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Force And Motion I
Dr. Venkat Kaushik Phys 211, Lecture 8
Sep 15, 2015
First Law
• Newton’s First Law § A body at rest remains at rest -- § A body in uniform linear motion (constant velocity) continues
to move with the same velocity (ie, same magnitude and direction on that straight line) --
§ UNLESS ACTED UPON BY A FORCE • Newton’s Laws do NOT hold good for all frames
§ need a reference frame does NOT accelerate (also called the inertial reference frame)
• First Law defines Inertia of an object • If the body is at rest OR in uniform linear motion
§ then the net (vector sum) of all the forces acting on the body is ZERO
Lecture 8
Second Law
• Newton’s Second Law § If a net (non-zero) force acts on a body, it accelerates in the direction
of the net force. Acceleration of the object is proportional to the net force
§ If the object has a mass m and acceleration a, the net force is given by
• If the acceleration is zero, then § The net force is zero § If there are say 5 forces, each may (or not) be zero, but their vector
sum is zero § The body could be at rest OR the body could be in a uniform linear
motion § F1, F2, F3 … FN are a SYSTEM of forces on that body
Lecture 8
T1 =
v0 sin ✓
g
H =
v20 sin2 ✓
2g
T =
2v0 sin ✓
g
R =
v20 sin 2✓
g
quantity 1D 2D 3D
position
#»r rx
ˆi rx
ˆi+ ry
ˆj rx
ˆi+ ry
ˆj + rz
ˆk
avg. velocity
�
#»r
�t
�rx
�tˆi
�rx
�tˆi+
�ry
�tˆj
�rx
�tˆi+
�ry
�tˆj +
�rz
�tˆk
inst. velocity
d #»r
dt
drx
dtˆi
drx
dtˆi+
dry
dtˆj
drx
dtˆi+
dry
dtˆj +
drz
dtˆk
avg. acceleration
�
#»v
�t
�vx
�tˆi
�vx
�tˆi+
�vy
�tˆj
�vx
�tˆi+
�vy
�tˆj +
�vz
�tˆk
inst. acceleration
d #»v
dt
dvx
dtˆi
dvx
dtˆi+
dvy
dtˆj
dvx
dtˆi+
dvy
dtˆj +
dvz
dtˆk
inst. acceleration
d2 #»r
dt2d2r
x
dt2ˆi
d2rx
dt2ˆi+
d2ry
dt2ˆj
d2rx
dt2ˆi+
d2ry
dt2ˆj +
drz
dt2ˆk
#»rPA
=
#»rPB
+
#»rBA
#»vPA
=
#»vPB
+
#»vBA
#»aPA
=
#»aPB
iff #»aBA
= 0
v = 2⇡rN/T
#»F 1 +
#»F 2 + · · ·+ #»
FN
=
i=NX
i=1
#»F
i
=
#»F
net
= m #»a
3
#»r = x
ˆ
i+ y
ˆ
j ,
#»r 0 = 0
#»v 0 = v0 cos ✓
ˆ
i+ v0 sin ✓ˆ
j
#»a = �g
ˆ
j
#»r =
#»r 0 +
#»v 0t�
1
2
#»a t
2
) x
ˆ
i+ y
ˆ
j = v0t cos ✓ˆ
i+ v0t sin ✓ˆ
j � 1
2
gt
2ˆ
j
x = v0t cos ✓
y = v0t sin ✓ �1
2
gt
2
quantity 1D 2D 3D
position
#»r
r
x
ˆ
i r
x
ˆ
i+ r
y
ˆ
j
r
x
ˆ
i+ r
y
ˆ
j + r
z
ˆ
k
avg. velocity
�
#»r
�t
�r
x
�t
ˆ
i
�r
x
�t
ˆ
i+
�r
y
�t
ˆ
j
�r
x
�t
ˆ
i+
�r
y
�t
ˆ
j +
�r
z
�t
ˆ
k
inst. velocity
d
#»r
dt
dr
x
dt
ˆ
i
dr
x
dt
ˆ
i+
dr
y
dt
ˆ
j
dr
x
dt
ˆ
i+
dr
y
dt
ˆ
j +
dr
z
dt
ˆ
k
avg. acceleration
�
#»v
�t
�v
x
�t
ˆ
i
�v
x
�t
ˆ
i+
�v
y
�t
ˆ
j
�v
x
�t
ˆ
i+
�v
y
�t
ˆ
j +
�v
z
�t
ˆ
k
inst. acceleration
d
#»v
dt
dv
x
dt
ˆ
i
dv
x
dt
ˆ
i+
dv
y
dt
ˆ
j
dv
x
dt
ˆ
i+
dv
y
dt
ˆ
j +
dv
z
dt
ˆ
k
inst. acceleration
d
2 #»r
dt
2
d
2r
x
dt
2ˆ
i
d
2r
x
dt
2ˆ
i+
d
2r
y
dt
2ˆ
j
d
2r
x
dt
2ˆ
i+
d
2r
y
dt
2ˆ
j +
dr
z
dt
2ˆ
k
1
#»r = x
ˆ
i+ y
ˆ
j ,
#»r 0 = 0
#»v 0 = v0 cos ✓
ˆ
i+ v0 sin ✓ˆ
j
#»a = �g
ˆ
j
#»r =
#»r 0 +
#»v 0t�
1
2
#»a t
2
) x
ˆ
i+ y
ˆ
j = v0t cos ✓ˆ
i+ v0t sin ✓ˆ
j � 1
2
gt
2ˆ
j
x = v0t cos ✓
y = v0t sin ✓ �1
2
gt
2
quantity 1D 2D 3D
position
#»r
r
x
ˆ
i r
x
ˆ
i+ r
y
ˆ
j
r
x
ˆ
i+ r
y
ˆ
j + r
z
ˆ
k
avg. velocity
�
#»r
�t
�r
x
�t
ˆ
i
�r
x
�t
ˆ
i+
�r
y
�t
ˆ
j
�r
x
�t
ˆ
i+
�r
y
�t
ˆ
j +
�r
z
�t
ˆ
k
inst. velocity
d
#»r
dt
dr
x
dt
ˆ
i
dr
x
dt
ˆ
i+
dr
y
dt
ˆ
j
dr
x
dt
ˆ
i+
dr
y
dt
ˆ
j +
dr
z
dt
ˆ
k
avg. acceleration
�
#»v
�t
�v
x
�t
ˆ
i
�v
x
�t
ˆ
i+
�v
y
�t
ˆ
j
�v
x
�t
ˆ
i+
�v
y
�t
ˆ
j +
�v
z
�t
ˆ
k
inst. acceleration
d
#»v
dt
dv
x
dt
ˆ
i
dv
x
dt
ˆ
i+
dv
y
dt
ˆ
j
dv
x
dt
ˆ
i+
dv
y
dt
ˆ
j +
dv
z
dt
ˆ
k
inst. acceleration
d
2 #»r
dt
2
d
2r
x
dt
2ˆ
i
d
2r
x
dt
2ˆ
i+
d
2r
y
dt
2ˆ
j
d
2r
x
dt
2ˆ
i+
d
2r
y
dt
2ˆ
j +
dr
z
dt
2ˆ
k
1
T1 =
v0 sin ✓
g
H =
v20 sin2 ✓
2g
T =
2v0 sin ✓
g
R =
v20 sin 2✓
g
quantity 1D 2D 3D
position
#»r rx
ˆi rx
ˆi+ ry
ˆj rx
ˆi+ ry
ˆj + rz
ˆk
avg. velocity
�
#»r
�t
�rx
�tˆi
�rx
�tˆi+
�ry
�tˆj
�rx
�tˆi+
�ry
�tˆj +
�rz
�tˆk
inst. velocity
d #»r
dt
drx
dtˆi
drx
dtˆi+
dry
dtˆj
drx
dtˆi+
dry
dtˆj +
drz
dtˆk
avg. acceleration
�
#»v
�t
�vx
�tˆi
�vx
�tˆi+
�vy
�tˆj
�vx
�tˆi+
�vy
�tˆj +
�vz
�tˆk
inst. acceleration
d #»v
dt
dvx
dtˆi
dvx
dtˆi+
dvy
dtˆj
dvx
dtˆi+
dvy
dtˆj +
dvz
dtˆk
inst. acceleration
d2 #»r
dt2d2r
x
dt2ˆi
d2rx
dt2ˆi+
d2ry
dt2ˆj
d2rx
dt2ˆi+
d2ry
dt2ˆj +
drz
dt2ˆk
#»rPA
=
#»rPB
+
#»rBA
#»vPA
=
#»vPB
+
#»vBA
#»aPA
=
#»aPB
iff #»aBA
= 0
v = 2⇡rN/T
#»F 1 +
#»F 2 + · · ·+ #»
FN
=
i=NX
i=1
#»F
i
=
#»F
net
= m #»a
#»F 12 +
#»F 21 = 0
(1)
3
T1 =
v0 sin ✓
g
H =
v20 sin2 ✓
2g
T =
2v0 sin ✓
g
R =
v20 sin 2✓
g
quantity 1D 2D 3D
position
#»r rx
ˆi rx
ˆi+ ry
ˆj rx
ˆi+ ry
ˆj + rz
ˆk
avg. velocity
�
#»r
�t
�rx
�tˆi
�rx
�tˆi+
�ry
�tˆj
�rx
�tˆi+
�ry
�tˆj +
�rz
�tˆk
inst. velocity
d #»r
dt
drx
dtˆi
drx
dtˆi+
dry
dtˆj
drx
dtˆi+
dry
dtˆj +
drz
dtˆk
avg. acceleration
�
#»v
�t
�vx
�tˆi
�vx
�tˆi+
�vy
�tˆj
�vx
�tˆi+
�vy
�tˆj +
�vz
�tˆk
inst. acceleration
d #»v
dt
dvx
dtˆi
dvx
dtˆi+
dvy
dtˆj
dvx
dtˆi+
dvy
dtˆj +
dvz
dtˆk
inst. acceleration
d2 #»r
dt2d2r
x
dt2ˆi
d2rx
dt2ˆi+
d2ry
dt2ˆj
d2rx
dt2ˆi+
d2ry
dt2ˆj +
drz
dt2ˆk
#»rPA
=
#»rPB
+
#»rBA
#»vPA
=
#»vPB
+
#»vBA
#»aPA
=
#»aPB
iff #»aBA
= 0
v = 2⇡rN/T
#»F 1 +
#»F 2 + · · ·+ #»
FN
=
i=NX
i=1
#»F
i
=
#»F
net
= m #»a
#»F 12 +
#»F 21 = 0
(1)
3
T1 =
v0 sin ✓
g
H =
v20 sin2 ✓
2g
T =
2v0 sin ✓
g
R =
v20 sin 2✓
g
quantity 1D 2D 3D
position
#»r rx
ˆi rx
ˆi+ ry
ˆj rx
ˆi+ ry
ˆj + rz
ˆk
avg. velocity
�
#»r
�t
�rx
�tˆi
�rx
�tˆi+
�ry
�tˆj
�rx
�tˆi+
�ry
�tˆj +
�rz
�tˆk
inst. velocity
d #»r
dt
drx
dtˆi
drx
dtˆi+
dry
dtˆj
drx
dtˆi+
dry
dtˆj +
drz
dtˆk
avg. acceleration
�
#»v
�t
�vx
�tˆi
�vx
�tˆi+
�vy
�tˆj
�vx
�tˆi+
�vy
�tˆj +
�vz
�tˆk
inst. acceleration
d #»v
dt
dvx
dtˆi
dvx
dtˆi+
dvy
dtˆj
dvx
dtˆi+
dvy
dtˆj +
dvz
dtˆk
inst. acceleration
d2 #»r
dt2d2r
x
dt2ˆi
d2rx
dt2ˆi+
d2ry
dt2ˆj
d2rx
dt2ˆi+
d2ry
dt2ˆj +
drz
dt2ˆk
#»rPA
=
#»rPB
+
#»rBA
#»vPA
=
#»vPB
+
#»vBA
#»aPA
=
#»aPB
iff #»aBA
= 0
v = 2⇡rN/T
#»F 1 +
#»F 2 + · · ·+ #»
FN
=
i=NX
i=1
#»F
i
=
#»F
net
= m #»a
#»F 12 +
#»F 21 = 0
(1)
3
Third Law • Defines “contact” force or the force of “interaction”
§ The forces of interaction between two bodies are equal in magnitude and opposite in direction.
§ Force of Action + Force of Reaction = 0 § The two forces form an “action/reaction pair” § 1 = Bottom Surface of the box § 2 = Top surface of a table (for example) § F12 = Force of 1 on 2 (Weight) § F21 = Force of 2 on 1 (Normal Force)
Lecture 8
Some Forces • Force
§ due to Earth’s gravitational pull – weight § due to surfaces in contact – normal force § due to relative motion between surfaces of contact – friction § due to rope/string holding (in suspension) a heavy object –
tension • Setting up and solving problems
§ Identify all objects in the system § Setup a reference frame § Identify forces acting on each object using a Free Body Diagram
(FBD) • remove/isolate the object from the system • indicate forces acting ON the object due to all other objects
§ Apply Newton’s Laws (usually 2nd and 3rd laws) § Solve for the unknown (forces, acceleration, mass etc)
Lecture 8
Example 1
• Assume § Friction between contact surfaces can be ignored, rope is
massless and does not slip on the pulley, pulley’s mass is negligible compared to 1 and 2 and 2 is falling (moving downward) with an acceleration
Lecture 8
#»rPA
=
#»rPB
+
#»rBA
#»vPA
=
#»vPB
+
#»vBA
#»aPA
=
#»aPB
iff #»aBA
= 0
v = 2⇡rN/T
#»F 1 +
#»F 2 + · · ·+ #»
FN
=
i=NX
i=1
#»F
i
=
#»F
net
= m #»a
#»F 12 +
#»F 21 = 0
T ˆi+ (N1 �m1g) ˆj = m1aˆi
(T �m2g) ˆj = �m2a ˆj
T = m1a , (N1 �m1g) = 0 , (T �m2g) = �m2a
) a =
m2g
(m1 +m2)T =
m1m2g
(m1 +m2)(1)
4
#»r = x
ˆ
i+ y
ˆ
j ,
#»r 0 = 0
#»v 0 = v0 cos ✓
ˆ
i+ v0 sin ✓ˆ
j
#»a = �g
ˆ
j
#»r =
#»r 0 +
#»v 0t�
1
2
#»a t
2
) x
ˆ
i+ y
ˆ
j = v0t cos ✓ˆ
i+ v0t sin ✓ˆ
j � 1
2
gt
2ˆ
j
x = v0t cos ✓
y = v0t sin ✓ �1
2
gt
2
quantity 1D 2D 3D
position
#»r
r
x
ˆ
i r
x
ˆ
i+ r
y
ˆ
j
r
x
ˆ
i+ r
y
ˆ
j + r
z
ˆ
k
avg. velocity
�
#»r
�t
�r
x
�t
ˆ
i
�r
x
�t
ˆ
i+
�r
y
�t
ˆ
j
�r
x
�t
ˆ
i+
�r
y
�t
ˆ
j +
�r
z
�t
ˆ
k
inst. velocity
d
#»r
dt
dr
x
dt
ˆ
i
dr
x
dt
ˆ
i+
dr
y
dt
ˆ
j
dr
x
dt
ˆ
i+
dr
y
dt
ˆ
j +
dr
z
dt
ˆ
k
avg. acceleration
�
#»v
�t
�v
x
�t
ˆ
i
�v
x
�t
ˆ
i+
�v
y
�t
ˆ
j
�v
x
�t
ˆ
i+
�v
y
�t
ˆ
j +
�v
z
�t
ˆ
k
inst. acceleration
d
#»v
dt
dv
x
dt
ˆ
i
dv
x
dt
ˆ
i+
dv
y
dt
ˆ
j
dv
x
dt
ˆ
i+
dv
y
dt
ˆ
j +
dv
z
dt
ˆ
k
inst. acceleration
d
2 #»r
dt
2
d
2r
x
dt
2ˆ
i
d
2r
x
dt
2ˆ
i+
d
2r
y
dt
2ˆ
j
d
2r
x
dt
2ˆ
i+
d
2r
y
dt
2ˆ
j +
dr
z
dt
2ˆ
k
1
#»r = x
ˆ
i+ y
ˆ
j ,
#»r 0 = 0
#»v 0 = v0 cos ✓
ˆ
i+ v0 sin ✓ˆ
j
#»a = �g
ˆ
j
#»r =
#»r 0 +
#»v 0t�
1
2
#»a t
2
) x
ˆ
i+ y
ˆ
j = v0t cos ✓ˆ
i+ v0t sin ✓ˆ
j � 1
2
gt
2ˆ
j
x = v0t cos ✓
y = v0t sin ✓ �1
2
gt
2
quantity 1D 2D 3D
position
#»r
r
x
ˆ
i r
x
ˆ
i+ r
y
ˆ
j
r
x
ˆ
i+ r
y
ˆ
j + r
z
ˆ
k
avg. velocity
�
#»r
�t
�r
x
�t
ˆ
i
�r
x
�t
ˆ
i+
�r
y
�t
ˆ
j
�r
x
�t
ˆ
i+
�r
y
�t
ˆ
j +
�r
z
�t
ˆ
k
inst. velocity
d
#»r
dt
dr
x
dt
ˆ
i
dr
x
dt
ˆ
i+
dr
y
dt
ˆ
j
dr
x
dt
ˆ
i+
dr
y
dt
ˆ
j +
dr
z
dt
ˆ
k
avg. acceleration
�
#»v
�t
�v
x
�t
ˆ
i
�v
x
�t
ˆ
i+
�v
y
�t
ˆ
j
�v
x
�t
ˆ
i+
�v
y
�t
ˆ
j +
�v
z
�t
ˆ
k
inst. acceleration
d
#»v
dt
dv
x
dt
ˆ
i
dv
x
dt
ˆ
i+
dv
y
dt
ˆ
j
dv
x
dt
ˆ
i+
dv
y
dt
ˆ
j +
dv
z
dt
ˆ
k
inst. acceleration
d
2 #»r
dt
2
d
2r
x
dt
2ˆ
i
d
2r
x
dt
2ˆ
i+
d
2r
y
dt
2ˆ
j
d
2r
x
dt
2ˆ
i+
d
2r
y
dt
2ˆ
j +
dr
z
dt
2ˆ
k
1
1
2 1
2
#»rPA
=
#»rPB
+
#»rBA
#»vPA
=
#»vPB
+
#»vBA
#»aPA
=
#»aPB
iff #»aBA
= 0
v = 2⇡rN/T
#»F 1 +
#»F 2 + · · ·+ #»
FN
=
i=NX
i=1
#»F
i
=
#»F
net
= m #»a
#»F 12 +
#»F 21 = 0
T ˆi+ (N1 �m1g) ˆj = m1aˆi
(T �m2g) ˆj = �m2a ˆj
T = m1a , (N1 �m1g) = 0 , (T �m2g) = �m2a
) a =
m2g
(m1 +m2)T =
m1m2g
(m1 +m2)(1)
4
#»rPA
=
#»rPB
+
#»rBA
#»vPA
=
#»vPB
+
#»vBA
#»aPA
=
#»aPB
iff #»aBA
= 0
v = 2⇡rN/T
#»F 1 +
#»F 2 + · · ·+ #»
FN
=
i=NX
i=1
#»F
i
=
#»F
net
= m #»a
#»F 12 +
#»F 21 = 0
T ˆi+ (N1 �m1g) ˆj = m1aˆi
(T �m2g) ˆj = �m2a ˆj
T = m1a , (N1 �m1g) = 0 , (T �m2g) = �m2a
) a =
m2g
(m1 +m2)T =
m1m2g
(m1 +m2)(1)
4
#»rPA
=
#»rPB
+
#»rBA
#»vPA
=
#»vPB
+
#»vBA
#»aPA
=
#»aPB
iff #»aBA
= 0
v = 2⇡rN/T
#»F 1 +
#»F 2 + · · ·+ #»
FN
=
i=NX
i=1
#»F
i
=
#»F
net
= m #»a
#»F 12 +
#»F 21 = 0
T ˆi+ (N1 �m1g) ˆj = m1aˆi
(T �m2g) ˆj = �m2a ˆj
T = m1a , (N1 �m1g) = 0 , (T �m2g) = �m2a
) a =
m2g
(m1 +m2)T =
m1m2g
(m1 +m2)(1)
4
#»rPA
=
#»rPB
+
#»rBA
#»vPA
=
#»vPB
+
#»vBA
#»aPA
=
#»aPB
iff #»aBA
= 0
v = 2⇡rN/T
#»F 1 +
#»F 2 + · · ·+ #»
FN
=
i=NX
i=1
#»F
i
=
#»F
net
= m #»a
#»F 12 +
#»F 21 = 0
T ˆi+ (N1 �m1g) ˆj = m1aˆi
(T �m2g) ˆj = �m2a ˆj
T = m1a , (N1 �m1g) = 0 , (T �m2g) = �m2a
) a =
m2g
(m1 +m2)T =
m1m2g
(m1 +m2)(1)
4
#»rPA
=
#»rPB
+
#»rBA
#»vPA
=
#»vPB
+
#»vBA
#»aPA
=
#»aPB
iff #»aBA
= 0
v = 2⇡rN/T
#»F 1 +
#»F 2 + · · ·+ #»
FN
=
i=NX
i=1
#»F
i
=
#»F
net
= m #»a
#»F 12 +
#»F 21 = 0
T ˆi+ (N1 �m1g) ˆj = m1aˆi
(T �m2g) ˆj = �m2a ˆj
T = m1a , (N1 �m1g) = 0 , (T �m2g) = �m2a
) a =
m2g
(m1 +m2)T =
m1m2g
(m1 +m2)(1)
4
#»rPA
=
#»rPB
+
#»rBA
#»vPA
=
#»vPB
+
#»vBA
#»aPA
=
#»aPB
iff #»aBA
= 0
v = 2⇡rN/T
#»F 1 +
#»F 2 + · · ·+ #»
FN
=
i=NX
i=1
#»F
i
=
#»F
net
= m #»a
#»F 12 +
#»F 21 = 0
T ˆi+ (N1 �m1g) ˆj = m1aˆi
(T �m2g) ˆj = �m2a ˆj
T = m1a , (N1 �m1g) = 0 , (T �m2g) = �m2a
) a =
m2g
(m1 +m2)T =
m1m2g
(m1 +m2)(1)
4
#»rPA
=
#»rPB
+
#»rBA
#»vPA
=
#»vPB
+
#»vBA
#»aPA
=
#»aPB
iff #»aBA
= 0
v = 2⇡rN/T
#»F 1 +
#»F 2 + · · ·+ #»
FN
=
i=NX
i=1
#»F
i
=
#»F
net
= m #»a
#»F 12 +
#»F 21 = 0
T ˆi+ (N1 �m1g) ˆj = m1aˆi
(T �m2g) ˆj = �m2a ˆj
T = m1a , (N1 �m1g) = 0 , (T �m2g) = �m2a
) a =
m2g
(m1 +m2)T =
m1m2g
(m1 +m2)(1)
4
