Lecture 4: Newton's Laws of Motion PDF

Lecture 4 Newton’s Laws of Motion By Dr. Ahmed M. Hassan Outlines 1- Force, Weight & Gravitational Mass 2- Newton’s 1st law 3- Newton’s 3rd law 4- Newton’s 2nd law 5- Some Examples of Newton’s Laws 6- Friction  INTRODUCTION - What is causing motio...

Lecture 4 Newton’s Laws of Motion By Dr. Ahmed M. Hassan Outlines 1- Force, Weight & Gravitational Mass 2- Newton’s 1st law 3- Newton’s 3rd law 4- Newton’s 2nd law 5- Some Examples of Newton’s Laws 6- Friction  INTRODUCTION - What is causing motion?  Force is the cause of change in the state of motion of a body or an object. - Although there are many kinds of forces in nature, the effects of any force are described accurately by three general laws first stated by Isaac Newton. - Newton’s laws work well for most things we see, but they don’t work as well for tiny particles or things moving very fast, like light. - But these laws work well for most things we do in science and engineering, like studying space, the human body, the Earth, and building things.  1 - FORCE, WEIGHT, AND GRAVITATIONAL MASS - If we push or pull an object, we are exerting a force on it. - Forces have both magnitudes and directions, so they are vector quantities. - Forces that happen only when two objects touch are called contact forces. - Some forces, like gravity, magnetism, and electric forces, can work between objects without them touching. These are called non-contact forces. - The SI force unit is the Newton (N).  WEIGHT, AND GRAVITATIONAL MASS An important force is gravity, which pulls things down. The gravitational force on something is called its weight (w). The weight of an object is related to its mass (m). We can find the mass by dividing the weight by the gravitational pull (g) of where it is: m = w / g Gravity pulls objects downward in the direction they would fall if nothing holds them up. So, the weight (w) points the same way as gravity (g), and we can also write: w=mg The unit for weight is: 1 Newton (N) = 1 kg * 1 m/s² For example, a man weighing 1000 N on Earth has a mass of: m = 1000 N / 9.8 m/s² = 102 kg  2- NEWTON’S FIRST LAW - Newton’s first law states that: An object will remain at rest or keep moving in a straight line at a constant speed unless an external force acts on it to change its state of motion. - An equivalent statement : (1) an object at rest remains at rest, and (2) an object in motion continues to move with constant velocity. - Equilibrium: It happens if two or more forces acting on an object add to zero or “balance.” Fnet = 𝐅=0 - Example, if you’re sitting still in your chair, the forces on you are balanced. Gravity pulls you down, but the chair pushes you up with an equal force. This balance of forces is called equilibrium. - The first law applies to objects in uniform motion as well as objects at rest. Example: A woman has a mass of 60 kg. She is standing on a floor and remains at rest. Find the normal force exerted on her by the floor. Solution: Givens; m = 60 kg, g = 9.8 m/s2 Unknown; N = ?? Apply Newton’s first law; Fnet = 0 N – w = 0  N = w N=mg Hence the normal force has a magnitude of N= 588 N  3- NEWTON’S THIRD LAW - The third law says that when two objects interact, they exert equal and opposite forces on each other. - For example, if you push against a pool wall, the wall pushes you back, moving you further into the pool. - The reaction force the wall exerts on you is opposite in direction to the (action) force you exert on the pool. - The third law is: If one object exerts a force F1 on a second, then the second object exerts an equal but opposite force -F2 on the first. F1 = - F2  4- NEWTON’S SECOND LAW - Newton’s second law states that: The acceleration of an object is directly proportional to the net force acting on it: Fnet = 𝐅 = m a - The mass of an object tells you how much matter it has, or in simpler terms, how much it resists changing its movement. - Example: A worker applies a constant horizontal force with magnitude 20 N to a box with mass 40 kg resting on a level floor with negligible frication. What is the acceleration of the box? (Assume friction is negligible.) Solution: Given; m = 40 kg, F = 20 N Unknow; a = ?? a = F/m = 20 / 40 = 0.5 m/s2 Example: A child pushes a sled across a frozen pond with a horizontal force of 20 N. Assume friction is negligible. a) If the sled accelerates at 0.5 ms-2, what is its mass? b) Another child with a mass of 60 kg sits on the sled. What acceleration, the same force produce now? Solution: a) Givens; F = 20 N, a = 0.5 m/s2 Unknown; m= ? m=a/F= b) The total mass of the passenger and the sled is 60 kg+40 kg=100 kg Thus the acceleration is now: The same force produces a small acceleration when applied to a more massive object.  5- SOME EXAMPLES OF NEWTON’S LAWS ₋ We use a step-by-step method to relate the acceleration of an object to the forces acting on it: 1.First, we draw a clear picture of the object and all the forces acting on it. 2.Next, we apply Newton’s second law, F = ma, to each object individually. If there are multiple forces, F1, F2,..., Fn, acting on the object, the total or net force (Fnet) is the sum of all the forces. This gives us: Fnet = F1 + F2 +... + Fn = ma ₋ In simpler terms, we can break this into two parts:  In the x-direction: F(net)x = F1x + F2x +... + Fnx = m ax  In the y-direction: F(net)y = F1y + F2y +... + Fny = m ay Example : A child pulls a train of two cars with a horizontal force F of 10 N. Car 1 has a mass m1 = 3 kg and car 2 has a mass m2 = 1 kg. The mass of the string connecting the cars is small enough so it can be set equal to zero, and friction can be neglected. a) Find the normal forces exerted on each car by the floor b) What is the tension in the string? c) What is the acceleration of the train? Solution: Givens; F = 10 N in horizontal direction, thus no motion in y-direction, Fy = 0 m1 = 3 kg, m2 = 1 kg a) Since Fy = 0, for car 1, N1 – w1 = 0 and N2 – w2 = 0 Thus the normal forces exerted by the floor are: b) Since F = 10 N moving to the right direction, thus for car 1, Fx = m1 a F – T = m1 a (i) Note that; T is the tension in the string Also for car 2, Fx = m2 a T = m2 a (ii) Equation (ii) gives a = T/m2. Substitute this in equation (i), F – T = m1 (T/m2)  F = T + m1 (T/m2) F = T (1+ m1 /m2) Solve for T, T = F / (1 + (m1/m2)) = 10 N / ( 1 + (3 kg /1 kg)) = 2.5 N c) Now we can find the acceleration a from equation (ii): a = T/m2 = 2.5 N / 1 kg = 2.5 m/s2  6- FRICTION - Friction is a force that always act to resist the motion of one object on another. - Frictional forces are very important, since they make it possible for a us to walk, use wheeled vehicles and hold books. - Frictional forces in fluids are called viscous forces. Example: A 50-N block is on a flat, horizontal surface. a) If a horizontal force T=20 N is applied and the block remains at rest; what is the frictional force? b) The block starts to slide when T is increased to 40 N. What is s ? c) The block continues to move at constant velocity if T is reduced to 32 N. What is K? Solution: a) Since the block remains at rest when the force T is applied, fr = T = 20 N b) Since the block just begins to slide when the applied force is increased to 40 N, the maximum frictional force must be: fr(max) = 40 N s = fr(max) / N = 40 N /50 N = 0.8 c ) Since the block moves with constant velocity when 32 N force is applied