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Understanding Dynamics: Forces & Friction in Physics

Learn about net force, weight, friction on level and inclined surfaces, statics, elevator problems, whiteboard examples, and more. Discover the forces involved in various scenarios and how they affect acceleration and motion.

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Understanding Dynamics: Forces & Friction in Physics

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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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