Forces on an inclined plane. And a little friction too

Transcription

1 Forces on an inclined plane And a little friction too

2 The Takeaway } You should be able to: } Identify the forces acting on an object } Forces on non-horizontal surfaces } Including Friction } State Newton s second law of motion. } Solve problems involving Newton s second law.

3 Inclined Plane Ø A crate of chop suey of mass m is set on a ramp with angle of inclination θ. The weight vector is straight down. Ø The parallel component (blue) acts parallel to the ramp and is the component of the weight pulling the crate down the ramp. Ø The perpendicular component (red) acts perpendicular to the ramp and is the component of the weight that tries to crush the ramp. m parallel component perpendicular component mg θ Note: red + blue = black

4 Geometry of Inclined Planes Ø Since The diagram blue and contains red are two perpendicular, right triangles. the angle α is between the angle red and between black must black also and be blue. θ. α + θ = 90 since they are both angles Ø of Imagine the right the triangle parallel on component the right. sliding down (dotted blue) to form a right triangle. Being opposite θ, we use sine. Red is adjacent to θ, so we use cosine. mg cos θ m α θ mg mg sin θ θ mg sin θ

5 Inclined Plane - Pythagorean Theorem m mg sinθ θ mg cosθ Let s show that the Pythagorean theorem holds for components on the inclined plane: mg (mg) 2 (sinθ + cosθ ) 2 (mg sinθ ) 2 + (mg cosθ ) 2 = (mg) 2 (sin 2 θ + cos 2 θ ) = (mg) 2 (1) = (mg) 2

6 Inclined Plane: Normal Force N = mg cosθ } The normal force is perpendicular to the contact surface. As long as the ramp itself isn t accelerating and no other forces are lifting the box off the ramp or pushing it into the ramp, N matches the perpendicular component of the weight. This must be the case, otherwise the box would be accelerating in the direction of red or green m mg sinθ θ mg cosθ mg

7 Net Force on a Frictionless Inclined Plane N = mg cosθ m mg sinθ With no friction, F net = mg + N = mg cosθ + mg sinθ + N = mg sinθ (mg cosθ + N = 0 since their magnitudes are equal but they re in opposite directions. That is, the perpendicular component of the weight and the normal cancel out.) mg cosθ mg θ Therefore, the net force is the parallel force in this case.

8 Net Force on a Frictionless Inclined Plane } Example: Determine the net force with no friction } Remember that N and mg cosθ cancel each other out } Only force leftover is mg sinθ N = mg cosθ 5 kg mg sinθ F net = mg sinθ = (5 kg)(9.8 ms -2 )(sin30 ) = 24.5 N 30 mg cosθ mg

9 Acceleration on a Frictionless Inclined Plane } Here F net = mg sinθ = m a. } So, a = g sinθ. Since sinθ has no units, a has the same units as g, as they should. Both the net force and the acceleration are down the ramp. 5 kg mg sinθ a = g sinθ = (9.8 ms -2 )(sin 30 ) a = 4.9 ms -2 30

10 Solid Friction Recall that friction acts opposite to the intended direction of motion, and parallel to the contact surface. Suppose we begin to pull a crate to the right, with gradually increasing force. We plot the applied force, and the friction force, as functions of time: Force tension friction static friction dynamic friction static Time dynamic f T

11 Solid friction During the static phase, the static friction force F s exactly matches the applied (tension) force. F s,max F s increases linearly until it reaches a maximum value F s,max. The friction force then almost instantaneously decreases to a constant value F d, called the dynamic friction force. Take note of the following general properties of the friction force: Force static Time dynamic 0 F s F s,max F d < F s,max F d = a constant F d tension friction

12 Solid friction So, what exactly causes friction? People in the manufacturing sector who work with metals know that the more you smoothen and polish two metal surfaces, the more strongly they stick together if brought in contact. In fact, if suitably polished in a vacuum, they will stick so hard that they cannot be separated. We say that the two pieces of metal have been cold-welded.

13 Solid friction At the atomic level, when two surfaces come into contact, small peaks on one surface cold weld with small peaks on the other surface. surface 1 Applying the initial sideways force, all of the cold welds oppose the motion. If the force is sufficiently large, the cold surface 11 welds break, and new peaks contact each other and cold weld. surface 2 cold welds If the surfaces remain in relative sliding motion, fewer welds have a chance to form. We define the unitless constant, called the coefficient of friction µ, which depends on the composition of the two surfaces, as the ratio of F f / R.

14 Describing solid friction by coefficients of friction Since there are two types of friction, static and dynamic, every pair of materials will have two coefficients of friction, µ s and µ d (µ k ) In addition to the "roughness" or "smoothness" of the materials, the friction force depends, not surprisingly, on the normal force N. The harder the two surfaces are squished together (this is what the normal force measures) the more cold welds can form. Here are the relationships between the friction force F f, the coefficients of friction µ, and the normal force N: F f µ s N static F f = µ d N dynamic friction

15 Describing solid friction by coefficients of friction EXAMPLE: A piece of wood with a coin on it is raised on one end until the coin just begins to slip. The angle the wood makes with the horizontal is θ = 15. What is the coefficient of static friction? θ = 15 F y = 0 F x = 0 R mg cos 15 = 0 R = mg cos 15 F f mg sin 15 = 0 F f = mg sin 15 F f = µ s N µ mg sin 15 = tan 15 s = mg sin 15 = µ s mg cos 15 mg cos 15 = Thus the coefficient of static friction between the metal of the coin and the wood of the plank is mg y R F f FBD, coin x 15

16 Incline with friction at equilibrium (cont.) } f s µ s N = µ s mg cosθ } Also, f s = mg sinθ (only because we have equilibrium) } So, mg sinθ µ s mg cos θ } Since the mg s cancel and tan θ = sinθ / cosθ, we have mg sinθ N = mg cosθ m f s = mg sinθ µ s tanθ θ mg cosθ mg

17 Friction } F f µ s N static } F f = µ d N dynamic (kinetic) F f = µ k N } For an inclined plane } F f = µ k N but because the N = mg cosθ so: } F f = µ k mg cosθ

18 Friction } You push a giant barrel o monkeys setting on a table with a constant force of 63 N. If µ k = 0.35 and µ s = 0.58, what is its acceleration? answer: Never, since this force won t even budge it! 63 < 0.58 (14.7) (9.8) 83.6 N Barrel o Monkeys 14.7 kg

19 Friction } Same as the last problem except with a bigger F A : You push the barrel o monkeys with a constant force of 281 N. µ k = 0.35 and µ s = 0.58, same as before. What is the acceleration? } f s, max = 0.58 (14.7) (9.8) 83.6 N } F A = 281N > f s, max. Thus, it budges this time. } Forget f s and calculate f k : f k = 0.35 (14.7) (9.8) = N

20 Friction Free body diagram while sliding: f k N F A mg } F net = F A f k = = N } a = F net / m = / 14.7 = m s -2 (16)