Download CH 4.1 * 4.2 - Seymour ISD

CH 4.1 – 4.2
Resistance in Mechanical and Fluid System
4.1 Objectives
 State Newton’s second law of motion and use it to
solve problems involving force, mass, acceleration
 Calculate an object’s weight given its mass
 Explain the difference between static and kinetic friction
 Explain how lubrication and rolling reduce friction
Newton’s 2nd Law of Motion
 The acceleration of an object is directly proportional to
the net force acting on the object and inversely
proportional to the mass of the object
a=F/m
F = ma
1 N = (1 kg ) (1 m/s2)
m=f/a
1 lb = (1 slug) (1 ft/s2)
Newton’s 2nd Law and a Car’s Motion
 In performance testing, a 1250-kg car accelerates from
0 to 100 km/hr in 8.2 seconds. What is the average net
force pushing the car during the test?
Example 1 Solution
 Convert velocity to m/s :
(100 km) (1000 m )
1hr
1 km
(1 hr)
= 27.78 m / s
3600 sec
 Determine the car’s acceleration :
a= (Vf – Vi) / t
a= (27.78 m/s – 0 m/s) / 8.2 s
a= 3.39 m/s2
 Calculate the force
F = ma
(1250-kg) (3.39 m/s2) = 4237.5 N
Weight and Mass
 The weight of an object is = (mass) (acceleration due to
gravity)
Fg = mg
Acceleration due to gravity:
9.80 m/s2
32.2 ft/s2
Example 2
 A 65-kg person stands on a bathroom scale. What
force does the scale exert on the person?
(Think of Newton’s 3rd Law of Motion)
Example 2 Solution
 Fscale = Fg
 Fg = m g
= (65 kg) (9.8 m/s2)
= 637 N
Example 3
 The same 65-kg person is placed on a scale in an
elevator. The elevator accelerates upward at a rate of
2.5 m/s2. What weight does the scale read while the
elevator is moving?
 Still consider Newton’s 3rd Law of Motion
Example 3 Solution
Fscale = Fnet + Fg
= ma + mg
= m(a + g)
= 65 kg (2.5 m/s2 + 9.80 m/s2)
= 65 kg (12.3 m/s2)
= 799.5 N
 What would the scale reading be if the elevator were
accelerating downward at the same rate?
Calculating Distance using Gravity
 Distance = ½ g t2
 Accelg = (2d) / t2
 Time = √ (2d) / a
Example 4
 Wiley coyote pushes a 0.32 kg stone from a cliff 123 ft
high. If the acceleration due to gravity is 30.1 ft/s2, how
long must the Road Runner “beep-beep” be standing
still below?
 Why is the acceleration due to gravity less than the
described value of 32.2 ft/s2 ?
FRICTION
 When a force is exerted on an object, inertia must be
overcome for there to be a change in the motion of an
object.
 Friction must also be overcome for a stationary object to
be moved.
 What happens to much of the energy applied to a system
when friction is present?
 What can be done to reduce friction in a system?
Types of Friction
 Static Friction
 The initial force of friction that must be overcome to start
movement
 Kinetic friction
 The second form of friction that must be overcome to
keep the object moving at a constant rate of speed
 Think of the coefficient of friction as how slippery the
surface of an object is.
 Which joint has the most friction? What do is this
condition known as?