Friction and Motion. Prof. Paul Eugenio 13 Sep Friction (cont.) Motion: kinetics and dynamics Vertical jump Energy conservation

Transcription

1 Friction and Motion Friction (cont.) Motion: kinetics and dynamics Vertical jump Energy conservation Ukulele means jumping flea Prof. Paul Eugenio 13 Sep 2018 Lecture 5

2 Reactive, Normal, and Friction Forces force diagram FN FR W Ff Ff FR FN Static Mass on an Inclined Surface W

3 Reactive, Normal, and Friction Forces force diagram FN Not STATIC! Fnet ΣF = Fnet = ma a = Fnet/m W Fnet a = W sin(θ)/m = g sin(θ) θ W FN θ Mass on a Frictionless Inclined Surface

4 Forces on incline with static friction θ θ Ff Static [v=0] µ s m g cos (θ) F =0 F f F g =0 µ s = tan(θs ) x F f =m g sin (θ) Slidding starts at θ=θs F f = µ s m g cos(θs ) µ s m g cos(θs ) = m g sin(θs ) valid for θ={0 o... θs } Independent of m and g!

5 Sliding versus toppling: sailor on listing deck l F = b cos (ϕ) R cm b W b cos(ϕ) Short and Squat Slides Tall and Thin Topples R L ϕ F L FR FR We found in HW1 that the critical angle for toppling is given by L = 0, or tanϕt = b/2h From the previous slide we saw that the critical angle for sliding is given by tanϕs = µs So, for toppling to happen at a smaller angle than sliding, we must have: b/2h < µs

6 Friction in the hip joint The reaction force from the socket is approximately normal to the ball surface. For each step the ball rotates about 60o ~ 1 radian (either forward or backward), so if the ball radius is about 3 cm, the surfaces slide about s = 3cm relative to each other. The friction force, Ff = FR µk is parallel to the ball surface. The work done against the friction is W = Ff s. Consider a person of weight mg = 600 N, and assume that there is no lubrication, so µk = 0.3. Then the work done for each step is W = 13 J. At a leisurely pace of one step per second, this is a power of 13 Watts the power of a small incandescent light bulb! With lubrication, µk = 0.003, for W = 0.13 J and a corresponding power of 0.13 Watts. This amount of heat can easily be conducted away. FR = 2.4mg Socket (Ball)

7 Motion

8 Forms of Motion Most human movement is general motion: A complex combination of Translational (linear) and Rotational (angular) motion

9 Linear Motion Motion along a line that may be straight or curved, with all parts of the body moving in the same direction at the same speed, also known as translational motion Rectilinear along a straight line Curvilinear along a curved line

10 Rectilinear motion Speed v = Δx / Δt (m/s) Acceleration a = Δv / Δt (m/s2)

11 Kinematic equations for constant acceleration: v = vo + at x = vot + ½ at2 Eliminating t from these two equations yields: v2 = vo2 + 2ax

12 Dynamics: NEWTON S LAWS I Newton s First Law (N1) Zero Force, Zero Acceleration II Newton s Second Law (N2) F = ma III Newton s Third Law (N3) F12 = - F21 Force-on-M-from-E = - Force-on-E-from-M

13 Example: vertical jump Kinematic treatment lift off flight thrust Feet on ground Feet in air Forces on: Forces on: Jumper Earth Jumper Earth

14 Example: vertical jump (cont.) While feet still on the ground, the acceleration is, by N2: a1 = (F-mg)/m = F/m g This yields the take-off velocity: vto2 = v02 + 2ac = 2(F/m g)c since v0=0

15 Example: vertical jump (cont.) After the jumper leaves the ground, the acceleration is a2 = g At the top of the trajectory, y=h, v=0, so 0 = vto2 2gH 2(F/m g)c 2gH = 0 H = [F/(mg) 1]c Assuming F = αmg, we get H = (α - 1)c If α < 1, the jumper will collapse to the ground.

16 WORK Work = displacement * parallel force component = force * parallel displacement component W = d*f *cos θ Units: N*m = J (Joule) F F θ θ d For θ = 0, W = Fd

17 ENERGY Energy is ability to do work. Unit: J Different forms : mechanical, thermal, electrical, etc. Conservation during transformations MECHANICAL ENERGY KINETIC ENERGY POTENTIAL ENERGY

18 KINETIC ENERGY Due to motion EK = ½ mv2 (linear) EK = ½ Iω2 (rotational) POTENTIAL ENERGY Stored energy EP = mgh (gravity) EP = ½ kx2 (spring)

19 EP Example: vertical jump by energy conservation EK ETot = EP + EK mg(h+c) mgc 0 0 ½mv12 0

20 vertical jump: Energy Conservation Since the kinetic energy is zero, both before the jump and at the top of the trajectory: Work done by F = change in potential energy Fc = mg(c+h) or: H = [F/(mg) 1]c Same as before!

21 vertical jump: Energy Conservation of the Leg Thrust While the feet are on the ground, the force F does a work: W = Fc Energy while crouched before jump: E0 = EK0 + EP0 = 0 Energy at point of leaving ground: E1 = EK1 + EP1 = ½mv12 + mgc The energy change equals the work done: E1 E 0 = W take off velocity: v12 = 2(F/m g)c

22 vertical jump: Energy Conservation of the Flight At the top of the trajectory the velocity is 0, so the energy is : E2 = EK2 + EP2 = 0 + mg(h +c) After the feet leave the ground, energy is conserved, so E2 = E 1 mg(h+c) = ½ m v12 + mgc mgh = ½mv12 H = v12 / 2g = [F/(mg) 1]c As before

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24 Block on Wall with Friction A wooden block of mass 1 kg is held against a wall with a horizontal force. If the coefficient of static friction is µs = 0.2, what force is required to keep the block static? m F

25 Block on Wall with Friction A wooden block of mass 1 kg is held against a wall with a horizontal force. If the coefficient of static friction is µs = 0.2, what force is required to keep the block static? Ff FN F m F W ΣFx = 0 F = FN known m = 1 kg W = 9.8 N µs = 0.2 Ff µ s F N ΣFy = 0 Ff = W W µs F W µ s F 9.8 N =49 N F 0.2 The force must be 49N or greater

26 Block on Wall with Friction A wooden block of mass 1 kg is held against a wall with a force applied at an angle θ. If the coefficient of static friction is µs = 0.2, what force is required to keep the box from moving down? m θ F