### 6-2 Using Newton*s Laws

```6-2 Using Newton’s Laws
Mass and Weight
 Objects at different weights will still fall at the same rate
 Weight force
 The force of gravity acting upon an object (Fg)
 Ball falling through air (neglecting air resistance)
 Only force acting on it is Fg.
 Acceleration is g
 Newton’s 2nd Law becomes Fg = mg
 Both force and acceleration are downward
Scale
 What does a scale actually measure?
 Scale exerts an upward force on you
 Because you are not accelerating, Fnet = 0
 Ergo, the magnitude of the spring force (Fsp) is equal to your
weight, Fg.
Weighing self in an elevator
 Your mass is 75 kg. You stand on a bathroom scale in an
elevator. Going up!! Starting from rest the elevator
accelerates at 2.0 m/s2 for 2.0 sec then continues at constant
speed. What is the scale reading during the acceleration?
 m = 75 kg a = +2.0 m/s2
t = 2.0 sec
 Unknown = Fscale
Strategy
 Fnet = Fscale – Fg
 Rearrange: Fscale = Fnet + Fg = ma + mg
 Fscale = ma + mg or Fscale = m(a + g)
 = (75 kg)(2.0 m/s2 + 9.80 m/s2)
 = 890 N
Lifting a Bucket
 A 50 kg bucket is being lifted by a rope. The rope is
guaranteed not to break is the tension is 500 N or less. The
bucket started at rest, and after being lifted 3.0 m, it is
moving at 3.0 m/s. Assuming the acceleration is constant, is
the rope in danger of breaking?
 m = 50 kg, v = 3.0 m/s, vo = 0.0 m/s, d = 3.0 m
 Unknown = FT = ?
Strategy
 Net force is the vector sum of FT (positive) and Fg (negative)
 Fnet = FT – Fg
 Rearrange: FT = Fnet + Fg = ma + mg
 Because vo is zero, a = v2 / 2d
 FT = m(a + g) = m(v2/2d + g)
 = (50 kg)((9.0m2/s2)/(2x3.0m) + 9.80 m/s2)
 FT = 570 N
Friction Force
 Friction is minimized in solving force and motion problems
 But friction is everywhere!!!!!
 Static Friction
 Force exerted on one surface by the other when there is no
relative motion between the two
 Static FF acts in response to other forces
 Kinetic Friction
 Force exerted on one surface by the other when the surfaces are
in motion
Model Friction Forces
 Friction forces put into an equation
 Kinetic Friction
 Ff,kinetic = µk FN
 Usually called the coefficient of kinetic friction
 Static Friction
 0 ≤ Ff,static ≤ µsFN
 Equation tells you that Static friction force can be anywhere
from 0 to the maximum force is before movement
Typical Coefficients of Friction
Surface
µs
µk
Rubber on concrete
0.80
0.65
Rubber on Wet concrete
0.60
0.40
Wood on Wood
0.50
0.20
Steel on Steel (dry)
0.78
0.58
Steel on Steel (oil)
0.15
0.06
Teflon on Steel
0.04
0.04
Balanced Friction Forces Problems
 You push a 25 kg wooden box across the wooden floor at a
constant speed of 1.0 m/s. How much force do you exert on
the box?
Unbalanced Friction forces
 If the force you exerted on the box is doubled, what is the





resulting acceleration on the box?
m = 25 kg
v = 1.0 m/s
µk = 0.20
Fp = 2(49 N) = 98 N
Unknown
 a = ???
If the force you exerted on the box is doubled, what is the resulting
acceleration on the box?
 y-direction: FN = Fg = mg; FN = mg
 x-direction: FP – Ff = ma
 Ff = µkFN = µkmg
 Fnet = ma
 a = Fnet / m
 = FP - µkmg/ m
 = (FP/m) - µkg
 a = (98 N / 25 kg) – (0.20)(9.8 m/s2)
 a = 2.0 m/s2
 A boy exerts a 36 N horizontal force as he pulls a 52 N sled
across a cement sidewalk at a constant speed. What is the
coefficient of kinetic friction between the sidewalk and the
metal sled runners? (Ignore air resistance)
 Suppose the sled runs on packed snow. The coefficient of
friction is now only 0.12. If a person weighting 650 N sits on
the sled, what force is needed to pull the sled across the snow
at a constant speed?
Air drag and terminal velocity
 This force depends upon…
 speed of the motion, becoming larger as speed increases
 Size and Shape of the object
 Density
 Kind of fluid
 Terminal Velocity
 Velocity of an object in free fall becomes equal to the drag force
on the object
Periodic Motion
 Playground swing, guitar string, pendulum
 Each object has one position where F Net = 0
 In equilibrium
 When object is pulled away from equilibrium, the net force
becomes nonzero and pulls it back toward equilibrium
 If the force that restores the object to its eq. position is
directly proportional to displaecment, motion resulted is
Simple Harmonic Motion
 Period: time to complete one cycle of motion (T)
 Amplitude: maximum distance object moves from equilibrium
Mass on a Spring
Pendulum
 Bob and suspended string (l)
 String exerts a tension force (FT)
 Gravity exerts the weight force (Fg)
 Vector sum of the two forces produces the net force
 Period of a pendulum of length l
 T = 2л√(l/g)
 Notice period is only determined by length of string and gravity
not mass or amplitude
Resonance
 How do you get a swing going?
```
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