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**General Physics I: Day 11 Fundamental Forces, Drag, Ropes & Circular Motion**

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Fundamental Forces We now understand (nearly) every object and behavior in the universe in terms of four fundamental forces: Gravity: The weakest of the four, but the most prominent due to its range. Electromagnetism: Responsible for every day-to-day force that isn’t gravity. Not generally noticed because most things are electrically neutral. Strong Nuclear Force: Holds the nucleus together. Strongest force, but has a tiny range. Weak Nuclear Force: Key to a few really important processes. (sunlight!)

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**WarmUp: Woman, rope & pulley**

A 500 N woman sits in a seat that is suspended from a rope. The rope passes over a pulley suspended from the ceiling, and the woman holds the other end of the rope in her hands. What can we say about the magnitudes of the force of tension in the rope and the force that the seat exerts on the woman? (Hint: Draw a free-body diagram for the woman.)

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**WarmUp: Woman, rope & pulley**

~17% → The tension is 250 N, the force of the seat on the woman is 250 N. ~17% → The tension is 500 N, the force of the seat on the woman is 250 N. ~13% → The tension is 250 N, the force of the seat on the woman is 500 N. ~54% → The tension is 500 N, the force of the seat on the woman is 500 N.

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Ropes & Pulleys Because of Newton’s 3rd Law, we can say: The tension in a stationary continuous piece of rope is exactly the same everywhere in the rope. If it is accelerating the force on the front end of the rope will have to be larger. But, if we assume the rope is of negligible mass… The massless rope approximation lets us say that the tension is the same, even if there is acceleration. Similarly, if a pulley is involved, it will not change the forces in the rope if it is massless and frictionless.

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The Levi Strauss trademark shows two horses trying to pull apart a pair of pants. Suppose Levi had only one horse and attached the other side of the pants to a sturdy tree. Using only one horse would cut the tension on the pants in half. not change the tension on the pants at all. double the tension on the pants.

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**at their current heights. at the same height. **

Two blocks with the same mass are connected by a lightweight cord that runs through an ideal pulley. They are held in the positions shown. When released, the blocks will end up at their current heights. at the same height. with block 1 on the ground. with block 2 on the ground. 1 2

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**Warm-Up: Broken Symmetry**

Say you throw a baseball straight upward with speed v. What can you say about the ball's speed when it returns to your hand a) in the absence of air resistance? ~75% → Returns at the same speed (symmetric!) ~25% → Didn’t answer this question/confused b) in the presence of air resistance? ~17% → Returns at the same speed ~75% → Returns at a smaller speed ~8% → Didn’t answer this question/confused

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**Warm-Up: Broken Symmetry**

“a. with out air drag to affect the ball the speed of the ball should still be increased to around 7m/s when it returns to your hand. b. with air resistance he balls speed is slowed down with the force of the air pushing right back on the ball so its speed would be lower than 7m/s when it returned to your hand.” “[…] With air resistance, the ball's speed will be slightly lower than when it left your hand because the ball has to travel and push its way through all those little air molecules, which requires more force.”

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Fluid Resistance Also called air resistance, air drag, drag, viscosity Unlike kinetic friction, fluid resistance depends very strongly on the speed of the object, and the dependence even changes with speed! Small objects, very slow Bigger than golf ball, throwing speed or faster k incorporates shape, size and fluid properties D incorporates shape, size and density of fluid

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**Air Resistance In air it is easy to reach the high speed regime:**

Crucial concept: Air resistance increases very quickly as an object goes faster! Eventually it equals Fg… terminal speed! Example: Calculate terminal speed for a penny falling flat (unrealistic). Assume 𝐷= kg/m 15.6 m/s, not bad, but a bit high (tumbling!). Diameter = .019 m, so A = π (0.0095)2 = ?? m2 mass = 2.5 grams = kg ρ = 1.29 kg/m3 I get a terminal speed of 16.4 m/s… not bad! Something closer to 11 m/s is more realistic because the penny tumbles.

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**Challenge: Drop with Drag**

Sketch plots of position, velocity and acceleration vs. time for a ball dropped from the top of a very tall building in the presence if air resistance. Use down as positive.

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Falling in Air

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Sample Problem A baseball (m = 145 g, r = 3.68 cm) is hit towards left field in a game in New York. At the top of the ball’s motion it is moving at 35 m/s. Find the acceleration of the ball at that moment. Assume 𝐷= kg/m. F = ¼ ρ A v2

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**The net force on a moving object suddenly becomes zero**

The net force on a moving object suddenly becomes zero. The object then: stops abruptly. stops during a short time interval. changes direction. continues at constant velocity. slows down gradually.

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**Centripetal Acceleration**

Centripetal acceleration always points towards the center of the circle What matters is speed and turning radius: If you only go faster, acceleration is larger. If you only tighten your turn, acceleration is larger. Centripetal acceleration is the amount of acceleration required for an object to go in a circle of radius r at speed v.

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**Worked-Example: Circling Car**

v FT R Find the maximum tension in the car as it goes around the circle at top speed.

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**Worked-Example: Circling Car**

Explain why the tension in the string is the greatest when the car is going the fastest. When would the tension be the least?

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**Worked-Example: Circling Car**

In this problem the car is traveling on a horizontal, level track, so the only force examined was the tension. What other forces act on the car and why were they neglected? If the car was instead traveling on a vertical loop would these forces need to be included in the calculations?

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**Worked-Example: Circling Car**

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**Worked-Example: Circling Car**

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Centripetal Force? Lets apply Newton’s 2nd Law to this… This net force is sometimes called the centripetal force, but it is not a new force! The net force (centripetal or not) can come from any force… friction, tension, gravity, the normal force, etc., depending on the situation. My advice: Avoid the phrase and definitely do not put it into your equations.

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You’re a passenger in a racecar approaching a turn after a straight section. As the car turns left, you are pressed against the car door. Which is true during the turn? A centripetal force pushes you against the door. There is no force that pushes you against the door. The frictional force of the ground pushes you against the door. No force in the radial direction is acting on you. You cannot analyze this situation in terms of the forces on you since you are accelerating.

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**Sample Problem - Banked Curve**

A car drives around a curve with radius 410 m at a speed of 32 m/s. The road is banked at 5.0°. The mass of the car is 1400 kg. What is the frictional force on the car? At what speed could you drive around this curve so that the force of friction is zero? This question gets messy! Start with FBD. Discuss how friction could be up, down or zero. Choose up (the “too slow” version). Do NSL for x & y, COMPACTLY. Both have F_f and F_N in them. 2 eq. 2 unknowns Combining them is ugly, so plug values in first! Should come up with F_f = N

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Coming up… Thursday (9/25) → Review for Exam 1 Tuesday (9/30) → Exam 1 Homework #5 due Saturday the 27th Exam 1 is a next Tuesday (Deal offer expires!) Sample questions posted online! Do these in “Test-like conditions” as diagnostic Extra credit (+3% on Exam 1) posted Wednesday

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