Chapter 5:  Laws of Motion including friction
5.1 The Concept of Force                 (external force, equilibrium) 5.2 Newton's First Law and Inertial Frames 5.3 Mass 5.4 Newton's Second Law					5.5 The Gravitational Force and Weight 5.6 Newton's Third Law 5.7 Some Applications of Newton's Laws                                (tension, equilibrium) 5.8 Forces of Friction
5.1 The Concept of Force
Net force is the summation of all forces on an object         If net force is zero…then acceleration is zero.         If mass increases…and net force is zero, then acceleration decreases. Gottfried Leibniz (calculus) was born on 1646 on July 1^st, published his finding on calculus in 1684.         Sir Isaac Newton developed calculus in 1665, 19 years earlier         than Leibniz, but didn’t publish results until 1687. Source www.prenhall.com/physics Physics 3/E daily calendar by James S. Walker
5.2 Newton's First Law and Inertial Frames
Newton's 1st Law In absence of external forces an object at rest remains at rest and an object in motion continue in motion with a constant velocity, when viewed from an inertial reference frame. If an object doesn't interact with other objects, it's possible to identify a reference frame in which the object has zero acceleration
Inertial Frames Have you ever been at a red light and thought the driver next to you was moving very slowly backwards? And it turned out your foot had moved ever so slightly off the break. Since it was such a small acceleration your body notice, thus didn’t signal your brain that you were moving forward, thus you thought the car (or sign, etc) was moving backwards.
Example 1:  A boy in the back seat of a moving SUV throws a ball straight up.  For the observer who is at a bus stop, which the SUV passes, the ball appears to travel in a ballistic path.	Explanation Once again you throw a ball up (vi = 4 m/s) in the moving car (vcar = 20 m/s) To you the ball travels directly up and then returns directly down. As you drive past your friend standing at his/her house...they see this ball moving along with your car. So to them it appears the path the ball follows a ballistic path with both with vx = 20 m/s vy = 4 m/s. One interesting note. The y-component of the velocity did have a force upon it...GRAVITY! Gravity only affect components in the vertical direction
The x-component of the velocity was constant (no forces acting upon it).                     *  If the vx has no net forces acting upon it, vx will be constant.     This holds true for low velocities for most objects when air friction is neglible.     Does this hold true for a feather?     Obviously not, as you will see when we discuss air friction which is dependant upon                         * Exposed surface area                           * shape                         * ρair (density)                                          * ò v dv = ½v² Important to remember: must use SI units (mks) (or be consistant; i.e. cgs, english, etc).
Demo:  Newtonian Table Cloth:  ME-F-NT Demo: Hoop & Tack:  ME-F-HT
5.3 Mass
Mass is a property of the object, just like density or volume.                    What a coincidence…there is even a formula linking these                    three properties together.											r = m / V
Since mass is a property of a material and changes as easily as the volume (thus not easy), mass is essentially a constant over a limited amount of time (at non-relativistic speeds) So…you weigh 100 kg.         You jump on the space shuttle…and in orbit you are weightless, so you weigh 0 kg.         Hmmm…interesting.          How can 100 kg = 0 kg? So if you ever tell someone you weight 100 kg…then you are saying that you can substitute 0 in any equation where there is a value of 100.         Result…can ANYONE weight 100 kg?         Ans:  NO…  kilograms is a unit of mass, which is a property of the body. Gravity gives an object weight.  So weight can only be measured in Newtons, pounds, dynes, or other force units.
5.4 Newton's Second Law
Newton's 2nd Law The acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass ΣF = ma (when viewed from a inertial reference frame)
Demo:  Rope and 3 Students:  ME-J-RS
5.5 The Gravitational Force and Weight
F = ma FW = m ag FW = mg So weight is mg			(g is the local acceleration due to gravity, about 9.8 m/s^2 for earth…about 1.6 m/s^2 for the moon)
5.6 Newton's Third Law
Newton's 3rd Law If two objects interact, the force exerted by the first object (on the second object) is equal (in magnitude) and opposite (in direction) to the force exerted by the second object (on the first objected.) ·  No isolated force can exist. Forces Always occur in pairs.
Demo: Hopper Popper:  ME-H-AR
5.7 Some Applications of Newton's Laws (tension, equilibrium)
Example: A Big Bird walking a tight rope A bird likes to walk on a tight rope. This is a very large bird, about 60 kg. If he starts 5 meters from the left pole and the total distance across is 20 meters, how far from the other end will he get before the tight rope breaks. The tight rope can support a maximum of 500 N. Assume vertical deflection at any time will be 0.8 meters below horizontal.				Since the tight rope stationary at any instance between steps we know: Fx = 0     and     ΣFy = 0         Σ Fx         = 0 N sinβ T1 - sinθ T2       = 0 N T1 = sinθ T2 / sinβ             ΣFy              = 0 N cosβ T1                 + cosθ T2 - 600N = 0 cosβ (sinθ T2 / sinβ) + cosθ T2 = 600 N cotβ sinθ T2              + cosθ T2 = 600 N T2 = 600 N / (cotβ sinθ + cosθ) Input the measured values for β and θ; solve for tension of T2 and then T1
Atwoods Machine In the Atwoods Machine two masses are attached on both ends of the end of a string which is stretched over a pulley.  If the pulley if Frictionless, the tension in both sides MUST be EQUAL (it’s the same string and no force is applied at the pulley).  Let’s calculate the tension caused by block A and by block B IF THE pulley is held stationary.
Think About It You enter an elevator.  The elevator begins to rise.  You “feel” heavier… Next…you enter an elevator…and it begins its descent.  During initial 1 to 2 seconds during descent you “feel” lighter.                 What’s going on? You are accelerating in the same direction of gravity or you’re accelerating in a direction opposing gravity. · If “g” = 10 m/s^2 and you accelerate in the direction of gravity…at 1 m/s^2 you will feel 10% lighter · And at 3 m/s^2 you will feel 30% lighter · If you accelerate down (in the direction of gravity) at a rate of 10 m/s^2 you will feel 100% lighter, which is called “free fall”. We’ll use this concept to calculate the tension in the elevator cable for this problem.						TA = m1 g TA = 0.9kg * 10m/s² TA = 9 N				TB = m2 g TB = 1.1kg * 10m/s² TB = 11 N
						Next the now Frictionless pulley is released.  Thus the tension in both sides MUST BE EQUAL. We also know the heavier side accelerates down and the lighter side accelerates up.
						Important note Acceleration due to gravity is about 10 m/s² Rarely I use agravity = 9.8 m/s² Many books use "g" instead of agravity As in Fweight = mg Actually many books use    W(eight) = mg      I prefer Fweight = mg
So you noticed when the pulley was stopped the tension in the lines was not equal, but what happens when you release the pulley so that the weights fall toward the side of Fnet. It's labelled as T1 & T2 as if the tension in the rope is different. Is this true? NO!!! This is ONE string...a pulley redirect ONLY. If T1 ≠ T2 you have a problem. This is the check. Let's try it. Above Fnet = 0.2 Newtons KEY QUESTION...HOW MUCH MASS IS MOVING? Total mass = m1 + m2 = (0.9 + 1.1) kg Newtons 2nd law ΣF = mtotal a 0.2 + = 0.2 kg * a a = 1 N/kg a = 1 m/s² Now calculate the tension for both sides In the drawing, T1 = m1 (g+a) T1 = 0.9 (10+1) T1 = 9.9 N and T2 = m2 (g-a) T2 = 1.1 (10-1) T2 = 9.9 N		Check Does T1 = T2? Yes, so you know the results are correct. Did you notice something else? The average Tension when at rest was 10 N ((9+11)/2). The tension in the line while accelerating will always be less than the average while at rest. Side note What is the basic units of a Newton? ΣF = ma/ N = kg m/s² You may be wondering why T1 = m1 (g+a) T2 = m2 (g-a) ALWAYS use your personal experience as a guideline. Everyone one of you knows the answer Many years ago...one of my friend's had a little sister that was afraid of elevators (she was 14 at the time) We loved going to the unversity center and sticking her in an elevator and then we ran up the steps to meet her at the top. So even she knew that as soon as the elevator started rising that she felt heavier (g+a) This I believe is one of the reason for her fear. Conversely, if you are at the top and the elevator starts lowering you feel lighter (g-a) (until it acheived constant velocity). So apply this to the Atwood's machine.
Box Pulled by weight over Pulley Apply the same logic as in Atwoods Machine example mtotal = M + m mtotal = 8kg + 2kg mtotal = 10 kg. Fnet = mg Fnet = 2kg * 10 m/s² Fnet = 20 N ΣF = ma 20 N = 10 kg * a a = 2 m/s² Once again the Tension in the rope better be equal for all involved. Big Box on top of table ΣF = ma ΣF = 8kg * 2 m/s² ΣF = 16 N Hanging Weight ΣF = ma ΣF = 2kg * (10 - 2) m/s² ΣF = 16 N
Boxed Pulled by weight over Pulley w/Friction Remember this... Friction is F=μN Ffriction = μ FNormal Normal force = cos θ mg So apply this to the above case. 8 kg block is resting on the surface whre cos θ= cos(0°) cos(0°) = 1 Ff = μ mg = 0.2 * 80 N = 16 N Which direction is Friction? ALWAYS opposes motion....thus the Ff vector must be drawn in to the left as shown Fnet = mg - μMg Fnet = 20 - 16 Fnet = 4 N How much mass is being moved? ΣF  = mtotala 4 N = 10 kg * a a = 0.4 m/s²
Downhill ramp (w/Friction) while weight hangs over a Pulley Coefficient of kinetic friction = 0.6 It's actually better to show the normal force originating at the table top (ramp) and rising perpendicular to the surface of the ramp. (instead of below as shown) I just want to show how to Fweight is broken into its respective components. Remember the weight must be the largest value (hypotenuse) when you are going to calculate the Normal Force and the Force in the direction of the motion (due to the angle of the ramp.) Two forces (component of weight & 2kg block) accelerate this block system down the ramp, and one resistive force (friction, μ FNormal) opposes this motion. ΣF = Fin dir   + Fhanging block- Ffriction ΣF = sinθ Mg +       mg           - μcosθ Mg Let θ = 30° ΣF = 40 N + 20 N - 41.6 N ΣF = 18.4 N ΣF = 18.4 N = (8kg + 2kg) * a a = 1.84 m/s²
Tension Between Block with Friction and Weight Over Pulley
One question: what is the tension in the line of a block on an incline attached to a weight suspended by a pulley where Ff = μ N   & N = cos30°(4kg)10m/s² = 34.6 N I believe we said that m1 was 4 kg and m2 was 2 kg, thus Fnet = mtotala Fnet = m2g         + sinθm1g -       Ff        =      mtotal     a Fnet = 20N +  ½ 40N - 0.7*34.6N = (2kg +4kg)*a Fnet = 15.8 N = 6 kg * a a = 2.6 m/s² And you were asked to calculate the tension in T1 This is where I simplify the problem.
T1 = T2 = m2*(g-a) = 2kg*(10-2.6) = 14.8 N This is where estimating g = 10 m/s² may cause problems. For instance if a = 9.6 m/s²...using g = 10 m/s² results in a 50% error!!!   (g-a) = 10 - 9.6 = 0.4 m/s² (g-a) = 9.8 - 9.6 = 0.2 m/s² So when ever you are subtracting you must be careful of your estimations. A very similar situation arises with uncertainty in your lab...when you subtract uncertainties you can end up with uncertainty associated with your result being larger than your result itself.
5.8 Forces of Friction
Static (not sliding) and Kinetic (moving/sliding) Friction Demo: Static Equilibrium Balls:  ME-K-SE Demo: Static vs Kinetic Incline Plane:  ME-K-SK Think back to when you had to push (slide across a floor) something heavy (sofa, desk, etc). Was it easier to start it moving or was it easier to keep it moving once you started it to slide In my experience...once it started moving in a small area you have to be careful not to hit a wall. BECAUSE IT SLIDES EASIER ONCE IT STARTS TO MOVE!!! This is called kinetic friction which is lower than static frictional Coefficient We will also use this example to introduce Ch 7. ·  The work you do is the force you apply to an object through a displacement Note: simplified definition