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.

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Presentation transcript:

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

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

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

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

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

.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:

-.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

-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

-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

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

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

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

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

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:

-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:

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)

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)

+1640 N W 52.0 kg F F = ma, a = m/s/s (from kinematics) wt = 509.6N downward = (52.0 kg)( m/s/s) F = N A falling 52.0 kg rock climber hits the end of the rope going 13.5 m/s, and is stopped in a distance of 4.20 m. What was the average force exerted to stop them?

+61.6 N W 65.0 kg F F = ma, a = m/s/s (from kinematics) wt = 637 downward = (65.0 kg)( m/s/s) F = N A 65.0 kg dumbwaiter is going up at 5.80 m/s and is brought to rest in a distance of 1.90 m What is the tension in the cable supporting it as it is stopping?

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

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)