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Slide 16 of 36
Vertical Motion
Use Newton’s 2nd law with vertical forces:
where:
FA = applied force in N or kg•m/s2
Fnet = FA + Fg
Fg = weight (Fg = mg) in N or
kg•m/s2
Fnet = net force (Fnet = ma) in N or
kg•m/s2
MEMORIZE THIS FORMULA!!
Dulku – Physics 20 – Unit 2 (Dynamics) – Topic H
Dulku – Physics 20 – Unit 2 (Dynamics) – Topic H
Newton’s Laws and Vertical Motion
Newton’s Laws and Vertical Motion
Slide 7 of 18
Dulku – Physics 20 – Unit 2 (Dynamics) – Topic H
Specific Outcome:
i. I can apply Newton’s laws of motion to solve, algebraically, linear motion
problems in horizontal, vertical and inclined planes near the surface of
Earth, ignoring air resistance.
ii. I can calculate the resultant force, or its constituents, acting on an object
by adding vector components graphically and algebraically.
iii. I can analyze a graph of empirical data to infer the mathematical
relationships among force, mass and acceleration.
Vertical Motion
Fg = gravitational force on the object
FA = force applied OR force exerted OR
tension force on the object
When there is no acceleration, FA = -Fg
When there is free-fall, FA = 0 N
Vector directions (up = +, down = -) are still
being used!
Dulku – Physics 20 – Unit 2 (Dynamics) – Topic H
1
There are six different situations for an
object in vertical motion:
Slide 5 of 18
Slide 5 of 18
Vertical Motion
Vertical Motion
There are six different situations for an
object in vertical motion:
1) at rest,
FA = -Fg (since a = 0 m/s2, Fnet = 0 N)
4) moving up with positive acceleration
(since a > 0 m/s2, Fnet > 0 N)
2) moving up at constant speed,
FA = -Fg (since a = 0 m/s2, Fnet = 0 N)
5) moving down with negative acceleration
(since a < 0 m/s2, Fnet < 0 N)
3) moving down at constant speed,
FA = -Fg (since a = 0 m/s2, Fnet = 0 N)
6) free fall, accelerating with gravity only
(FA = 0 N, so Fnet = Fg)
Vertical Motion
ex. 1: Find the net force if an upward force of
100 N is applied to a 7.0 kg mass.
Fg = mg = (7.0 kg)(-9.81 m/s2)
FA
•
Fg
= -68.67 N
Dulku – Physics 20 – Unit 2 (Dynamics) – Topic H
Slide 20 of 36
Slide 19 of 36
Dulku – Physics 20 – Unit 2 (Dynamics) – Topic H
Vertical Motion
ex. 2: What is the acceleration of the object in
the previous example?
+31.33 N
Fnet
a=
m
=
7.0 kg
= +4.5 m/s2
= 4.5 m/s2 [up]
Fnet = FA + Fg = (100 N) + (-68.67 N)
= +31.33 N
= 31 N [up]
Dulku – Physics 20 – Unit 2 (Dynamics) – Topic H
Dulku – Physics 20 – Unit 2 (Dynamics) – Topic H
2
Slide 21 of 36
Slide 21 of 36
Vertical Motion
ex. 3: Find the force needed to accelerate a 2.0
kg object upward with an acceleration of
5.0 m/s2.
Fg = mg = (2.0 kg)(-9.81 m/s2) = -19.62 N
Fnet = ma = (2.0 kg)(5.0 m/s2) = +10 N
Fnet = FA + Fg
FA = Fnet - Fg
ex. 4: Will exerts a force of 450 N to jump up at
3.0 m/s2. Determine his mass.
Fnet = ma
FA
Fg
•
m=
FA
a-g
=
+450 N
(+3.0 m/s2) – (-9.81 m/s2)
= 35 kg
Dulku – Physics 20 – Unit 2 (Dynamics) – Topic H
Slide 12 of 18
Dulku – Physics 20 – Unit 2 (Dynamics) – Topic H
Slide 11 of 18
Fg = mg
Fnet = FA + Fg becomes ma = FA + mg
FA = 10 N – (-19.62 N) = 30 N [up]
Vertical Motion
ex. 5: A 62 kg wrestler is standing on a spring
scale in an elevator. What does the scale
read when:
a) the elevator is accelerating up at 2.0 m/s2?
Fg = mg = (62 kg)(-9.81 m/s2) = -608.22 N
Fnet = ma = (62 kg)(2.0 m/s2) = +124 N
FA
Fnet = FA + Fg
Fg
FA = Fnet - Fg
Vertical Motion
•
Vertical Motion
ex. 5: A 62 kg wrestler is standing on a spring
scale in an elevator. What does the scale
read when:
b) the elevator moves down at 2.0 m/s2?
Fnet = ma = (62 kg)(-2.0 m/s2) = -124 N
FA
Fnet = FA + Fg
Fg
FA = Fnet - Fg
•
FA = -124 N – (-608.22 N) = 4.8 x 102 N [up]
FA = 124 N – (-608.22 N) = 7.3 x 102 N [up]
Dulku – Physics 20 – Unit 2 (Dynamics) – Topic H
Dulku – Physics 20 – Unit 2 (Dynamics) – Topic H
3
Slide 14 of 18
Slide 13 of 18
Vertical Motion
ex. 5: A 62 kg wrestler is standing on a spring
scale in an elevator. What does the scale
read when:
c) the elevator moves down at a constant
speed of 2.5 m/s?
Fnet = 0 N
F
A
FA = -Fg = -(-608.22 N)
Fg
Vertical Motion
ex. 5: A 62 kg wrestler is standing on a spring
scale in an elevator. What does the scale
read when:
d) the elevator falls (the cable broke!)?
FA = 0 N
Fg
•
•
= 6.1 x 102 N [up]
Dulku – Physics 20 – Unit 2 (Dynamics) – Topic H
Dulku – Physics 20 – Unit 2 (Dynamics) – Topic H
4
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