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Net Force Contents: What is the Net force Using Newton’s Second law with more than one forcePowerPoint Presentation

Net Force Contents: What is the Net force Using Newton’s Second law with more than one force

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- 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

Net Force – Example 1 Finding acceleration

17.0 N

9.0 N

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/s/s

TOC

Net Force – Example 2 Finding an unknown force

F = ??

450. N

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 - 450. N = -135 N (to the left)

TOC

7.0 N

3.0 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/s/s

W

.80 m/s/s

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 = -.1739 = -.17 m/s/s

W

-.17 m/s/s

F = ??

452 kg

67.3 N

a = .12 m/s/s

F = ma

<67.3 N + F> = (452 kg)(.12 m/s/s)

<67.3 N + F> = 54.24 N

F = 54.24 N - 67.3 N

F = -13.06 = -13 N

W

-13 N

F ???

2100 kg

580 N

125 N

a = .15 m/s/s

F = ma

<580 N - 125 N + F> = (2100 kg)(-.15 m/s/s)

455 N + F = -315 N

F = -770 N (To the LEFT)

W

-770 N

-49 N

Net Force – Example 3 Using Weight

5.0 kg

Draw a Free Body Diagram:

Don’t Forget the weight:

F = ma = 5.0*9.8 = 49 N

TOC

-49 N

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/s/s

TOC

F = ma,

weight = (8.0 kg)(9.80 N/kg)

= 78.4 N down

Making up +

<100. N - 78.4> = (8.0kg)a

21.6 N = (8.0kg)a

a = 2.7 m/s/s

100. N

8.0 kg

W

2.7 m/s/s

F = ma,

wt = (15.0 kg)(9.8 N/kg) = 147 N down

<120. N - 147 N> = (15.0kg)a

-27 N = (15.0kg)a

a = -1.8 m/s/s

It accelerates down

120. N

15.0 kg

W

-1.8 m/s/s

F = ma,

wt = (16 kg)(9.8 N/kg) = 156.8 N down

<F – 156.8 N> = (16.0 kg)(+1.5 m/s/s)

F – 156.8 N = 24 N

F = 180.8 N = 180 N

F

16 kg

a = 1.5 m/s/s

(upward)

W

180 N

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 = -4.50 m/s/s

(DOWNWARD)

W

636 N

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

- 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 = -4.50 m/s/s

(DOWNWARD)

W

F = ma,

wt = m(9.80 m/s/s) downward

<13.6 – m(9.80 m/s/s)> = m(-1.12 m/s/s)

13.6 N = m(9.80 m/s/s) - m(1.12 m/s/s)

13.6 N = m(9.80 m/s/s-1.12 m/s/s)

13.6 N = m(8.68 m/s/s)

m = 1.5668 kg = 1.57 kg

13.6 N

m

a = 1.12 m/s/s

(downward)

W

1.57 kg

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