Transcript What is the normal force for a 500 kg object resting on a

What is the normal force for a 500 kg
object resting on a horizontal surface if
a massless rope with a tension of 150 N
is acting at a 45oangle to the normal
force?
The mass of the object is too large to be moved
by the vertical component of the rope’s force, so
it is not in motion.  Fy = 0, where
 Fy = FN – mg + Fy
Fy = FN – (500kg)(9.8 m/s2) + (150N)(sin45o)
Free Body Diagrams
• A mass at the origin is pulled by four perpendicular
ropes
ˆ , (8N ) x
ˆ, (15N )  x
ˆ, (18N )  y
ˆ
(10N ) y
1. Find the magnitude of the resultant force.
2. Find the angle at which the force is acting.
Now each of the ropes is pulled at 30o counterclockwise
from the given axes.
3. Draw a free-body diagram and find the horizontal and
vertical components of the new resultant force.
4. At what angle is the resultant force acting with respect
to the x-axis?
Newton’s First Law
• An object at rest will remain at rest and an
object in motion will keep moving with
constant velocity if the net force on the object
is zero.
• Often called the law of inertia. Inertia is the
natural tendency of an object to resist a
change in motion, measured quantitatively by
its mass.
• What happens if the net force is not zero?
Newton’s Second Law
• Acceleration depends on the net force and on the
mass of the object 

 Fnet

a
or Fnet  ma
m
• Newton’s law is a vector equation.
• Acceleration is in the direction of the net force,
not necessarily the direction of the velocity.
• Include all forces acting on an object to
determine the net force.
• Do not include balanced forces (such as weight
and normal force) because they cancel each
other
Newton’s Second Law of Motion
The units of force are called newtons.
1 N = 1 kg . m/s2.
Example 4.2
• A 5.0 N horizontal force pulls a 20 kg box on a
horizontal surface. A 3.0N friction force slows
the motion. What is the acceleration of the
object?
1. draw a free-body diagram of the box
2. write the given information
3. Apply the equation F = ma to solve the
problem
Example 4.3
• Find the magnitude of the weight of a 3.50 kg
object on the surface of the earth where
w = mg
Newton’s Second Law of Motion
Newton’s second law applies separately to
each component of the force.
Newton’s Third Law
• If object 1 exerts force on object 2, then
object 2 exerts an equal and opposite force on
object 1.
• The two forces are acting on two different
objects, therefore even though the two forces
are equal and opposite, they do not
necessarily cancel each other.
• Often called the law of action and reaction.
4.4 Newton’s Third Law of Motion
For every force (action), there is an equal and
opposite force (reaction).
Note that the action and reaction forces act on
different objects.
This image shows how a
block exerts a downward
force on a table; the
table exerts an equal and
opposite force on the
block, called the normal
force N.
Example 4.4
• A large truck collides head-on with a small car
and causes a lot of damage to the small car.
Explain why there is more damage to the
small car than to the large truck.
Homework 10/27/08
• P. 131 -132; 2, 3, 5, 6, 13, 14, 15, 16, 19
• Finish questions from each of the four lab
stations