Rutgers University Department of Physics & Astronomy. 01:750:271 Honors Physics I Fall Lecture 19. Home Page. Title Page. Page 1 of 36.

Transcription

1 Rutgers University Department of Physics & Astronomy 01:750:271 Honors Physics I Fall 2015 Lecture 19 Page 1 of 36

2 12. Equilibrium and Elasticity How do objects behave under applied external forces? Under what conditions can they remain static or stationary? Under what conditions do objects deform and what are the effects of their deformations? Page 2 of 36

3 Equilibrium An object is in equilibrium if: The linear momentum P of its center of mass is constant. Its angular momentum about its center of mass, or about any other point, is also constant. Page 3 of 36 P, L constant

4 An object is in static equilibrium if P = 0, L = 0 No translation, no rotation. Page 4 of 36

5 Static equilibrium is: Stable if the body returns to the state of static equilibrium after having been displaced from that state by a small force. Unstable if any small force can displace the body and end the equilibrium. Page 5 of 36

6 Page 6 of 36 (a) unstable static equilibrium (c), (d) stable static equilibrium

7 Page 7 of 36

8 Conditions for equilibrium d P dt = F net P constant Fnet = 0 d L dt = τ net L constant τnet = 0 1. The vector sum of all the external forces that act on the body must be zero. Page 8 of The vector sum of all external torques that act on the body, measured about any possible point, must also be zero.

9 Conditions for equilibrium Page 9 of 36

10 A ladder of length L = 12m and mass m = 45kg leans against a frictionless wall. Its upper end is at height h = 9.3m above the pavement. The ladder s center of mass is L/3 from the lower end. A firefighter of mass M = 72kg climbs the ladder until her center of mass is L/2 from the lower end. What then are the magnitudes of the forces on the ladder from the wall and the pavement? Page 10 of 36

11 Forces on the ladder: Ladder s weight: m g = mgĵ Firefighter s weight: M g = Mgĵ Normal to wall: N w = N w î Page 11 of 36 Normal to pavement: N p = N p ĵ Static friction: f s = f s î

12 Force balance: x axis : N w f s = 0 y axis : N p Mg mg = 0 Torque balance about O: Mg(a/2) + mg(a/3) N w h = 0 Page 12 of 36 a = length of projection of ladder onto pavement a = L 2 h 2

13 N w = ag h ( M 2 + m 3 ) f s = N w N p = (M + m)g Page 13 of 36

14 Center of gravity Gravitational force acting on a rigid body: F g = i ( m i ) g = M g provided that the gravitational field is uniform i.e. g is the same for all mass elements m i Page 14 of 36

15 Torque of gravitational force acting on a rigid body: τ Fg = i ( m i ) r i g = r com (M g) provided that the gravitational field is uniform i.e. g is the same for all mass elements m i Page 15 of 36

16 Conclusions: 1. The gravitational force F g on a body effectively acts at a single point, called the center of gravity (cog) of the body. 2. If g is the same for all elements of a body, then the body s center of gravity (cog) is coincident with the body s center of mass (com). 3. If g is not the same for all mass elements COG COM Page 16 of 36

17 Page 17 of 36

18 Page 18 of 36

19 Rigid objects actually deform under applied external forces. Elastic deformations: the object returns to its original shape when the force is removed. Page 19 of 36

20 Stress: unit area force per Strain: deformation per unit length Elastic deformations: stress = modulus strain Page 20 of 36 The modulus is an intrinsic property of the material.

21 Page 21 of 36 Tensile stress: the force stretches the cylinder. Shearing stress: the deformation is perpendicular to main axis.

22 Page 22 of 36 Hydraulic stress: uniform applied force from all sides

23 Tension and compression Suppose the applied force is to the face of the object. It can stretch or compress the object. Strain: L L Young s modulus: F A = E L L Page 23 of 36

24 Small stress: deformations. elastic Yield strength: magnitude of stress causing permanent deformations. Ultimate strength: magnitude of stress tearing the object apart. Page 24 of 36

25 i-clicker same steel, same diameter, same applied force. Compared to the first rod the second rod has: A) more stress and more strain. B) the same stress and more strain. C) the same stress and less strain. D) less stress and less strain. E) the same stress and the same strain. Page 25 of 36

26 i-clicker same steel, same diameter, same applied force. Compared to the first rod the second rod has: A) more stress and more strain. B) the same stress and more strain. C) the same stress and less strain. D) less stress and less strain. E) the same stress and the same strain. Page 26 of 36

27 i-clicker same steel, same applied force. longer rod has greater diameter Compared to the first rod the second rod has: A) more stress and more strain. B) the same stress and more strain. C) the same stress and less strain. D) less stress and less strain. E) the same stress and the same strain. Page 27 of 36

28 i-clicker same steel, same applied force. longer rod has greater diameter Compared to the first rod the second rod has: A) more stress and more strain. B) the same stress and more strain. C) the same stress and less strain. D) less stress and less strain. E) the same stress and the same strain. Page 28 of 36

29 Shearing Applied force vector lies in the same plane as the face of the object. Stress: area Strain: force per unit F A x L Shear modulus: F A = G x L Page 29 of 36

30 Hydraulic Stress Force applied uniformly from all sides. Stress = pressure = force per unit area p Strain: change in volume per unit volume V V Bulk modulus: p = B V V Page 30 of 36

31 15. Oscillations Periodic or harmonic motion: periodic in time that is motion that repeats itself in time. Page 31 of 36

32 Frequency: f = number of oscillations per unit time. Units : 1 hertz = 1 Hz = 1 oscillation per second = 1 s 1 Period: T = time needed to complete one oscillation T = 1 f. Page 32 of 36

33 Projection of uniform circular motion Consider a particle P on a circular trajectory of radius x m with constant angular speed ω. The projection P of the particle on the x-axis moves according to the law: Page 33 of 36 x(t) = x m cos(ωt + φ)

34 Page 34 of 36 v P x = v P x v P x = ωx m sin(ωt + φ). a P x = a P x a P x = ω 2 x m cos(ωt + φ).

35 Simple harmonic motion (SHM) SHM is the projection of uniform circular motion on a diameter of the circular trajectory. One dimensional motion of a point particle given by x(t) = x m cos(ωt + φ). Page 35 of 36

36 x m amplitude = maximum displacement t time ω angular frequency φ phase constant or phase angle Page 36 of 36