Lesson 1: Newton’s First Law Unit 4: Newton’s Laws Lesson 1: Newton’s First Law

Newton’s First Law (The Law of Inertia) An object remains at rest or in uniform motion (constant speed in a straight line) unless acted upon by a net unbalanced force. Example 1: A car approaches a curve covered in black ice, which reduces friction to zero. What will happen next? FREE BODY DIAGRAM: 10 min normal force The car will move in a straight line since there are no unbalanced forces on it. force of gravity

Example 2 (Try it!): A box is pulled along a rough floor by a horizontal force of 15 N at a constant speed. Draw a free body diagram and find the friction force. FREE BODY DIAGRAM: Normal force P = 15 N f (Force of friction) 10 min Force of gravity f = 15 N (must be balanced)

Inertia Inertia is the tendency of objects to resist changes in their states of motion (i.e. their tendency to follow Newton’s 1st Law). The greater the mass of an object, the more inertia it has.

Brain Break!

Think-Pair-Share! Does 2 kg of apples have twice the inertia or half the inertia of 1 kg of apples? If the pen on your desk is at rest, can you say that no forces are acting on it? If the forces acting on a pen are balanced, is it correct to say that the pen is at rest? 10 min

Think-Pair-Share! If you place a ball in the centre of a wagon and then quickly push the wagon forward, in what direction does the ball appear to go? You are travelling in a school bus. The driver has to apply the brakes suddenly to prevent an accident. Describe how your body would move in response to this sudden braking. 10 min

Think-Pair-Share! While travelling in Africa, you are chased by a very large elephant. Would it make more sense to run in a straight line to get away, or in a zigzag motion? A hockey puck moving across an ice rink eventually comes to a stop. Does this prove that Newton’s 1st Law does not apply to all situations? 10 min

Lesson 2: Newton’s Second Law Unit 4: Newton’s Laws Lesson 2: Newton’s Second Law

Newton’s Second Law If a net force “F” acts on an object of mass “m”, the object accelerates in the direction of the net force such that F=ma Example 1: A sled of mass 50 kg is pushed across a frictionless ice surface with a force of 220 N. If it starts from rest, find the acceleration and the velocity after 10 seconds. N Unbalanced! Fnet = 220 N 220 N 10 min mg

Example 2: A 561 kg rocket has an upward thrust force of 16500 N Example 2: A 561 kg rocket has an upward thrust force of 16500 N. It starts from rest on the ground. Find its height after 10 seconds. T 5 min mg

Example 3 (Try it!): A 50 kg sled is pulled with a force of 250 N and accelerates at 1.5 m/s2. Find the force of friction. N f 250 N mg 10 min

Brain Break!

Example 4 (Try it!): A 530 kg snowmobile is pushed forwards by a force of 2500 N. There is a friction force of 750 N and air resistance of 300 N. Find the acceleration. N f P = 2500 N A 10 min mg

Practice Time! Do #1-3 on the Newton’s Second Law Problem Set. 20 min

Lesson 3: Practice with Newton’s Second Law Unit 4: Newton’s Laws Lesson 3: Practice with Newton’s Second Law

there is a 50 N friction force  =0.3 Practice #1: Find the acceleration of a 12 kg sled pulled with a force of 120 N if there is no friction there is a 50 N friction force  =0.3 N f 120 N mg 20 min

Brain Break!

Find: a) the thrust required. Practice #2: A 7000 kg jet accelerates forwards at 2.0 m/s2 against a 25000N drag force. Find: a) the thrust required. b) the acceleration if the thrust force is 75000 N. L = lift T = thrust D = drag mg 15 min

Practice #3: A helicopter moving upwards at 10 m/s slows down and stops in 2.5 seconds. Find the tension in the rope supporting the 20 kg package during this time. 20kg T = tension 20kg mg 10 min

Homework Do #6 and 7 on the Newton’s Second Law Problem Set. 10 min

Lesson 4: More Practice with Newton’s Second Law Unit 4: Newton’s Laws Lesson 4: More Practice with Newton’s Second Law

b) the reading if the acceleration is 2.0 m/s2 upwards. Practice #4: Being a curious individual, Jacob Z. decides to conduct an experiment by riding an elevator while standing on a scale. Jacob has a mass of 75 kg and his scale is calibrated in Newtons. Find: a) the scale reading if the elevator is moving upwards at a constant speed of 1.0 m/s. b) the reading if the acceleration is 2.0 m/s2 upwards. c) the reading if the elevator is moving downwards and speeding up at 1.0 m/s2. d) the acceleration if the reading is 650 N. e) the acceleration if the reading is 800 N. N (scale reading) 30 min mg = (75)(9.8) = 735 N

Either going up and slowing down, or going down and speeding up 5 min

Brain Break!

Practice #5: A sled of mass 15 kg accelerates from rest to 15 m/s over 30 m. The pulling force is 75 N; find the coefficient of friction. N P = 75 N f mg N = mg 10 min

Homework Do #4, 5, and 8 on the Newton’s Second Law Problem Set. (You have now been assigned all of the questions.) 5 min

Lesson 5: Newton’s Third Law Unit 4: Newton’s Laws Lesson 5: Newton’s Third Law

Newton’s 3rd Law “Every action force is accompanied by an equal and opposite reaction force.” In other words, if A pushes or pulls B, then B pushes or pulls A in the opposite direction with the same amount of force. 5 min

The ball pushes back on the bat with 250 N [East]. Try This: A baseball player hits a ball with a bat. The force on the ball during contact is 250 N [West]. What is the reaction force? The ball pushes back on the bat with 250 N [East]. 5 min

Try This: Your head slowly sinks into a soft pillow Try This: Your head slowly sinks into a soft pillow. Show all the action-reaction pairs in a diagram. Fp Fg Rp Rg 10 min

Brain Break!

Action/Reaction Problems Example: Two boxes (10 kg and 15 kg) are tied together by a string. The 15 kg box is pulled forwards by a force of 50 N. If there is no friction, find: the acceleration of the boxes. the force exerted by a the 10 kg box on the 15 kg box. 15 kg 10 kg 50 N b) 10 kg S F = ma S = (10)(2) = 20 N 15 min OR a) 25 kg 15 kg F = ma 50-S = (15)(2) S = 20 N 50 N F = ma 50 = 25a a = 2 m/s2 S 50 N

the force exerted by a the 10 kg box on the 15 kg box. Your turn! Repeat the previous problem, but with a coefficient of friction of 0.1 this time… 15 kg 10 kg 50 N Find: the acceleration. the force exerted by a the 10 kg box on the 15 kg box. a) N f = (mg) = (0.1)(25)(9.8) = 24.5 N F = ma 50-24.5 = 25a a = 1.02 m/s2 25 kg f 50 N mg 20 min b) N F = ma S - 9.8 = (10)(1.02) S = 20 N f = (mg) = (0.1)(10)(9.8) = 9.8 N 10 kg f S mg

Unit 4: Newton’s Laws Lesson 6: Review