More situations with Newton’s Second Law

1. More situations with Newton’s Second Law Consider a problem that includes an air resistance term: Fair resistance = bv2 , where b is a coefficient that depends on the shape of the object and the density of the air. The direction of the air resistance term is opposite the motion.

2. Air Resistance and Gravity Consider an object dropped from a high place: • we have Fgravity = W = mg directed down, and with air resistance (AR) • we have FAR = bv2 directed up. Newton’s Second Law gives (for 1-D): -mg + bv2 = ma , or -mg + bv2 = dv/dt which is a differential equation with v(t) being the solution.

3. Air Resistance and Gravity -mg + bv2 = dv/dt This differential equation can be solved, but it is beyond the level of this course. We can get an approximate solution, however, by using a numerical (discrete) technique rather than the differential equation (continuous function) technique.

4. Air Resistance and Gravity -mg + bv2 = dv/dt The first step is to calculate the forces based on initial conditions. Let’s say we drop an object that has a mass of 2 kg and an air resistance coefficient 0.03 Nt-s2/m2 from a helicopter 1,000 meters above the ground. Fgravity = mg = 2 kg * 9.8 m/s2 = 19.6 Nt. FAR = bv2 = 0.03 Nt-s2/m2 * [0 m/s]2 = 0 Nt. Using Newton’s Second Law, we can now calculate the initial acceleration:

5. Air Resistance and Gravity F = ma gives: -19.6 Nt + 0 Nt = 2 kg * a, or a = -9.8 m/s2; we now assume that the acceleration does not change much over the first second, and calculate the velocity at one second: v1 = vo + ao*(1 sec) v1 = 0 m/s + (-9.8 m/s2)*(1 sec) = -9.8 m/s. x1 = xo + ½(vo+v1)*(1 sec) x1 = 1,000 m + ½ (0 m/s + -9.8 m/s)*(1 sec) = 995.1 m.

6. Air Resistance and Gravity We now repeat the whole process with the new position and velocity providing new initial conditions: Fgravity = mg = 2 kg * 9.8 m/s2 = 19.6 Nt. FAR = bv2 = 0.03 Nt-s2/m2 * [-9.8 m/s]2 = 2.88 Nt.

7. Air Resistance and Gravity F = ma  -19.6 Nt + 2.88 Nt = 2 kg * a, or a = -8.36 m/s2; we now assume that the acceleration does not change much over the second second, and calculate the velocity at two seconds: v2 = v1 + a1*(1 sec) v2 = -9.8 m/s + (-8.36 m/s2)*(1 sec) = -18.16 m/s. x2 = x1 + ½ (v1+v2)*(1 sec) x2= 995.1 m + ½(-9.8 m/s + -18.16 m/s)*(1 sec) = -981.12 m.

8. Air Resistance and Gravity We simply keep this process up until x becomes zero. Normally this would be a lot of steps, but we can use either a computer program to do this or a spreadsheet. We can then plot the graph of either v versus t or x versus t to see what the motion looks like. (See the Excel spreadsheet FallAR.xls which can be downloaded from my PHYS 201 web page.)

9. Air Resistance and Gravity To see the effects of different air resistance coefficients, we simply change the value of b and run the program again or recalculate the spreadsheet. To see the effects of different air resistance functions, such as F=-bv, simply change the F-air resistance function in cell E-5 and copy this change into all the following E cells.

10. Terminal Velocity As we can see from the spreadsheet, the velocity does not continue to increase (as it would without air resistance), but instead it approaches a terminal velocity. This is because as the speed increases due to gravity, the air resistance also increases which acts against gravity. As the air resistance approaches the strength of gravity, the net force approaches zero and so the acceleration also approaches zero, and the velocity levels off.

11. Terminal Velocity We can calculate the terminal velocity without having to set up the spreadsheet by recognizing that there will be no acceleration and hence no increase in speed when the air resistance equals the gravity: mg = bv2; solving the terminal speed gives: vterminal = [mg/b]1/2. Using the numbers we used earlier, vterminal = [2 kg * 9.8 m/s2 / .03 Nt-s2/m2]1/2= 25.56 m/s = 57 mph.

12. Ramps and Pulleys We use ramps and pulleys to make it easier to do certain things. We’ll first look at ramps, and then pulleys.

13. Ramps Why is it easier to push something up a ramp than it is to lift it? Let’s look at a picture of the situation: Fc P W=mg

14. Ramps In the static case with no friction, Newton’s 2nd law gives: F// = P - mg sin() = ma// F = Fc - mg cos() = ma = 0 Fc P W=mg 

15. Ramps In the absence of friction, to balance gravity we only need the Pull, P to be Pbalance = mg sin(). Since sin() is always less than or equal to 1, Pbalance will always be less than or equal to mg: Pbalance mg Fc P W=mg

16. Ramps If we now include friction, Newton’s 2nd law gives: F// = P - mg sin() - Ff = ma// F = Fc - mg cos() = ma  = 0 Fc P Ff =Fc W=mg

17. Ramps From the perpendicular equation, we see that Fc = mg cos(), so the parallel equations, with a// = 0 (just balancing weight and friction) gives: P = mg sin() +  mg cos() Fc P Ff =Fc W=mg

18. Ramps P = mg sin() +  mg cos() As long as  and  are such that sin() +  cos() < 1, the pulling force, P, will be less than the weight, which means that the ramp will make it easier to raise the object up.

19. Pulleys Pulleys can do two different things for us: 1. Pulleys can change the direction of a force - sometimes it is easier to pull down than to lift up. 2. Pulleys can “add ropes” to an object to reduce the tension in the rope, and hence reduce the pull we need to apply.

20. Pulleys Consider the single pulley below: All the pulley does is change the direction of the rope so we can pull down on the rope instead of lifting up on it. One of the effects of the pulley is that there are essentially two ropes coming from it - one to the weight and one to us. This will cause it’s attachment to the ceiling to have twice the force on it! P W = mg P = Tension = W

21. Pulleys Consider the single pulley in this diagram: In this case, the single pulley essentially adds a second rope to the weight. We still have to pull up, but now the weight is split between the two ropes, so we only have to pull with half the weight! P T T P=T, 2T = W; so P=W/2 W

22. Pulleys We now consider putting two pulleys together as in the diagram below: The left pulley essentially adds a second rope to the weight, reducing the tension in the rope by half. The right pulley simply redirects the tension so the pull is down instead of up. P

23. Pulleys Consider this combination of pulleys: The rightmost pulley just redirects the tension so we pull down. The other four pulleys combine to provide five ropes to hold the weight, thus reducing the tension in the rope by a factor of 5. P

24. Another Example of Newton’s 2nd Law:Car going around a turn Consider first a car making a right turn on a level road. To make the turn, the car must go in a circle (for a 90o turn, the car must go in a circle for 1/4 of the complete circle). This means that there will be an acceleration towards the center of the circle, which is to the right for a right turn. From the circular motion relations, we know: a = w2r, and v=wr; or a = v2/r. That says that we have a bigger acceleration the faster we go and the shorter the radius (sharper the turn).

25. Car going around a turn What are the forces that cause this circular acceleration? Gravity (weight) acts down Contact Force acts up (perpendicular to the surface) Friction - which way does it act (left or right)? Fc a=v2/r W=mg Car is heading into the screen.

26. Car going around a turn Without friction the car will NOT make the turn, it will continue straight - into the left ditch! Therefore, friction by the road must push on the car to the right. (Friction by the car on the road will be opposite - to the left.) The fastest the car can go around the turn without sliding is when the friction is maximum: Ff = mFc. Fc Ff mFc a=v2/r W=mg

27. Car going around a turn We now apply Newton’s Second Law - in rectangular components: SFx = Ff = ma = mv2/r To make the car go around the turn fastest, we need the maximum force of friction: Ff = mFc SFy = Fc - W = 0 which says Fc = mg, so mmg = mv2/r, or vmax = [mgr] Fc Ff= mFc a=v2/r W=mg

28. Car going around a turn vmax = [mgr] Note that the maximum speed (without slipping) around a turn depends on the coefficient of friction, the amount of gravity (not usually under our control), and the sharpness of the turn (radius). If we go at a slower speed around the turn, friction will be less than the maximum: Ff < mFc. There is one other thing we can do to go faster around the turn - bank the road! How does this work?

29. Banked turn By banking the road, we have not added any forces, but we have changed the directions of both the contact force and the friction force! Have we changed the direction of the acceleration? No - the car is still travelling in a horizontal circle. Fc a=v2/r Ff q W=mg

30. Banked turn Since the acceleration is still in the x direction, we will again use x and y components (rather than // and ) SFx = Fc sin(q) + Ff cos(q) = mv2/r SFy = Fc cos(q) - Ff sin(q) - mg = 0 If we are looking for the max speed, we will need the max friction: Ff = mFc . This gives 3 equations for 3 unknowns: Ff, Fc, and v. Fc a=v2/r Ff q W=mg

31. Banked turn SFx = Fc sin(q) + Ff cos(q) = mv2/r SFy = Fc cos(q) - Ff sin(q) - mg = 0 Ff = mFc . Using the third equation, we can eliminate Ff in the first two: Fc sin(q) + mFc cos(q) = mv2/r Fc cos(q) - mFc sin(q) - mg = 0 We can now use the second equation to find Fc: Fc = mg / [cos(q) - m sin(q)], and use this in the first equation to get: v = [gr {sin(q)+m cos(q)} / {cos(q) - m sin(q)} ]1/2

32. Banked Turn v = [g r {sin(q)+m cos(q)} / {cos(q) - m sin(q)} ]1/2 Notice that the mass cancels out. This means that the mass of the car does not matter! (Big heavy trucks slip on slippery streets just like small cars. When going fast, big heavy trucks flip over rather than slide off the road; little cars don’t flip over like big trucks. But flipping over is not the same as slipping! We’ll look at flipping in Part 4 of the course.) Note also that when q = 0, the above expression reduces to the one we had for a flat road: vmax = [mgr]1/2 .

33. Banked Turn v = [g r {sin(q)+m cos(q)} / {cos(q) - m sin(q)} ]1/2 Note that as m sin(q) approaches cos(q), the denominator approaches zero, so the maximum speed approaches infinity! What force really supports such large speeds (and so large accelerations)? As the angle increases, the contact force begins to act more and more to cause the acceleration. And as the contact force increases, so does friction. Actually, there is a limit on the maximum speed because there is a limit to the contact force.

34. Banked turn - Minimum Speed Is there a minimum speed for going around a banked turn? Consider the case where the coefficient of friction is small and the angle of bank is large. In that case the car, if going too slow, will tend to slide down (to the right) so friction should act to the left. Can you get an equation for the minimum speed necessary? What changes in what we did for max speed? Fc Ff a=v2/r q W=mg

35. Example of Torque Consider two people carrying a board with a weight on it – see the diagram below. The weight is 100 lb, and is located 7 feet away from the left end. The board is 8 feet long and we neglect the weight of the board. How much force does each person have to exert? FL = ? FR = ? W = 100 lb 7 feet 8 feet

36. Example of Torque We recognize this as a statics problem, and a one dimensional problem (in y). This gives us one relation: SF = 0. We identify the force the left person exerts, FL, the force the right person exerts, FR, and the Weight. SF = +FL + FR – 100 lb = 0 . FL = ? FR = ? W = 100 lb 7 feet 8 feet

37. Example of Torque SF = +FL + FR – 100 lb = 0 . This is one equation in two unknowns. There are lots of ways of dividing up the weight among the left and right persons. What does determine how much each has to exert? FL = ? FR = ? W = 100 lb 7 feet 8 feet

38. Example of Torque SF = +FL + FR – 100 lb = 0 . What is important is where the weight is placed! To get our second equation, we recognize this as a torque situation: St = 0 where each of the three forces has a distance (moment arm). But where do we measure each moment arm from – since there is no actual rotation about any point? FL = ? FR = ? W = 100 lb 7 feet 8 feet

39. Example of Torque SF = +FL + FR – 100 lb = 0 . Since there is no obvious point about which the board rotates, we are free to choose any point. Since the diagram indicates the distances measured from the left end, we can initially choose that point. Also for definiteness sake, let’s choose a counter-clockwise (CCW) rotation as being positive. FL = ? FR = ? W = 100 lb 7 feet 8 feet

40. Example of Torque SF = +FL + FR – 100 lb = 0 . From the choices of the previous slide, we see that xL = 0 ft, xR = 8 ft, and xW = 7 ft, and the torque due to FL is zero, the torque due to FR is CCW and so is positive, and the torque due to W is CW and so is negative: St = {FL * 0 ft} + {FR * 8 ft} - (100 lb * 7 ft) = 0. FL = ? FR = ? W = 100 lb 7 feet 8 feet

41. Example of Torque SF = +FL + FR – 100 lb = 0 . St = {FL * 0 ft} + {FR * 8 ft} - (100 lb * 7 ft) = 0. We now have two equations for two unknowns, and we can solve the problem. From the torque equation we can solve for FR: FR = (100 lb * 7 ft) / 8 ft = 87.5 lb. And from the force equation: FL= 100 lb – 87.5 lb = 12.5 lb. FL = ? FR = ? W = 100 lb 7 feet 8 feet

42. Example of Torque FR = (100 lb * 7 ft) / 8 ft = 87.5 lb, andFL= 100 lb – 87.5 lb = 12.5 lb. Note that the further the weight is from the person, the lighter is the necessary force. Also note that by choosing the measuring point at the left force location, we effectively eliminated the FL unknown from the torque equation making it easier to solve. FL = ? FR = ? W = 100 lb 7 feet 8 feet