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Published byDeshawn Schooling Modified about 1 year ago

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

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Net Force In F = ma m = mass a = acceleration F = TOC

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Net Force – Example 1 Finding acceleration F = ma Making to the right + = (5.0kg)a 8.0 N = (5.0kg)a a = (8.0 N)/(5.0kg) = 1.6 m/s/s TOC 5.0 kg 17.0 N 9.0 N

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Net Force – Example 2 Finding an unknown force F = ma Making to the right + = (35.0kg)(+9.0 m/s/s) 450. N + F = 315 N F = 315 N N = -135 N (to the left) TOC 35.0 kg 450. N F = ?? a = 9.0 m/s/s Some other force is acting on the block

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Whiteboards: Net Force 1 11 | 2 | 3 | 4234 TOC

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.80 m/s/s W 5.0 kg 7.0 N 3.0 N F = ma Making to the right + = (5.0kg)a 4.0 N = (5.0kg)a a =.80 m/s/s Find the acceleration:

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

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-13 N W 452 kg 67.3 N F = ?? F = ma = (452 kg)(.12 m/s/s) = N F = N N F = = -13 N Find the other force: a =.12 m/s/s

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

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Net Force – Example 3 Using Weight TOC 5.0 kg 35 N Find the acceleration (on Earth)

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Net Force – Example 3 Using Weight Draw a Free Body Diagram: TOC 5.0 kg 35 N Don’t Forget the weight: F = ma = 5.0*9.8 = 49 N -49 N

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Net Force – Example 3 Using Weight F = ma 35 N – 49 N = (5.0 kg)a -14 N = (5.0 kg)a a = -2.8 m/s/s TOC 5.0 kg 35 N -49 N

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Whiteboards: Using Weight 11 | 2 | 3 | 4 | TOC

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2.7 m/s/s W 8.0 kg 100. N F = ma, weight = (8.0 kg)(9.80 N/kg) = 78.4 N down Making up + = (8.0kg)a 21.6 N = (8.0kg)a a = 2.7 m/s/s Find the acceleration:

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

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

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

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16,900 N W 120. kg F First, suvat: s = m, u = m/s, v = 0, a = ? use v 2 = u 2 + 2as, a = m/s/s F = ma, wt = 1176 N downward = (120. kg)( m/s/s) F – 1176 N = N F = = 16,900 N This box is going downwards at 22.0 m/s and is stopped in a distance of 1.85 m. What must be the upwards force acting on it to stop it?

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

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1.57 kg W m 13.6 N F = ma, wt = m(9.80 m/s/s) downward = 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 = kg = 1.57 kg Find the mass: a = 1.12 m/s/s (downward)

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