1. Net Force Contents: What is the Net force
Using Newton’s Second law with more than one force
Whiteboard Net Force
Applying weight
Whiteboards with weight

2. Net Force In F = ma m = mass a = acceleration
F = <The vector sum of all the forces>
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3. Net Force – Example 1 Finding acceleration
5.0 kg
F = ma
Making to the right +
<+17.0 N – 9.0 N> = (5.0kg)a
8.0 N = (5.0kg)a
a = (8.0 N)/(5.0kg) = 1.6 m/s2
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4. Net Force – Example 2 Finding an unknown force
35.0 kg
Some other force is acting on the block
a = 9.0 m/s/s
F = ma
Making to the right +
<+450. N + F> = (35.0kg)(+9.0 m/s/s)
450. N + F = 315 N
F = 315 N N = -135 N (to the left)
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5. Examples:
Net Force 1
1 | 2 | 3 | 4
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6. F = ma Making to the right + <7.0 N – 3.0 N> = (5.0kg)a
Find the acceleration:
3.0 N
7.0 N
5.0 kg
F = ma
Making to the right +
<7.0 N – 3.0 N> = (5.0kg)a
4.0 N = (5.0kg)a
a = .80 m/s2 W
.80 m/s/s

7. F = ma <5.0 N – 3.0 N – 6.0 N> = (23.0kg)a -4.0 N = (23.0kg)a
Find the acceleration:
5.0 N
3.0 N
23.0 kg
6.0 N
F = ma
<5.0 N – 3.0 N – 6.0 N> = (23.0kg)a
-4.0 N = (23.0kg)a
a = = -.17 m/s2 W
-.17 m/s/s

8. <67.3 N + F> = (452 kg)(.12 m/s2) <67.3 N + F> = 54.24 N
Find the other force:
F = ??
452 kg
67.3 N
F = ma
<67.3 N + F> = (452 kg)(.12 m/s2)
<67.3 N + F> = N
F = N N
F = = -13 N
a = .12 m/s2 W
-13 N

9. <580 N - 125 N + F> = (2100 kg)(-.15 m/s2) 455 N + F = -315 N
Find the other force:
F ???
2100 kg
580 N
125 N
a = .15 m/s/s
F = ma
<580 N N + F> = (2100 kg)(-.15 m/s2)
455 N + F = -315 N
F = -770 N (To the LEFT) W
-770 N

10. Net Force – Example 3 Using Weight
Find the acceleration
(on Earth)
5.0 kg
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11. Net Force – Example 3 Using Weight
5.0 kg
Draw a Free Body Diagram:
-49 N
Don’t Forget the weight:
F = ma = 5.0*9.8 = 49 N
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12. Net Force – Example 3 Using Weight
5.0 kg
F = ma
35 N – 49 N = (5.0 kg)a
-14 N = (5.0 kg)a
a = -2.8 m/s2
-49 N
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13. Examples:
Using Weight
1 | 2 | 3 | 4 | 5
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14. Find the acceleration:
F = ma,
weight = (8.0 kg)(9.80 N/kg)
= 78.4 N down
Making up +
<100. N > = (8.0kg)a
21.6 N = (8.0kg)a
a = 2.7 m/s2
100. N
8.0 kg W
2.7 m/s/s

15. Find the acceleration:
F = ma,
wt = (15.0 kg)(9.8 N/kg) = 147 N down
<120. N N> = (15.0kg)a
-27 N = (15.0kg)a
a = -1.8 m/s2
It accelerates down
120. N
15.0 kg W
-1.8 m/s/s

16. Find the force: F 16 kg a = 1.5 m/s/s (upward) F = ma,
wt = (16 kg)(9.8 N/kg) = N down
<F – N> = (16.0 kg)(+1.5 m/s/s)
F – N = 24 N
F = N = 180 N F
16 kg
a = 1.5 m/s/s
(upward) W
180 N

17. Find the force: F 120. kg a = -4.50 m/s/s (DOWNWARD) F = ma,
wt = 1176 N downward
<F – 1176 N> = (120. kg)(-4.50 m/s/s)
F – 1176 N = -540 N
F = 636 N F
120. kg
a = m/s/s
(DOWNWARD) W
636 N

18. Find the force: F 120. kg a = -4.50 m/s/s (DOWNWARD) F = ma,
wt = 1176 N downward
<F – 1176 N> = (120. kg)(-4.50 m/s2)
F – 1176 N = -540 N
F = 636 N F
120. kg
Relationship between tension, weight and acceleration
Accelerating up = more than weight (demo, elevators)
Accelerating down = less than weight (demo, elevators, acceleration vs velocity)
Climbing ropes
a = m/s/s
(DOWNWARD) W

19. Find the mass: 13.6 N m a = 1.12 m/s/s (downward) F = ma,
wt = m(9.80 m/s2) downward
<13.6 – m(9.80 m/s2)> = m(-1.12 m/s2)
13.6 N = m(9.80 m/s2) - m(1.12 m/s2)
13.6 N = m(9.80 m/s m/s2)
13.6 N = m(8.68 m/s2)
m = kg = 1.57 kg
13.6 N m
a = 1.12 m/s/s
(downward) W
1.57 kg