Transcript ppt - HRSBSTAFF Home Page

Unit Two: Dynamics
Section 4: Forces and
acceleration
Newton’s Second Law

Newton’s first law states that an object does
not accelerate unless a net force is applied to
the object.

This relates back to the first law (an object
will continue with the same velocity unless a
force acts upon it).
Newton’s Second Law

But how much will an object accelerate when there
is a net force?



The larger the force the larger the acceleration.
Therefore acceleration is directly proportional to mass.
Acceleration also depends on mass.



The larger the mass, the smaller the acceleration.
Therefore acceleration is inversely proportional to mass.
We say that a massive body has more INERTIA than a less
massive body.
Newton’s Second Law:
Summary



A net force is needed to accelerate an object.
A larger force means more acceleration.
A larger mass means less acceleration as it is
harder to move.
Newton’s Second Law
- Newton’s Law of Motion
(Summary)

Force = mass x acceleration

Fnet = ma

***The acceleration is in the same direction
as the net force.
Newton’s Second Law
Examples

Ex. 1: What net force is required to
accelerate a 1500. kg race car at +3.00m/s2?

Draw a FBD to show the net force.
Practice Problems

Page 163, Questions 1, 2, 3
Putting it All Together


Now that we have considered Newton’s
Second Law, you can use that to analyze
kinematics problems with less information
than we have used previously
We can either use dynamics information to
then apply to a kinematic situation or vice
versa
Newton’s Second Law
Examples

Ex. 2: An artillery shell has a mass of 55 kg.
The shell is fired from a gun leaving the
barrel with a velocity of +770 m/s. The gun
barrel is 1.5m long. Assume that the force,
and the acceleration, of the shell is constant
while the shell is in the gun barrel. What is
the force on the shell while it is in the gun
barrel?
Example 2.5

A 25kg crate is slid from rest across a floor
with an applied force 72N applied force. If
the coefficient of static friction is 0.27,
determine:



The free body diagram. Include as many of the
forces (including numbers) as possible.
The acceleration of the crate.
The time it would take to slide the crate 5.0m
across the floor.
FBD
FN=250N
Fa=72N
Ff=?
Fg=-250N
Use the frictional force equation to
determine the magnitude of the
frictional force
F f  FN
F f  (.27)( 250 N )
F f  66 N

F f  66 N
The net force is the sum of the forces
(acting parallel or anti-parallel)

Fnet

Fnet

Fnet

Fnet

  Fi


 F f  Fa
 66 N  72 N
 5.8 N
Use Newton’s Second Law to solve
for the acceleration


Fnet  ma

5.8 N  (25kg)a

2
a  0.23m / s
Use kinematics to solve for the time
taken to cross the floor

2



at
d (t ) 
 v0t  d 0
2
2 2
0.23m / s t
5 .0 m 
2
2(5.0m)
t
2
0.23m / s
t  6 .6 s
Practice Problems

Page 168, questions 4 to 8
Example 3



A baby carriage with a mass of 50. kg is
being pushed along a rough sidewalk with
an applied horizontal force of 200. N and it
has a constant velocity of 3.0 m/s.
A) What other horizontal force is acting on
the carriage and what is the magnitude and
direction of that force?
B) What value of applied horizontal force
would be required to accelerate the carriage
from rest to 7.0 m/s in 2.0 s?
Example 3





A) Force of friction must be equal in
magnitude (so 200.N) in the opposite
direction to the force applied (of the baby
carriage moving forward).
B) First find a using a = (vf – vi)/t = 2.0m/s2
Now find Fnet = ma = 50x2 = 100N
Now use Fnet = Fapp – Ff
Fapp = 300N = 3.0 x 102 N
Example 4

A horizontal force of 50. N is required to pull
an 8.0 kg block of aluminum at a uniform
velocity across a horizontal wooden desk.
What is the coefficient of kinetic friction?
Example 4





You know that Force of friction is equal and
opposite to Force applied. Therefore, Ff =
50.N.
You know that Force of friction = coefficient of
friction x normal force.
Normal Force = - Force of gravity = -mg
Fn = 78.48N
Sub in and rearrange to find that coefficient =
0.64 (no units)
Example 5





A 75 kg man stands in an elevator. Draw a free body
diagram and determine what the force the elevator
exerts on him will be when
A) the elevator is at rest
B) the elevator is moving upward with a uniform
acceleration of 2.0 m/s2
C) the elevator is moving downward with a uniform
velocity of 2.0 m/s
D) the elevator is moving downward with a uniform
acceleration of 2.0 m/s2
Example 5




A)
B)
C)
D)
Fnet = 0N, Fapp = 0N
Fnet = 150N [up]
Fnet = 0N
Fnet = 150N [down]