Physics 2210 Fall smartphysics 05 (continued) Newton s Universal Gravitation 06 Friction Forces 09/21/2015

Transcription

1 Physics 2210 Fall 2015 smartphysics 05 (continued) Newton s Universal Gravitation 06 Friction Forces 09/21/2015

2 Exam 1 Statistics Average: / 30 Std. Dev.: 6.85 Median: 22

3 Reminder about Interactive Examples Some homework problems are designated Interactive Examples. By clicking the Help button below the answer box, OR the IE Outline tab, you can get a step-by-step tutorial (including multiple-choice conceptual questions) for the problem.

4 Example 5.4: orbit of Earth (3/3) Earth has mass m = kg, and orbits the Sun (mass M = kg) in approximately a circular orbit, at a distance of r = km. Find the orbital period of the Earth in days. Universal Gravitational Constant G = Nm 2 kg 2 Magnitude of the force F G = GGG r 2 F G r M a c m T 2 = 4π2 GG r3 Note the period does not depend on the mass of the planet, just the mass of the Sun and the distance to the planet. (%i1) G: 6.67e-11; (%o1) 6.67E-11 (%i2) M: 1.99e30; (%o2) 1.99E+30 (%i3) /* convert r to meters by multiplying 1=1000m/km */ r: 1.50e8*1000; (%o3) 1.5E+11 (%i4) T2: 4*(%pi)^2/G/M*r^3, numer; (%o4) E+15 (%i5) /* this is the square of the period */ T: sqrt(t2); (%o5) E+7 (%i6) /* convert to days by muitplying 1=1day/24h and then by 1=1h/3600s */ T/24/3600; (%o6)

5 Unit 06

6 Unit 06

7 Video excerpt from Pulling with a (approx.) constant force Resulting Motion???

8 Digitized every 15 frames (30 f/s) using tpsdig2 Length units: pixels

9 Video Analysis Process and Tools Convert video (Cornell videos are in.mov format) using Any Video Converter Free Digitize landmarks from frames using tpsdig2 Linked from Edit/view tps files using notepad++ Cut and paste to Excel (data text-to-columns) Insert chart (xy plot) + trend line

10 Data Analysis t (s) x (pix) x = vv + x 0 v = 4.84t v 2s = 34.4 pix/s Good Fit to Constant Velocity Even Better Fit to A Small Constant Acceleration x = 1 2 at2 + v 0 t + x 0

11 Simple Model of kinetic friction Kinetic friction force occurs when one surface slides against another force acts parallel to the surface to reduce the relative speed. Once relative motion stops kinetic friction force stops acting (static friction may take over) Kinetic Friction force is proportional to: Pressure from normal force between surfaces (pressure is force per unit area perpendicular to the surfaces) Area of contact Area x pressure = normal force Result (magnitude): f k = μ k N Coefficient of Kinetic Friction (no units: ratio of force magnitudes)

12 Same surface characteristics -> same μ k, + same normal force -> same f k From

13 Run 1: Digitized Landmarks from From

14 Run 2: Digitized Landmarks from From

15 t (s) x (pix) ½ at 2 Same Acceleration! Run 1 1 st run had higher initial speed v 0 t (s) x (pix) Run 2

16 Poll A block slides on a table pulled by a string attached to a hanging weight. In case 1 the block slides without friction and in case 2 there is kinetic friction between the sliding block and the table. In which case is the tension in the string the biggest? A. Same B. Case 1 C. Case 2

17 A block with mass m 1 = 8.8 kg is on an incline with an angle θ = 39 with respect to the horizontal. (a) When there is no friction, what is the magnitude of the acceleration of the block? Example 6.1 (1/2) (%i1) /* For this full problem we take +x to be down the ramp and +y to be perpendicularly away from the ramp So in the y-direction we have */ Fy: N - m*g*cos(theta); (%o1) N - g m cos(theta) (%i2) /* and so N = m*g*cos(theta) since ax=fx/m = 0 NOw in the x direction we have (without friction) only the x-component of the gravitational force (weight) */ Fx: m*g*sin(theta); (%o2) (%i3) ax: Fx/m; (%o3) (%i4) deg39: 39*%pi/180, numer; g m sin(theta) g sin(theta) (%o4) (%i5) ax, m=8.8, g=9.81, theta=deg39; (%o5) Answer (a) 6.17 m/s^2

18 m 1 = 8.8 kg, incline with an angle θ = 39 Coefficient of kinetic friction μ k = 0.37 (b) When there is kinetic friction, what is the magnitude of the acceleration of the block? Example 6.1 (2/2) (%i6) /* with friction, we need to know the normal force */ soln2: solve(fy/m=0, N); (%o6) (%i7) N2: rhs(soln2[1]); (%o7) [N = g m cos(theta)] g m cos(theta) (%i8) /* kinetic friction force magnitude (in -x direction) */ Ffk: mu_k*n2; (%o8) (%i9) Fx: m*g*sin(theta)-ffk; (%o9) (%i13) ax: Fx/m; g m mu_k cos(theta) g m sin(theta) - g m mu_k cos(theta) g m sin(theta) - g m mu_k cos(theta) (%o13) (%i14) ax: expand(ax); (%o14) m g sin(theta) - g mu_k cos(theta) (%i15) ax, g=9.81, theta=deg39, mu_k=0.37; (%o15) Answer: (b) with kinetic friction the acceleration is 3.35 m/s^2

19 Video excerpt from

20 Simple Model of static friction Static friction force develop in response to other forces: and the friction develops as much force as required to PREVENT the two surfaces involved from sliding against one another There is however a limit to the magnitude of the friction force that can be developed The Maximum Static Friction Force is proportional to: Pressure from normal force between surfaces (pressure is force per unit area perpendicular to the surfaces) Area of contact Area x pressure = normal force Result (magnitude): f s μ s N Coefficient Static Friction (no units: ratio of force magnitudes) Once relative motion/sliding starts static friction force stops acting (kinetic friction takes over).

21 Poll Which of the following diagrams best describes the static friction force acting on the box?

22 Mass m 1 = 8.8 kg is on an incline with θ = 39 to the horizontal. the coefficients of static friction is μ s = To keep the mass from accelerating, a spring is attached. What is the minimum spring constant of the spring to keep the block from sliding if it extends x = 0.12 m from its unstretched length. Example 6.2 (1/2) (%i19) /* with the spring in the minimum spring constant case, the friction force still points in the -x direction, and the spring force Fs also points in the -x direction */ Fs: k*dl; (%o19) Dl k (%i21) /* static friction force maxed out... in this case */ Ffs_max: mu_s*n2; (%o21) g m mu_s cos(theta) (%i22) Fx: m*g*sin(theta) - Ffs_max - Fs; (%o22) g m sin(theta) - g m mu_s cos(theta) - Dl k (%i26) /* now we set the acceleration in the x direction: ax = Fx/m to zero and solve for the spring constant */ soln3: solve(fx/m=0, k); g m sin(theta) - g m mu_s cos(theta) (%o26) [k = ] Dl... continued

23 m 1 = 8.8 kg, θ = 39 to the horizontal, μ s = To keep the mass from accelerating, a spring is attached. What is the minimum spring constant of the spring to keep the block from sliding if it extends x = 0.12 m from its unstretched length. Example 6.2 (1/2) (%i27) k_min: rhs(soln3[1]); g m sin(theta) - g m mu_s cos(theta) (%o27) Dl (%i29) k_min, g=9.81, m=8.8, g=9.81, mu_s=0.407, theta=deg39, Dl=0.12; (%o29) Answer : minimum spring constant is 225 N/m

24 Video excerpt from Static friction force increases with (increasing) applied force until the maximum static friction force is exceeded. After that kinetic friction takes over

25 Mass m 1 = 8.8 kg is on an incline with θ = 39 to the horizontal. the staic coefficients of friction is μ s = A second mass m 2 = 15.8 kg is attached to the first block by a cord. m 2 is made of a different material and has a greater coefficient of static friction. What minimum value for the coefficient of static friction is needed between the new block and the plane to keep the system from accelerating? Example 6.3 (1/3) (%i1) /* for part (d) we first note that the normal forces in each case is given by m*g*cos(theta) */ deg39: 39*%pi/180, numer; (%o1) (%i2) /* Forces in the x direction on block 1, friction force is maxed out */ N1: m1*g*cos(theta); (%o2) g m1 cos(theta) (%i3) Ffs1: mu_s1*n1; (%o3) g m1 mu_s1 cos(theta) (%i4) Fx1: m1*g*sin(theta) - Ffs1 - T; (%o4) - T + g m1 sin(theta) - g m1 mu_s1 cos(theta) (%i5) /* we set ax1=fx1/m1=0 to solve for the tension force T */ soln4: solve(fx1/m1=0, T); (%o5) [T = g m1 sin(theta) - g m1 mu_s1 cos(theta)]... continued

26 m 1 = 8.8 kg, θ = 39 μ s = A second mass m 2 = 15.8 kg is attached to the first block by a cord. m 2 is made of a different material and has a greater coefficient of static friction. What minimum value for the coefficient of static friction is needed between the new block and the plane to keep the system from accelerating? Example 6.3 (2/3) (%i6) T: rhs(soln4[1]); (%o6) g m1 sin(theta) - g m1 mu_s1 cos(theta) (%i7) /* note T is now in terms of known quantities, now we look at forces on block 2 */ N2: m2*g*cos(theta); (%o7) g m2 cos(theta) (%i8) Ffs2: mu_s2*n2; (%o8) g m2 mu_s2 cos(theta) (%i9) Fx2: T + m2*g*sin(theta) - Ffs2; (%o9) g m2 sin(theta) + g m1 sin(theta) - g m2 mu_s2 cos(theta) - g m1 mu_s1 cos(theta) (%i10) /* set ax2-fx2/m2 to zero and solve for mu_s2 */ soln5: solve(fx2/m2=0, mu_s2); (m2 + m1) sin(theta) - m1 mu_s1 cos(theta) (%o10) [mu_s2 = ] m2 cos(theta)... continued

27 m 1 = 8.8 kg, θ = 39 μ s = A second mass m 2 = 15.8 kg is attached to the first block by a cord. m 2 is made of a different material and has a greater coefficient of static friction. What minimum value for the coefficient of static friction is needed between the new block and the plane to keep the system from accelerating? (%i11) mu_s2: rhs(soln5[1]); (m2 + m1) sin(theta) - m1 mu_s1 cos(theta) (%o11) m2 cos(theta) (%i12) mu_s2, m2=15.8, m1=8.8, theta=deg39, mu_s1=0.407; (%o12) Answer: Minimum coefficient of static friction on m2 is 1.03 Example 6.3 (3/3)