Physics 2210 Fall 2015

smartPhysics 05 (continued) Newton’s Universal Gravitation

06 Friction Forces 09/21/2015

Exam 1 Statistics Average: 21.08 / 30 Std. Dev.: 6.85 Median: 22

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.

Example 5.4: orbit of Earth (3/3) Earth has mass m = 5.97×1024 kg, and orbits the Sun (mass M = 1.99×1030 kg) in approximately a circular orbit, at a distance of r = 1.50 ×108 km. Find the orbital period of the Earth in days.

Magnitude of the force

𝐹𝐺 = 𝐺𝐺𝐺𝑟2

FG

M

m 𝑟

Universal Gravitational Constant G = 6.67×10−11 Nm2kg−2

𝑇2 =4𝜋2

𝐺𝐺𝑟3

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) 1.003817132248245E+15

(%i5) /* this is the square of the period */

T: sqrt(T2);

(%o5) 3.1683E+7

(%i6) /* convert to days by muitplying

1=1day/24h and then by 1=1h/3600s */

T/24/3600;

(%o6) 366.702236

ac

Unit 06

Unit 06

http://www.physics.utah.edu/~jui/2210_s2015/friction/kinetic01.avi Video excerpt from

http://courses2.cit.cornell.edu/physicsdemos/secondary.php?pfID=41

Pulling with a (approx.) constant force Resulting Motion???

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

http://www.physics.utah.edu/~jui/2210_s2015/friction/kinetic01.avi

Video Analysis Process and Tools • Convert video (Cornell videos are in .mov format)

using “Any Video Converter Free” http://www.any-video-converter.com/products/for_video_free/

• Digitize “landmarks” from frames using tpsdig2 http://life.bio.sunysb.edu/morph/morphmet/tpsdig2w32.exe Linked from http://life.bio.sunysb.edu/morph/soft-dataacq.html

• Edit/view “tps” files using notepad++ http://notepad-plus-plus.org/

• Cut and paste to Excel (datatext-to-columns) – Insert chart (xy plot) + trend line

Even Better Fit to A Small Constant Acceleration

𝑥 = 12� 𝑎𝑡2 + 𝑣0𝑡 + 𝑥0

Good Fit to Constant Velocity

𝑥 = 𝑣𝑡 + 𝑥0

t (s) x (pix) 0.0 75 0.5 87 1.0 103 1.5 119 2.0 135 2.5 150 3.0 170 3.5 192 4.0 213

Data Analysis

→ 𝑣 = 4.84𝑡 + 24.7

→ 𝑣 2s = 34.4 pix/s

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): 𝑓𝑘 = 𝜇𝑘𝑁

Coefficient of Kinetic Friction (no units: ratio of force magnitudes)

Same surface characteristics -> same 𝜇𝑘, + same normal force -> same 𝑓𝑘 From http://courses2.cit.cornell.edu/physicsdemos/secondary.php?pfID=92

Run 1: Digitized Landmarks from From http://courses2.cit.cornell.edu/physicsdemos/secondary.php?pfID=92

Run 2: Digitized Landmarks from From http://courses2.cit.cornell.edu/physicsdemos/secondary.php?pfID=92

t (s) x (pix) 0.000 581 0.067 500 0.133 428 0.200 344 0.267 297 0.333 245 0.400 219 0.467 204 0.500 198

t (s) x (pix) 0.000 573 0.067 506 0.133 447 0.200 380 0.267 345 0.333 314 0.400 302 0.467 299

Run 1

Run 2

1st run had higher initial speed |v0|

Same Acceleration ! ½ at2

Poll 09-21-01

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

Example 6.1 (1/2) A block with mass m1 = 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?

(%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) g m sin(theta)

(%i3) ax: Fx/m;

(%o3) g sin(theta)

(%i4) deg39: 39*%pi/180, numer;

(%o4) 0.6806784082777885

(%i5) ax, m=8.8, g=9.81, theta=deg39;

(%o5) 6.173633036198905

Answer (a) 6.17 m/s^2

m1 = 8.8 kg, incline with an angle 𝜃 = 39° Coefficient of kinetic friction μk = 0.37 (b) When there is kinetic friction, what is the magnitude of the acceleration of the block? (%i6) /* with friction, we need to

know the normal force */

soln2: solve(Fy/m=0, N);

(%o6) [N = g m cos(theta)]

(%i7) N2: rhs(soln2[1]);

(%o7) g m cos(theta)

(%i8) /* kinetic friction force magnitude (in -x direction) */

Ffk: mu_k*N2;

(%o8) g m mu_k cos(theta)

(%i9) Fx: m*g*sin(theta)-Ffk;

(%o9) g m sin(theta) - g m mu_k cos(theta)

(%i13) ax: Fx/m;

g m sin(theta) - g m mu_k cos(theta)

(%o13) ------------------------------------

m

(%i14) ax: expand(ax);

(%o14) g sin(theta) - g mu_k cos(theta)

(%i15) ax, g=9.81, theta=deg39, mu_k=0.37;

(%o15) 3.352826339898537

Answer: (b) with kinetic friction the acceleration is 3.35 m/s^2

Example 6.1 (2/2)

Video excerpt from http://courses2.cit.cornell.edu/physicsdemos/secondary.php?pfID=41

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): 𝑓𝑠 ≤ 𝜇𝑠𝑁

• Once relative motion/sliding starts static friction force stops acting (kinetic friction takes over).

Coefficient Static Friction (no units: ratio of force magnitudes)

Poll 09-21-02

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

Mass m1 = 8.8 kg is on an incline with θ = 39° to the horizontal. the coefficients of static friction is μs = 0.407. 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. (%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

Example 6.2 (1/2)

m1 = 8.8 kg , θ = 39° to the horizontal, μs = 0.407 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.

(%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) 225.1880158196901

Answer : minimum spring constant is 225 N/m

Example 6.2 (1/2)

Video excerpt from http://courses2.cit.cornell.edu/physicsdemos/secondary.php?pfID=41

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

Mass m1 = 8.8 kg is on an incline with θ = 39° to the horizontal. the staic coefficients of friction is μs = 0.407. A second mass m2 = 15.8 kg is attached to the first block by a cord. m2 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? (%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) 0.6806784082777885

(%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

Example 6.3 (1/3)

m1 = 8.8 kg , θ = 39° μs = 0.407. A second mass m2 = 15.8 kg is attached to the first block by a cord. m2 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? (%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

Example 6.3 (2/3)

m1 = 8.8 kg , θ = 39° μs = 0.407. A second mass m2 = 15.8 kg is attached to the first block by a cord. m2 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) 1.034119444088428

Answer: Minimum coefficient of static friction on m2 is 1.03

Example 6.3 (3/3)