Introduction to Mechanics Dynamics Forces Applying Newton s Laws

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Introduction to Mechanics Dynamics Forces Applying Newton s Laws Lana heridan De Anza College Feb 21, 2018

Last time force diagrams Newton s second law examples

Overview Newton s second law examples Newton s third law action-reaction pairs of forces kinds of forces and problem solving gravity and weight

, and time: Example s 2 (5.4) e is the pound (lb). A force of 1 lb is oduces an acceleration of 1 ft/s 2 : /s 2 Consider a 0.3 kg hockey puck on frictionless ice. Find its acceleration. y F 2 F 1 = 5.0 N F = 8.0 N 2 60 oving s suro two 2. 20 F 1 x hapter 3, predict the approximate n the same direction. s problem is categorized as one that

, and time: Example s 2 (5.4) e is the pound (lb). A force of 1 lb is oduces an acceleration of 1 ft/s 2 : /s 2 Consider a 0.3 kg hockey puck on frictionless ice. Find its acceleration. y F 2 F 1 = 5.0 N F = 8.0 N 2 F net = F 1 + F 2 = (F 1 cos( 20) + F 2 cos(60)) i +(F 1 sin( 20) + F 2 sin(60)) j = 8.70 i + 5.21 j N 60 oving s suro two 2. 20 F 1 x hapter 3, predict the approximate n the same direction. s problem is categorized as one that

, and time: Example s 2 (5.4) e is the pound (lb). A force of 1 lb is oduces an acceleration of 1 ft/s 2 : /s 2 Consider a 0.3 kg hockey puck on frictionless ice. Find its acceleration. y F 2 F 1 = 5.0 N F = 8.0 N 2 F net = F 1 + F 2 = (F 1 cos( 20) + F 2 cos(60)) i +(F 1 sin( 20) + F 2 sin(60)) j = 8.70 i + 5.21 j N oving s suro two 2. 60 20 F 1 hapter 3, predict the approximate n the same direction. s problem is categorized as one that x a = F net m 8.70 N i + 5.21 N j = 0.3 kg = 29.0 i + 17.4 j ms 2

, and time: Example s 2 (5.4) e is the pound (lb). A force of 1 lb is oduces an acceleration of 1 ft/s 2 : /s 2 Consider a 0.3 kg hockey puck on frictionless ice. Find its acceleration. y F 2 a = 29.0 i + 17.4 j ms 2 F 1 = 5.0 N F = 8.0 N 2 a = 29.0 2 + 17.4 2 = 34 ms 2 oving s suro two 2. 60 20 F 1 hapter 3, predict the approximate n the same direction. s problem is categorized as one that x at an angle ( ) 17.4 θ = tan 1 29.0 = 31 above the horizontal (x-axis).

Newton s Third Law Newton s Third Law is commonly stated as For every action, there is an equal and opposite reaction. However it is more precisely stated: Newton III If two objects (1 and 2) interact the force that object 1 exerts on object 2 is equal in magnitude and opposite in direction to the force that object 2 exerts on object 1. F 1 2 = F 2 1

Newton s Third Law Main idea: you cannot push on something, without having it push back on you. If object 1 pushes on (or interacts with) object 2, then the force that object 1 exerts on object 2, and the force that object 2 exerts on object 1 form an action reaction pair.

us isolate only those forces Newton s Third Law: Action Reaction analysis. Pairs ewton s Third Law 119 n F tm s 2 F 12 F 21 r - t s - F 12 F 21 Figure 5.5 Newton s third law. The force F 12 exerted by object 1 on object 2 is equal in magnitude and opposite in direction to the force F 21 exerted by object 2 1 a F mt F g F me F Em

5.6a. The gravita- Pitfall Prevention 5.6 Defining a ystem Consider5.6 these Newton s particles Third which Law exert a force119 on each other: n two objects, we ted by a on b. The erts on object 2 is ect 1 is called the rthermore, either e terms for convet objects and must, the force acting by the Earth on 2 Figure 5.5 Newton s third law. agnitude They are of attracted. this The Each force will F12 accelerate exerted by object toward 1 the other. xerted by the procelerate the Earth on object 2 is equal in magnitude and opposite in direction to the force F21 exerted by object 2 projectile toward on object 1. s acceleration due F 12 F 12 F 21 F 21 1

Defining a ystem Consider5.6 these Newton s particles Third which Law exert a force119 on each other: n two objects, we ted by a on b. The erts on object 2 is ect 1 is called the rthermore, either e terms for convet objects and must, the force acting by the Earth on agnitude of this xerted by the procelerate the Earth projectile toward s acceleration due 2 F 12 F 12 F 21 F 21 Figure 5.5 Newton s third law. They are attracted. The Each force will F12 accelerate exerted by object toward 1 the other. on object 2 is equal in magnitude But wait: do the forces and opposite cancel? in direction to the force F21 exerted by object 2 F on object 1. 1 2 = F 2 1 F 1 2 + F 2 1 = 0 Is the net force zero? How can they each accelerate? 5.6a. The gravita- Pitfall Prevention 5.6 1

Defining a ystem Consider these 5.6 particles Newton s Third which Law exert a force 119 on each other: between two objects, we e exerted by a on b. The t 1 exerts on object 2 is on object 1 is called the ms; furthermore, either se these terms for conveifferent objects and must ample, the force acting xerted by the Earth on Is the net force zero? the magnitude of this orce exerted by the proust accelerate the Earth tes the projectile toward ver, its acceleration due 2 F 12 F 12 igure The 5.6a. only The force gravitao this accelerates. force is the force n Does Not Always Equal mg In on particle Pitfall Prevention 1 is F 2 1 5.6, so the net force is not zero: it onitor does not acceler- the situation shown in Figure 5.6 F 21 Figure 5.5 Newton s third law. The force F12 exerted by object 1 on object 2 is equal in magnitude No! The forces act on and different opposite in objects. direction to To find if particle 1 accelerates, we find the force net Fforce 21 exerted onby particle object 2 1. We do not on object 1. consider forces on particle 2. F 21 1

Action and Reaction Why when we fire a cannon does the cannon ball move much faster forward than the cannon does backwards? Why when we drop an object does it race downwards much faster than the Earth comes up to meet it?

Action and Reaction Why when we fire a cannon does the cannon ball move much faster forward than the cannon does backwards? Why when we drop an object does it race downwards much faster than the Earth comes up to meet it? The masses of each object are very different! From Newton s second law a = F m If m is smaller, a is bigger. If m is very, very big (like the Earth), the acceleration is incredibly small.

ject, which we will model as a particle. Therefore, a free-body dia isolate Forceonly Diagrams those forces on the object and eliminate the other forc alysis. Question. Do the two forces shown in the diagram that act on the monitor form an action-reaction pair under Newton s third law? n F tm n F tm n F tm F mt F g F me (A) Yes. (B) No. F Em Figure 5 F g F Em F g F Em b c the force the gravi exerted b force F m diagram s shows th

ject, which we will model as a particle. Therefore, a free-body dia isolate Forceonly Diagrams those forces on the object and eliminate the other forc alysis. Question. Do the two forces shown in the diagram that act on the monitor form an action-reaction pair under Newton s third law? n F tm n F tm n F tm F mt F g F me (A) Yes. (B) No. F Em Figure 5 F g F Em F g F Em b c the force the gravi exerted b force F m diagram s shows th

ome types of forces Gravitation The force that massive objects exert on one another. Newton s Law of Universal Gravitation F G = Gm 1m 2 r 2 for two objects, masses m 1 and m 2 at a distance r. G = 6.67 10 11 Nm 2 kg 2. (Challenge: check the units of G.)

ome types of forces: Gravitation For the moment, we will care about this force in that it gives objects weight, W. F g = W = mg and g = GM Earth R 2 Earth The force F g or W, acts downwards towards the center of the Earth.

Gravitation: Measurement of G G was first measured by Henry Cavendish using a torsion balance. 1 Diagram from Wikipedia by Chris Burks.

Gravitation: Mass of the Earth Determining G allowed for a the mass of the Earth M E to be calculated. Christopher Columbus notwithstanding, the radius of the Earth was known fairly accurately from ancient times: R E = 6.37 10 6 m G = 6.67 10 11 Nm 2 kg 2. F G = Gm 1m 2 r 2 Figure out the mass of the Earth, M E.

Gravitation: Mass of the Earth

ummary Newton s third law action-reaction pairs of forces types of forces: gravity Homework Walker Physics: PREV: Ch 5, onward from page 138. Questions: 8, 11, 13, 23; Problems: 11, 16 & 17, 19, 33