Introduction to Mechanics Friction Examples Friction Springs

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Introduction to Mechanics Friction Examples Friction Springs Lana Sheridan De Anza College Mar 7, 2018

Last time kinetic and static friction friction examples

Overview one more friction example springs

Friction Question #5, page Hopping into your Porsche, you floor it and accelerate at 12 m/s 2 without spinning the tires. Determine the minimum coefficient of static friction between the tires and the road needed to make this possible. 0 Walker, Physics.

Friction Question #5, page Hopping into your Porsche, you floor it and accelerate at 12 m/s 2 without spinning the tires. Determine the minimum coefficient of static friction between the tires and the road needed to make this possible. Sketch a free body diagram for the car. (What force causes the car s forward acceleration?) 0 Walker, Physics.

Friction Question #5, page Hopping into your Porsche, you floor it and accelerate at 12 m/s 2 without spinning the tires. Determine the minimum coefficient of static friction between the tires and the road needed to make this possible. Sketch a free body diagram for the car. (What force causes the car s forward acceleration?) Hypothesis: coefficients of friction are usually between 0 and 1. Car tires are designed not to slip on asphalt. µ s should be high, but we are looking for the minimum it could be. Guess: 0.5. 0 Walker, Physics.

Friction Question Hopping into your Porsche, you floor it and accelerate at 12 m/s 2 without spinning the tires. Determine the minimum coefficient of static friction between the tires and the road needed to make this possible. Strategy: Newton s 2nd law. F net = ma F net = f s. If we want to find the minimum coefficient of static friction, assume that we are getting the max possible force from that coefficient: f s = f s,max = µ s n. µ s mg = ma µ s = a g µ s = 12 m/s2 9.81 m/s 2 = 1.2

Friction Question Hopping into your Porsche, you floor it and accelerate at 12 m/s 2 without spinning the tires. Determine the minimum coefficient of static friction between the tires and the road needed to make this possible. µ s = 1.2 Reasonable?: Woah! This is not only much bigger than my guess, it is bigger than 1! This would mean that it requires less force to pick up the entire Porsche and move it to one side than it does to push it along the ground starting from rest.

Friction Question Hopping into your Porsche, you floor it and accelerate at 12 m/s 2 without spinning the tires. Determine the minimum coefficient of static friction between the tires and the road needed to make this possible. µ s = 1.2 Reasonable?: Woah! This is not only much bigger than my guess, it is bigger than 1! This would mean that it requires less force to pick up the entire Porsche and move it to one side than it does to push it along the ground starting from rest. Research Apparently, yes! This is the sort of number you can get for high-performance racing tires. Cool.

Some types of forces Elastic Forces Springs exert forces as they are being compressed or extended. They have a natural length, at which they remain if there are no external forces acting. Hooke s Law gives F spring = kx where k is a constant. x is the amount of displacement of one end of a spring from it s natural length. (The amount of compression or extension. 1 Figure from CCRMA Stanford Univ.

Elasticity The force that the spring exerts to restore itself to its original length is proportional to how much it is compressed or stretched. This is called Hooke s Law: F = kx where x is the distance that the spring is stretched or compressed by and k is a constant that depends on the spring itself. (The spring constant ). If a very large force is put on the spring eventually it will break: it will not return to its original shape. The elastic limit is the maximum distance the spring can be stretched so that it still returns to its original shape.

Spring example If a 2 kg painting is hung from a spring, the spring stretches 10 cm. What if instead a 4 kg painting is hung from the spring? How far will it stretch? (A) 10 cm (B) 20 cm (C) 30 cm (D) None of the above.

Spring example If a 2 kg painting is hung from a spring, the spring stretches 10 cm. What if instead a 4 kg painting is hung from the spring? How far will it stretch? (A) 10 cm (B) 20 cm (C) 30 cm (D) None of the above.

Spring example If a 2 kg painting is hung from a spring, the spring stretches 10 cm. What if instead a 4 kg painting is hung from the spring? How far will it stretch? We don t know the spring constant, but we can work it out from the information about the first 2 kg painting. The force on the spring is just the weight of the painting. k = F g x = (2 kg)g 0.1 m = 196.2 N/m x = F k = (4 kg)g (196.2 N/m) = 0.2 m If you put on twice the force, you stretch the spring twice as far!

Spring example If a 2 kg painting is hung from a spring, the spring stretches 10 cm. What if instead a 4 kg painting is hung from the spring? How far will it stretch? We don t know the spring constant, but we can work it out from the information about the first 2 kg painting. The force on the spring is just the weight of the painting. k = F g x = (2 kg)g 0.1 m = 196.2 N/m x = F k = (4 kg)g (196.2 N/m) = 0.2 m If you put on twice the force, you stretch the spring twice as far!

Spring example If a 2 kg painting is hung from a spring, the spring stretches 10 cm. Now suppose a 6 kg painting is hung from the same spring. How far does it stretch?

Spring example If a 2 kg painting is hung from a spring, the spring stretches 10 cm. Now suppose a 6 kg painting is hung from the same spring. How far does it stretch? (A) 10 cm (B) 20 cm (C) 30 cm (D) None of the above.

Spring example If a 2 kg painting is hung from a spring, the spring stretches 10 cm. Now suppose a 6 kg painting is hung from the same spring. How far does it stretch? (A) 10 cm (B) 20 cm (C) 30 cm (D) None of the above.

18. A 50.0-kg person takes a nap in a backyard hammock. Both ropes supporting the hammock are at an angle of 15.0 above the horizontal. Find the tension in the ropes. Spring-Friction Example f the applied is 0.45. p. When the eased to 23, e coefficient (b) At what bled? st to a speed of static friche track? (b) t (a). A backpack full of books weighing 52.0 N rests on a table in a 19. IP A backpack full of books weighing 52.0 N rests on a table in physics laboratory a physics laboratory classroom. classroom. A spring A spring with a force constant of 150 N/m is of 150 attached N/m is attached to theto the backpack and pulled horizontally, as as indicated in Figure 6 22. (a) If the spring is pulled until it shown. (a) stretches If the2.00 spring cm and isthe pulled pack remains until it at rest, stretches what is the 2.00 force cm and the pack remains of friction at rest, exerted what on the isbackpack the force by the of table? friction (b) Does exerted your on the answer to part (a) change if the mass of the backpack is doubled? backpack by the Explain. table? (b) Does your answer to part (a) change if the mass of the backpack is doubled? Explain. er car while. When she F th the coffee tatic friction 0.24, what is out causing ce. (b) What n can accele coffee cup 1 FIGURE 6 22 Problems 19 and 20 rink, a puck 1 Walker, page 179, #24. f s

Spring-Friction Example (a) If the spring is pulled until it stretches 2.00 cm and the pack remains at rest, what is the force of friction exerted on the backpack by the table? Sketch a free-body diagram for the backpack. Hypothesis:

Spring-Friction Example (a) If the spring is pulled until it stretches 2.00 cm and the pack remains at rest, what is the force of friction exerted on the backpack by the table? Sketch a free-body diagram for the backpack. Hypothesis: Will point to the right, units, Newtons. Magnitude will equal kx, since F net = 0. Much less than the weight, perhaps about 5 N.

Spring-Friction Example (a) If the spring is pulled until it stretches 2.00 cm and the pack remains at rest, what is the force of friction exerted on the backpack by the table? Strategy: use F net = 0, analyze horizontal and vertical directions. [ Fnet,y = n mg = 0 n = mg ] F net,x = f s F = 0 f s = F f s = kx = (150 N)(0.02 m) = 3.00 N

Spring-Friction Example (b) Does your answer to part (a) change if the mass of the backpack is doubled? Explain.

Spring-Friction Example (b) Does your answer to part (a) change if the mass of the backpack is doubled? Explain. The solution did not use the mass of the backpack, so no, the answer will not change if the mass is doubled. The static friction force is always just big enough to counteract the applied force on an object (unless the applied force exceeds the max static friction force).

Summary more practice with friction springs Quiz Thursday in class. Homework Walker Physics: Ch 6, onward from page 177. Problems: 19, 21, 25, 101