Transcript Chapter 4 Notes-part 2

Chapter 4 Sec
6-8
Weight, Vector
Components, and
Friction
The Force of Gravity, Weight,
and the Normal Force
•
The force causing all objects to accelerate toward
earth’s is called the force of gravity, g.
•
Applying Newton’s Second Law, F=ma, we obtain the
force of gravity on an object accelerating at 9.80m/s/s is
called Weight, Fg=mg and is measured in Newtons, N.
•
When two objects are in contact with each other a
contact force acts perpendicular to the surface of
contact and is called a Normal force, FN.
Do forces cancel?
• Newton’s third law of motion describes ‘action-
reaction’ pairs of forces. It is important to note
these forces are on DIFFERENT objects, so they
do NOT cancel. (Example: carry a ball: ball on
back force,F1, does not cancel back on ball
force,F2, because F1 is on the back and F2 is
on the ball.
• Normal force, FN on an object is not cancelled
by Earth’s Fg on the same object.
Example: Consider a box on a
table 
 A friend gives you a gift box with a mass of
10.0 kg. The box rests on a horizontal table.
Determine the weight of the box and the
normal force on it. If your friend pushes
down on the box with a force of 40.0 N,
determine the new normal force on the box.
Be sure to draw free body diagrams to show
the forces acting.
Solution
 Given: m = 10.0 kg, g = 9.80m/s/s, FN = ?
 Fg = mg = 10.0 kg * 9.80m/s/s = 98 N acting
downward. The only other force on the box
is the normal force, FN and acts upward.
F
y
 FN  mg  0
 Since the box is at rest and the upward
direction is positive, the sum of the y forces
=0 so
 FN = mg.
 The normal force on the box, exerted by the
table, is 98.0N upward, and has a
magnitude equal to the box’s weight.
Solution part B
 If your friend pushes down with 40.0N of
force, the normal force will change to keep
the net force = 0 (assuming the box doesn’t
move)
 Now there are 3 forces on the box.
 The weight is still 98.0 N (mg). The net force,
ΣFy=0, so FN-mg-40.0N = 0 and FN = 40.0N +
mg = 98.0N + 40.0 N = 138.0 N so the table
pushes back with more force than before.
Example 2 Accelerating the box
 What happens when a person pulls upward
on the box in the previous problem with a
force equal to or greater than the box’s
weight, say Fp = 100.0 N?
Fy  Fp  mg
 The net force is now
Fy  100.0N  98.0N
Fy  2.0N
 With a net force larger than zero, the box
will accelerate upward.
 a  Fy  2.0N  0.20m /s2
y
m 10.0kg
Combining Force Vectors
 The net force on an object is the vector sum
of all forces acting on the object. (Also
called the net force, Fnet or ΣF).
 Forces add together like vectors according
to the rules we’ve learned previously.
 We can use Pythagorean theorem and Law
of Cosines to add forces at angles.
 We can find force components, Fx and Fy
using trig ratios.
Pulley Example
 Muscleman is trying to lift a piano (slowly) up
to a second story apartment. He is using a
rope looped over two pulleys as shown in Fig
4-24. How much of the piano’s 2000N
weight does he have to pull on the rope?
Pulley Solution
 Looking at the forces on the lower pulley,
we have 2 Tension force sections opposing
the weight of the piano. Since the piano
moves up slowly (a = 0m/s/s), then using
Newton’s second law: 2FT –mg = ma.
 To move the piano requires a tension in the
cord of FT = mg/2. Muscleman exerts a
force equal to half the piano’s weight.
 We say this pulley has given a mechanical
advantage of 2, since we exert 2x less force.
Example 4-13: Vector
components at angles
 Look at this example on page 95 and
follow through the solution.
Your turn to Practice
 Please do Chapter 4 Review pgs
104-105 #s 6, 7, 8, 10, 13, 16, 18,
and 28.