from a thin
string & can swing like a
pendulum. You attach it above the
window of your car as shown.
What angle does the string make
a. When the car accelerates at a
constant a = 1.20 m/s2?
b. When the car moves at
constant
velocity, v = 90 km/h?
Free Body Diagram
Example
= 300 N
Free Body
Diagram
FT1x = -FTcosθ
FT1y = -FTsinθ
FT2x = FTcosθ
FT2y = -FTsinθ
Inclined Plane Problems
The tilted coordinate
System is convenient,
but not necessary.
Engineers & scientists
MUST understand these!
a
Understand ∑F = ma & how to resolve it into x,y
components in the tilted coordinate system!!
Example: Sliding Down An Incline
A box of mass m is placed on a smooth (frictionless!) incline that
makes 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
Free Body
Diagram
Example 5.10
Inclined Plane, 2 Connected Objects