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Review for Dynamics test Page 1 - Net force with weight Page 2 - Friction on a level surface

Review for Dynamics test Page 1 - Net force with weight Page 2 - Friction on a level surface Page 3 - Inclined Plane Page 4 - Statics. Page 1 - elevator problem. Net Force – Example 3 Using Weight. 35 N. Find the acceleration (on Earth). 5.0 kg. TOC. 35 N. -49 N.

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Review for Dynamics test Page 1 - Net force with weight Page 2 - Friction on a level surface

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  1. Review for Dynamics test Page 1 - Net force with weight Page 2 - Friction on a level surface Page 3 - Inclined Plane Page 4 - Statics

  2. Page 1 - elevator problem

  3. Net Force – Example 3 Using Weight 35 N Find the acceleration (on Earth) 5.0 kg TOC

  4. 35 N -49 N Net Force – Example 3 Using Weight 5.0 kg Draw a Free Body Diagram: Don’t Forget the weight: F = ma = 5.0*9.8 = 49 N TOC

  5. 35 N -49 N Net Force – Example 3 Using Weight 5.0 kg F = ma 35 N – 49 N = (5.0 kg)a -14 N = (5.0 kg)a a = -2.8 m/s/s TOC

  6. Whiteboards: Using Weight 1 | 2 | 3 | 4 | 5 TOC

  7. Find the acceleration: F = ma, weight = (8.0 kg)(9.80 N/kg) = 78.4 N down Making up + <100. N - 78.4> = (8.0kg)a 21.6 N = (8.0kg)a a = 2.7 m/s/s 100. N 8.0 kg W 2.7 m/s/s

  8. Find the acceleration: F = ma, wt = (15.0 kg)(9.8 N/kg) = 147 N down <120. N - 147 N> = (15.0kg)a -27 N = (15.0kg)a a = -1.8 m/s/s It accelerates down 120. N 15.0 kg W -1.8 m/s/s

  9. Find the force: F = ma, wt = (16 kg)(9.8 N/kg) = 156.8 N down <F – 156.8 N> = (16.0 kg)(+1.5 m/s/s) F – 156.8 N = 24 N F = 180.8 N = 180 N F 16 kg a = 1.5 m/s/s (upward) W 180 N

  10. Find the force: F = ma, wt = 1176 N downward <F – 1176 N> = (120. kg)(-4.50 m/s/s) F – 1176 N = -540 N F = 636 N F 120. kg a = -4.50 m/s/s (DOWNWARD) W 636 N

  11. This box is going downwards at 22.0 m/s and is stopped in a distance of 1.85 m. What must be the upwards force acting on it to stop it? F First, suvat: s = -1.85 m, u = -22.0 m/s, v = 0, a = ? use v2 = u2 + 2as, a = +130.81 m/s/s F = ma, wt = 1176 N downward <F – 1176 N> = (120. kg)(+130.81 m/s/s) F – 1176 N = 15697 N F = 16873.2973 = 16,900 N 120. kg W 16,900 N

  12. Page 2 - Friction on the level

  13. Force of Friction in N Normal Force - Force exerted by a surface to maintain its integrity Coefficient of Friction. 0 < < 1 (Specific to a surface) - in your book (Table 4-2) Usually the weight (level surfaces) FFr = FN

  14. Kinetic Friction - Force needed to keep it going at a constant velocity. • FFr = kFN • Always in opposition to velocity • (Demo, example calculation) • Static Friction - Force needed to start motion. • FFr<sFN • Keeps the object from moving if it can. • Only relevant when object is stationary. • Always in opposition to applied force. • Calculated value is a maximum • (Demo, example calculation, examples of less than maximum)

  15. Whiteboards: Friction 1 | 2 | 3 | 4 | 5 | 6 TOC

  16. What is the force needed to drag a 12 kg chunk of rubber at a constant velocity across dry concrete? F = ma, FFr = kFN FN = weight = mg = (12 kg)(9.8 N/kg) = 117.6 NFFr = kFN = (.8)(117.6 N) = 94.08 N = 90 N W 90 N

  17. What is the force needed to start a 150 kg cart sliding across wet concrete from rest if the wheels are locked up? F = ma, FFr<sFN FN = weight = mg = (150 kg)(9.8 N/kg) = 1470 NFFr = kFN = (.7)(1470) = 1029 N = 1000 N W 1000 N

  18. v 72 N FFr 8.5 kg s = .62, k = .48 What is the acceleration if there is a force of 72 N in the direction an 8.5 kg block is already sliding? FFr = kFN, FN = mg, FFr = kmg FFr = (.48)(8.5 kg)(9.8 N/kg) = 39.984 N F = ma <72 N - 39.984 N> = (8.5 kg)a, a = 3.77 = 3.8 ms-2 W 3.8 m/s/s

  19. v=12m/s FFr 22 kg s = .62, k = .48 A 22 kg block is sliding on a level surface initially at 12 m/s. What time to stop? FFr = kFN, FN = mg, FFr = kmg FFr = (.48)(22 kg)(9.8 N/kg) = 103.488 N F = ma < -103.488 N> = (22 kg)a, a = -4.704 ms-2 v = u + at, v = 0, u = 12, a = -4.704 ms-2, t = 2.55 s = 2.6 s W 2.6 s

  20. v a = 3.2 ms-2 F = ? FFr 6.5 kg s = .62, k = .48 A 6.5 kg box accelerates and moves to the right at 3.2 m/s/s, what force must be applied? FFr = kFN, m = 6.5 kg FFr = (.48)(6.5 kg)(9.8 N/kg) = 30.576 N F = ma < F - 30.576 N > = (6.5 kg)(3.2 ms-2), F = 51.376 = 51 N W 51 N

  21. v=12m/s F = ? FFr 22 kg s = .62, k = .48 A 22 kg block is sliding on a level surface initially at 12 m/s stops in 2.1 seconds. What external force is acting on it besides friction?? v = u + at, v = 0, u = 12 m/s, t = 2.1 s, a = -5.7143 ms-2 FFr = kFN, FN = mg, FFr = kmg FFr = (.48)(22 kg)(9.8 N/kg) = 103.488 N F = ma < -103.488 N + F> = (22 kg)(-5.7143 ms-2), F = -22 N (left) W -22 N (to the left)

  22. a = 1.2 ms-2 v 35 N FFr m s = .62, k = .48 A force of 35 N in the direction of motion accelerates a block at 1.2 m/s/s in the same direction What is the mass of the block? FFr = kFN, FN = mg, FFr = kmg FFr = (.48)m(9.8 ms-2) = m(4.704 ms-2) F = ma < 35 N - m(4.704 ms-2) > = m(1.2 ms-2) 35 N = m(4.704 ms-2) + m(1.2 ms-2) = m(4.704 ms-2 + 1.2 ms-2) m = (35 N)/(5.904 ms-2) = 5.928 kg = 5.9 kg W 5.9 kg

  23. Page 3 - Inclined planes

  24. FN = mgcos (Causes friction) (FFr = kFN ) And the plane pushes back (It doesn’t break)  Fperp = mgcos mg Since we know the angle, we can calculate the components F|| = mgsin (Acts down the plane)

  25. Whiteboards: Inclines with friction 1 | 2 | 3 | 4 | 5 TOC

  26. 3.52 kg s = .82 k = .37  = 42.0o Find Fperp, F||, the kinetic and maximum static friction: F|| = mgsin() = (3.52 kg)(9.8 N/kg)sin(42.0o) = 23.08 N Fperp = mgcos() = (3.52 kg)(9.8 N/kg)cos(42.0o) = 25.64 N FFr(kinetic) = kFN = (.37)(25.64 N) = 9.49 N FFr(static)<sFN = (.82)(25.64 N) = 21.02 N W 26 N, 23 N, 9.5 N, 21 N

  27. + 3.52 kg s = .82 k = .37 -  = 42.0o • F|| = 23.08 N, FFr(kinetic) = 9.49 N, FFr(static)< 21.02 N • Will it stay on the plane if u = 0? • No, it will not stay. The maximum static (FFr(static)<21.02 N) friction is smaller than the gravity parallel to the plane (F|| =23.08 N) W No, Blue

  28. +9.49 N -23.08 N + 3.52 kg s = .82 k = .37 -  = 42.0o • F|| = 23.08 N, FFr(kinetic) = 9.49 N, FFr(static)< 21.02 N • What will be its acceleration down the plane if it is sliding down the plane? • Down the plane is negative, so we have the parallel force down (-) the plane, and kinetic friction up (+) the plane: • <-23.08 N + 9.49 N> = (3.52 kg)a, a = -3.86 m/s/s = -3.9 m/s/s W -3.9 m/s/s

  29. -23.08 N -9.49 N + 3.52 kg s = .82 k = .37 -  = 42.0o F|| = 23.08 N, FFr(kinetic) = 9.49 N, FFr(static)< 21.02 N The block is given an initial velocity of 5.0 m/s up the plane. What is its acceleration as it slides up the plane? How far up does it go? As it slides up the plane, the parallel force is down (-) the plane (always), and since the velocity is up the plane, the kinetic friction is now also down (-) the plane: <-23.08 N - 9.49 N> = (3.52 kg)a, a = -9.25 m/s/s = -9.3 m/s/s v = 0, u = 5.0 m/s, a = -9.25 m/s/s, v2=u2+2as, s = 1.35 m = 1.4 m W -9.3 m/s/s, 1.4 m

  30. F = ? -23.08 N -9.49 N + 3.52 kg s = .82 k = .37 -  = 42.0o F|| = 23.08 N, FFr(kinetic) = 9.49 N, FFr(static)< 21.02 N What force in what direction will make it accelerate and slide up the plane at 6.7 m/s/s? So now we have an unknown force F probably up the plane, and the parallel force as always down (-) the plane, and friction which is down (-) the plane, because the velocity is up the plane, as well as an acceleration that is up (+) the plane: <-23.08 N - 9.49 N + F> = (3.52 kg)(+6.7 m/s/s), F = 56.15 N = 56 N W 56 N

  31. a = -2.5 m/s/s +9.49 N F = ? -23.08 N + 3.52 kg s = .82 k = .37 -  = 42.0o F|| = 23.08 N, FFr(kinetic) = 9.49 N, FFr(static)< 21.02 N What force in what direction will make it accelerate and slide down the plane at 2.5 m/s/s? So now we have an unknown force F , and the parallel force as always down (-) the plane, and friction which is up (+) the plane, because the velocity is down the plane, as well as an acceleration that is down (-) the plane: <-23.08 N + 9.49 N + F> = (3.52 kg)(-2.5 m/s/s), F = 4.797 N = +4.8 N W 4.8 N

  32. Page 4 - Statics

  33. Force equilibrium: • Step By Step: • Draw Picture • Calculate weights • Express/calculate components • Set up a <sum of all forces> = 0 equation for x and another for the y direction • Do math.

  34. First - the box weighs (54 kg)(9.8 N/kg) = 529.2N This is a downward force 529.2 N If F1 is 185 N, what is F2 ? 54 kg F1 Next, since there are no forces in the x direction, and there are no components to make, let’s set up our Y equation: F1 + F2 - 529.2 = 0, but since F1 = 185 N 185 N + F2 - 529.2 = 0, so F2 = 344.2 N F2 (Two questions like this)

  35. How to set up torque equilibrium: • Pick a point to torque about. • Express all torques: • +rF +rF+rF… = 0 • + is CW, - is ACW • r is distance from pivot • Do math 5.25 N F = ? 2.15 m 5.82 m 1. Torque about the pivot point 2 and 3. (2.15 m)(5.25 N) - (5.82 m)F = 0 F = 1.94 N (mech. adv.)

  36. Whiteboards Simple Torque Equilibrium 1 | 2 | 3 (One question like this)

  37. Find the missing distance. Torque about the pivot point. 315 N 87.5 N 12 m r = ? (315 N)(12 m) - (87.5 N)r = 0 r = 43.2 m = 43 m W 43 m

  38. Find the missing force. Torque about the pivot point. (Be careful of the way the distances are marked) F = ? 1.5 m 6.7 m 34 N (34 N)(1.5 m) - (8.2 N)F = 0 F = 6.2 N W 6.2 N

  39. Find the missing Force. Torque about the pivot point. 512 N 481 N 2.0 m 3.1 m 4.5 m -(512 N)(2.0 m) - (481 N)(5.1 m) + F(9.6 m) = 0 F = 362 N = 360 N F = ? W 360 N

  40. To do a combined force and torque problem: • Force Equilibrium: • Draw picture • Calculate weights • Draw arrows for forces. • (weights of beams act at their center of gravity) • Make components • Set up sum Fx = 0, sum Fy = 0 • Torque Equilibrium: • Pick a Pivot Point • (at location of unknown force) • Express all torques: • +rF +rF+rF… = 0 • + is CW, - is ACW • r is distance from pivot • Do Math

  41. Whiteboards: Torque and force 2a | 2b | 2c (Pretty much kinda exactly this problem) TOC

  42. Beam is 18.0 m long, person is 5.0 m from the right side. Find the two tensions in the cables at either end. T1 T2 77 kg 52 kg Step 1 - Set up your vertical force equation T1 and T2 are up, and the beam and person weights are down: Beam: -(52 kg)(9.8 N/kg) = -509.6 N (down) Person: -(77 kg)(9.8 N/kg) = -754.6 N (down) T1 + T2 -509.6 N - 754.6 N = 0 T1 + T2 -509.6 N - 754.6 N = 0

  43. Beam is 18.0 m long, person is 5.0 m from the right side. Find the two tensions in the cables at either end. T1 T2 77 kg 52 kg 18.0 m 13.0 m 9.0 m 509.6 N 754.6 N Step 2 - Let’s torque about the left side Set up your torque equation: torque = rF T1 = 0 Nm torque, (r = 0) Beam: (9.0 m)(509.6 N) = +4586.4 Nm (CW) Person: (13.0 m)(754.6 N) = +9809.8 Nm (CW) T2: T2 at 18.0 m = -(18.0 m)(T2) (ACW) Finally:+4586.4 Nm + 9809.8 Nm - (18.0 m)(T2) = 0 +4586.4 Nm + 9809.8 Nm - (18.0 m)(T2) = 0

  44. Beam is 18.0 m long, person is 5.0 m from the right side. Find the two tensions in the cables at either end. T1 T2 77 kg 52 kg Step 3 - Math time. Solve these equations for T1 and T2: +4586.4 Nm + 9809.8 Nm - (18.0 m)(T2) = 0 T1 + T2 -509.6 N - 754.6 N = 0 +4586.4 Nm + 9809.8 Nm = (18.0 m)(T2), T2 = 799.8 N T1 + 799.8 N-509.6 N - 754.6 N = 0, T1 = 464.4 N T2 = 799.8 N T1 =464.4 N

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