Download Section 5.3 Force and Motion in 2 Dimensions

Mr. Borosky
Physics Section 5.3 Notes
Page 1 of 3
Section 5.3 Force and Motion in 2 Dimensions
Objectives
Determine the force that produces equilibrium when three forces
act on an object.
Analyze the motion of an object on an inclined plane with and
without friction.
Read intro paragraph p. 131
When Friction acts between 2 surfaces you must take into account
both the frictional force that is parallel to the surface and Normal
(Perpendicular) to the surface.
Now we will use our skills in adding vectors to analyze situations
in which the forces acting on the object are at angles other than
90°.
EQUILIBRIUM REVISITED
Read Section.
Recall that when the net force on an object is zero, the object is
in equilibrium.
According to Newton’s laws, the object will not accelerate because
there is no net force acting on it; an object in equilibrium is
motionless or moves with constant velocity.
It is important to realize that equilibrium can occur no matter how
many forces act on an object. As long as the resultant is zero, the
net force is zero and the object is in equilibrium.
Go over Figure 5-11 p. 131
Resultant Force – vector sum of 2 or more vectors.
Equilibrium – condition in which net force on an object is zero.
When the net force is zero the object is in equilibrium.
Equilibrant Force – force needed to bring an object into
equilibrium. Force that is applied to produce equilibrium. We will
use this for the lab. It is the single additional force that if
applied to the same point as the other forces, will produce
equilibrium.
Physics Principals and Problems © 2005 Started 2006-2007 School Year
Mr. Borosky
Physics Section 5.3 Notes
Page 2 of 3
To find the equilibrant find the Resultant Force. The equilibrant
force is equal in magnitude to the resultant but opposite in
direction. So add 180°.
Go over Figure 5.12 p. 131
MOTION ALONG AN INCLINED PLANE
Read Section.
Give equations from the old book.
*****
*****
F = Fg cos 
F|| = Fg sin 
When the incline becomes steeper F|| becomes greater and F
becomes smaller.
The inclined plane exerts an upward force perpendicular to its
surface, this is the NORMAL FORCE.
Since the trunk has no acceleration perpendicular to the plane
all forces in that direction must balance. Therefore we get
the following equations.
FN + F = 0
FN – W cos  = 0
FN = W cos 
If there is no friction between the trunk and the plane, the
only force on the trunk is the parallel component of its weight
and
F|| = W sin . So then according to Newton’s 2nd Law the
acceleration would be
a = F / m = W sin  / m
But W / m = g so the acceleration can be found using
*****
a = g sin 
And if there is friction then we get
*****
a = g(sin  - μ cos )
Do
F
F
F
Example Problem 5 p. 133
= Fg cos 
= 562 cos 30°
= 486.71 N
F|| = Fg sin 
F|| = 562 sin 30°
F|| = 281 N
Physics Principals and Problems © 2005 Started 2006-2007 School Year
Mr. Borosky
Physics Section 5.3 Notes
Page 3 of 3
Do Practice Problems p. 133 # 33-37
Do
Fg
Fg
Fg
Example Problem 6 p. 134
= mg
F = Fg cos 
= 62(9.8)
F = 607.6 cos 37°
= 607.6 N
F = 485.25 N
F|| = Fg sin 
F|| = 607.6 sin 37°
F|| = 365.66 N
Then
FF = μFN
FF = .15(485.25)
FF = 72.7875 N
FNet = F|| - FF
ma = 365.66 – 72.7875
62a = 292.8725
a = 4.72 m/s2
vF = vI + at
vF = 0 4.72(5)
vF = 23.6 m/s
OR
a
a
a
a
=
=
=
=
g(sin  - μ cos )
9.8(sin 37° - .15 (cos 37°))
9.8(.482)
4.72 m/s2
vF = vI + at
vF = 0 4.72(5)
vF = 23.6 m/s
Do Practice Problems p. 135 # 38-41
Do 5.3 Section Review p. 135 # 43-44 (Skip 42, 45, 46)
Physics Principals and Problems © 2005 Started 2006-2007 School Year