Unit 3 Friction & Unbalanced Forces

Transcription

1 Name: Unit 3 Friction & Unbalanced Forces Hr: Vocabulary Force: A push or a pull. When an unbalanced force is exerted on an object, the object accelerates in the direction of the force. The acceleration is proportional to the force and inversely proportional to the mass of the object. This is Newton s second law and it can be represented with an equation that says net force = (mass)(acceleration) or F net = ma The unbalanced force is called the net force, or resultant of all the forces acting on the system. The SI unit for force is the newton, with equals one kilogram meter per second squared (N = kg m/s 2 ). You can think of a newton as being about equivalent to the weight of a stick of butter. Vocabulary Friction: The force that acts to oppose the motion between two materials moving past each other. There are two types of friction that act between surfaces. They include: Static friction: The resistance force that must be overcome to start an object in motion Kinetic or sliding friction: The resistance force between two surfaces already in motion The force of sliding friction between two surfaces depends on the normal force pressing the surfaces together, and on the types of surfaces that are in contact with each other. The magnitude of this force is written as force of sliding friction = (coefficient of sliding friction)(normal force) or f = n If an object is sitting on a horizontal surface, the normal force is usually equal to the weight of the object. The symbol (pronounced mu ) is called the coefficient of sliding friction. A high coefficient of friction (in other words, a large number for ) means that the object is not likely to slide easily, while a low coefficient of friction (or a small ) is found between very slippery surfaces. Because the coefficient of sliding friction is simply a ratio of the force of sliding friction to the normal force, it has no units. Solved Examples Example 1) A 20-g sparrow flying toward a bird feeder mistakes the plane of glass in a window for an opening and slams into it with a force of 2.0 N. What is the bird s acceleration? Solution: Since the sparrow exerts 2.0 N of force on the window, the window must provide -2.0 N back in the opposite direction. Don t forget to convert g into kg before beginning. Since there are 1000 g in a kg, 20 g = 0.02 kg m = 0.02 kg Unknown: a =? F = -2.0 N Original equation: F net = ma Solve: a = F m = 2.0 N 0.02 kg = 100 m/s2 (about 10 g s!) Therefore, the bird experiences a very rapid negative acceleration, as the window brings the bird to a sudden stop. Ouch!

2 Example 2) A 30.0 g arrow is shot by William Tell through an 8.00 cm thick apple sitting on top of his son s head. If the arrow enters the apple at 30.0 m/s and emerges at 25.0 m/s in the same direction, with what force has the apple resisted the arrow? Solution: First, convert g to kg (there are 1000 g in a kg) and cm to m (there are 100 cm in a m) g = kg 8.00 cm = m Next, find the acceleration of the arrow before finding the force. v i = 30.0 m/s Unknown: a =? V f = 25.0 m/s 2 2 Original equation: v f - v i = 2ad d = m Solve: a = v f 2 - v i 2 2d = (25.0 m/s)2 - (30.0 m/s) 2 2 ( m) = 625 m2 /s m 2 /s m = m/s 2 The negative sign before the answer implies that the apple was causing the arrow to slow down. Now solve for the force exerted by the apple. m = kg Unknown: F net =? a = m/s 2 Original equation: F net = ma Solve: F net = ma = ( kg)(-1720 m/s 2 ) = N This is the force that the apple exerts on the arrow. It is negative because its direction is opposite to the arrow s direction of motion. Example 3) While redecorating her apartment, Kitty slowly pushes an 82-kg china cabinet across the wooden dining room floor, which resists the motion with a force of friction of 320 N. What is the coefficient of sliding friction between the china cabinet and the floor? Solution: The normal force is equivalent to the weight of the china cabinet because the cabinet is sitting on a horizontal surface. m = 82 kg g = 9.8 m/s 2 Unknown: W =? Original Equation: W = mg Solve: W = mg = (82 kg)(9.8 m/s 2 ) = 804 N, so n is also 804 N. n = 804 N f = 320 N Unknown: =? Original Equation: f = n Solve: = f = 320 N = 0.39 n 820 N Remember, has no units!

3 n f Example 4) At Sea World, a 900.-kg polar bear slides down a wet slide inclined at an angle of 25.0 to the horizontal. The coefficient of friction between the bear and the slide is What frictional force impedes the bear s motion down the slide? Solution: In this example, the polar bear is inclined at an angle to the horizontal, so the normal force is equal to W, instead of the weight. Remember, the normal force, n, always acts perpendicular to the point of contact between two objects. First, solve for weight. m = 900. kg g = 9.8 m/s 2 Unknown: W =? Original Equation: W = mg Solve: W = (900. kg) (9.8 m/s 2 ) = 8,820 N Next, knowing that the magnitude of the normal force is equal to W, solve for W using trig. W = 8,820 N θ = 25.0 o Unknown: W =? Original equation: Solve: cosθ= W Finally, solve for the friction force. W W = W cosθ = (8,820 N) cos(25.0 o ) = 7,994 N n = 7994 N = Unknown: f =? Original Equation: f = n Solve: f = n = (0.050)(7994 N) = 400 N W W W ll

4 Exercises Show ALL work, including FORCE DIAGRAMS FOR EACH PROBLEM, formulas, plugged-in numbers, units, and circled answers. 1) Claudia stubs her toe on the coffee table with a force of 100 N. a. What is the acceleration of Claudia s 1.80 kg foot? b. What is the acceleration of the table if it has a mass of 20.0 kg? (Ignore any frictional effects.) c. Why would Claudia s toe hurt less if the table had less mass? 2) A 500 kg boat accelerates from rest over a distance of 270 m in 12 seconds. a. What is the boat s acceleration? b. What is the net force acting on the boat?

5 3) A space shuttle has a total mass of 2.0 x 10 6 kg. At liftoff, the engines exert a thrust force of 3.5 x 10 7 N. a. What is the shuttle s weight? b. What is the shuttle s acceleration at liftoff? c. If the acceleration averages 13 m/s 2 over the first 10 minutes, what final velocity does the shuttle reach?

6 4) Unbeknownst to the students, every time the school floors are waxed, Mr. Wright, the principal, likes to slide down the hallway in his socks. Mr. Wright weighs 850. N and the coefficient of sliding friction between his socks and the floor is What is the force of friction that opposes Mr. Wright s motion down the hall? 5) Howard, the soda jerk at Bea s diner, slides a 0.60-kg root beer from the end of the counter to a thirsty customer. A force of friction of 1.2 N brings the drink to a stop right in front of the customer. a. What is the coefficient of sliding friction between the glass and the counter? b. If the glass stops in 1.35 seconds, with what initial velocity was the root beer slid?

7 6) Rather than taking the stairs, Martin gets from the second floor of his house to the first floor by sliding down the banister that is inclined at an angle of 30.0 to the horizontal. a. If Martin has a mass of 45 kg and the coefficient of sliding friction between Martin and the banister is 0.20, what is the force of friction impeding Martin s motion down the banister? b. If the banister is made steeper (inclined at a larger angle), will this have any effect on the force of friction? If so, what?

8 7) Carmen sleds down a 28 o hill. The combined mass of Carmen and her sled is 47 kg. a. What is the combined weight of Carmen and the sled? b. What is the normal force acting on the sled by the hill? c. If the coefficient of friction between the sled and the snow is 0.18, what is the friction force acting against the sled s motion? d. What is Carmen s acceleration as she sleds down the hill?