Physics 121

4. Motion and Force: Dynamics 4.1 Force 4.2 Newton’s First Law of Motion 4.3 Mass 4.4 Newton’s Second Law of Motion 4.5 Newton’s Third Law of Motion 4.6 The Force of Gravity and Weight 4.7 Applications of Newton’s Laws 4.8 Applications involving Friction 4.9 The art and science of problem solving

4.1 Force A force is a push or a pull A force is that which causes a body to accelerate

Example 4.1 You see an object moving towards the East with a constant speed of 3 m/s. What can you say about the force acting on it?

Solution 4.1 The force (or net force to be precise) must be zero. The keyword is acceleration. The acceleration of the object is given to be zero. (Neither speed nor the direction is changing). If there is no acceleration then there cannot be a net force. So you see, motion without force is possible but acceleration without force is impossible!

Example units 3 units The term “net force” simply means the resultant or unbalanced force. (a) What is the net force acting on the object shown? (b) Would you expect this object to accelerate?

Solution units 3 units (a) The net force is 4 units to the right (b) This object will accelerate towards the right

4.2 Newton’s First Law In the absence of external forces, an object at rest remains at rest and an object in motion continues in motion with a constant velocity (that is with a constant speed in a straight line)

Newton’s First Law (in English) When no force acts on an object, the acceleration of the object is zero

4.3 Mass The mass of an object depends on how much stuff is in the object. The mass of a given object is the same on Earth, under water, and on the moon! The tendency of an object to resist acceleration is called inertia. The greater the mass of an object the greater the inertia. In other words, the mass of an object is a measure of its inertia and vice- versa.

4.4 Newton’s Second Law The acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass F = ma

Newton’s Second Law... In English F = ma Acceleration is greater if the force is greater and Acceleration is smaller if the mass is greater

Example to 60 in 3 seconds! Explain two different ways to design a car in order for it to go from 0 to 60 in 3 seconds.

Solution to 60 in 3 seconds! There is a right way and a wrong way to go about this and, by golly, I am going to show you the right way! 1. Design a bigger engine in order to apply a greater force. The downside of this is that it will cost you! 2. Take a tin can (in order to reduce the mass) and paint it red

4.5 Newton’s Third Law When two objects interact, the force exerted by object 1 on object 2 is equal in magnitude but opposite in direction to the force exerted by object 2 on object 1 F 12 = -F 21

Newton’s Third Law... in English Action and reaction are equal and opposite Action: You stand on a skateboard and throw a ball Reaction: The ball pushes you back! Note Action and reaction act on different objects and so cannot cancel each other.

Example Head-on Collision A big truck slams into a Neon. Which statement is most nearly correct: A. Force of truck on neon equals the force of Neon on truck B. Force of truck on neon far exceeds the force of Neon on truck C. Force of truck on Neon is much less than the force of Neon on truck D. Neither experiences any force because the equal and opposite forces cancel each other out

Solution Head-on Collision A. Force of truck on neon equals the force of Neon on truck Note Do not confuse force with “damage” (acceleration). The same force will cause a tremendous amount of acceleration on the puny Neon.

4.6 The Force of Gravity and Weight The weight of an object is the attractive force of gravity acting on it Acceleration due to gravity Near the surface of the Earth, the attractive force of gravity accelerates objects at the rate of approximately 9.8 m/s 2

Example weight of a crate (a) What are the units of weight? (b) What is the weight of a 5 kg crate?

Solution weight of a crate (b) F = ma F g = mg (“g” stands for “a” due to gravity) F g = (5)(9.8) F g = 49 N (a) Weight is a force (of gravity) so the units would be newtons. A newton is a kg m / s 2. A force of 1 N will cause a 1 kg mass to accelerate at the rate of 1 m / s 2.

Example Moon walk Object P weighs the same on the Moon as object Q does on Earth. Identify the correct statement: A. Mass of P is more than the mass of Q B. Mass of P is less than the mass of Q C. Mass of P on the Moon is more than its mass on Earth D. Weight of P on the Moon is more than the weight of Q on Earth

Solution Moon walk A. Mass of P is more than the mass of Q. In fact, mass of P must be 6 times the mass of Q because g M = 1/6 g E Note: Mass never changes. Mass is mass is mass. (Unless you break the object in two!). Weight changes if the acceleration due to gravity changes.

4.7 Applications of Newton’s Laws We will present situations from daily life and apply Newton’s Laws to answer questions about forces and motion. This will aid us in comprehending the material learned so far. Practice makes perfect!

Example The sun also rises! 45 T T What is the tension in the string (T) if the mass of the Van Gogh painting is 2 kg and the price is 37 million dollars? A. 10 N B. 15 N C. 20 N D. 40 N

Solution The sun also rises! T sin 45 + T sin 45 = 20 N T = 14 N Free - body Diagram [FBD]

Example Slippery when wet What is “a” in m/s 2 if there is no friction? kg

Solution Slippery when wet F = ma mg sin  = ma a = g sin  a = 5 m/s 2  m mg mg sin  mg cos  Free - body Diagram [FBD] What is “a” in m/s 2 if there is no friction?

4.8 Applications involving Friction Frictional forces resist the motion of objects. They can arise when one surface rubs against another, or an object encounters resistance as it moves through air, water, and other media.

Example Calculate  You apply a 30 N force on a 5 kg crate to push it horizontally across a factory floor. The speed of the crate increases at the rate of 1.5 m/s every second. What is the coefficient of friction?

Solution Calculate  F net = ma F - f = (5)(1.5) 30 - f = 7.5 f = 22.5 N f =  n 22.5 = (  ) (49)  = 0.46 F = 30N mg = 49N n = 49N f = 22.5N Free - body Diagram [FBD]

All about friction... Static friction must be overcome in order to start the motion f s   s n Kinetic friction must be overcome in order to maintain the motion. f k =  k n n = normal reaction or the force that the surface exerts perpendicularly on the object  s and  k are the respective coefficients of friction and depend on the nature of the surfaces.  s is slightly greater than  k.

Example Who wins? kg 2 kg Calculate the acceleration and tension if  = 0.1

Solution Who wins? a = 0.5 m/s 2 T = 19 N Apply F = ma to each mass individually kg 2 kg 20-T =2a ….(1) T – 30 sin 30 – 30 cos 30(0.1) =3a …(2) Now solve (1) and (2) for a and T

That’s all folks!