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…as is the direction of the applied force

The distribution of material within an object can be as important in determining how it will respond to efforts to move it…. …as is the direction of the applied force. F. with respect to its center of mass. F. R. m. m. d. h. r. Work is performed by applying a force

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…as is the direction of the applied force

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  1. The distribution of material within an object can be as important in determining how it will respond to efforts to move it…

  2. …as is the direction of the applied force F with respect to its center of mass. F

  3. R m m d h r Work is performed by applying a force (down on this end) over the distance d. Which changes the potential energy of the load on this end. By how much? Conservation of energy requires: d h R r d r A. B. C. h d =? r R h r D. E.

  4. d R h r Raising the handlebars d raises the load h Again: Fd = mgh h d should be in the same proportion as 1. R/r 2. r/R 3. R/d 4. r/h So

  5. When an object is on a ramp its weight pulls it into the supporting surface and down along its surface. This component represents the unbalanced force that will accelerate the car downhill. This component of the weight is compensated by the normal force. Weight Gravity’s pull straight DOWN is both pulling both into and along the ramp at the same time!

  6. A E B C D At which position does the pedalist get the greatest acceleration by bearing down with his full weight?

  7. In both position A and D the force is exerted straight along a line toward the center of the axle which can do nothing in cranking the sprocket.

  8. B r r F F r When F and r are perpendicular we define the torque as F Notice the units on torque are Newton-meters.

  9. CENTER OF MASS This “off-center” push consists of a component of force directed along the line to the center (which will launch the book into an trajectory) and a component perpendicular to the line to center (the torque which will set it spinning as well).

  10. Together these two torques work to rotate this refrigerator out from its corner. R F1 R2 R1 F2 F2 is perpendicular to the line back to center. so it contributes a torque = F2R2 F1 is not perpendicular to R1, the line to center. Instead we use the “lever arm” R, the distance F1 misses the center by. This torque = F2R2

  11. The light turns green and you’re in a hurry! Will the car accelerate faster if you floor the pedal and “burn rubber” or if instead you accelerate so as to just avoid skidding your wheels? A. Skid your tires and burn rubber. B. Just barely avoid skidding the tires.

  12. Beginner skiers learn to ski-plow to slow down or stop. The skis dig into the snow, and do work in plowing the snow aside. This work comes at a cost: it depletes the skier’s kinetic energy…slowing her to a stop. But the snow also pushes back on the skier! Perhaps more tangible if the encounter was with a boulder rather than a pile of soft snow.

  13. How do you turn right when skiing downhill? You push LEFT against the snow! When snow-plowing you actually push out with your LEFT skit o turn right! In more advanced parallel skiing you lean right to dig the inside (right) edge of your skis into the snow…pushing LEFT against the hill!

  14. The tennis ball pushes into the racquet’s netting, stretching its strings. The net pushes against the tennis ball (see how it has been deformed?) slowing it to a stop, then sending it back. Contact always produces pairs of forces. The place-kicker’s toe pushes into the football. Note how it has been deformed! This force will send it flying. The kicker’s toe will feel the football. This force will ever-so-slightly slow his foot down.

  15. The cue ball traveling with speed v strikes a stationary billiard ball head-on. A. The cue ball rebounds backward, while its target is sent moving forward. B. The cue ball stops while its target continues forward with the speed v. C. The cue ball and target ball roll forward together with a speed <v. D. The cue ball comes to rest in place next to the target ball.

  16. Consider this rear end collision with a parked car. If left in neutral without the parking brake set, the force of impact will send this car rolling forward. The force of impact stops this car. If this involves speeds over 5-10 mph, both vehicles will sustain damage. Which one “feels” or experiences the force of impact? The cue ball loses its energy in this head-on collision. The force of the impact pushes back in stopping it. The force of impact pushes the target forward. The cue ball decelerates from v to 0 in the same fraction of a second (the time both balls in contact) that its target accelerates from 0 to v.

  17. A boxer’s right hook delivers a knock-out punch! The force this punch delivers to his opponent’s face/head/neck is A. greater than B. exactly equal to C. less than the force the boxer’s hand experiences from the blow.

  18. A fly is struck against by windshield of a car traveling 65mph down the highway. The force experienced by the fly on impact A. is greater than B. is equal to C. is less than the force of impact experienced by the windshield due to the impact.

  19. Nothing’s moving, but not from lack of trying! 2 1 5 6 4 3 1. Stranded motorist pushes on car. 2. Car pushes back on her. How do we know? 3. Because it is mired in sand, the car’s tires have a mound of sand to push up against. How do we know? 4. Sand pushes back on car. 5. With feet dug in, she pushes back into the sand. 6. The sand pushes back on her. This is what balances 2. What needs to be changed to get out?

  20. How do you walk? What are the forces involved that allow you to walk? As bracing yourself to push a car showed, you push back against the ground below you to propel yourself forward. Imagine trying to walk across a surface without friction!

  21. Micro-polished glass Smooth plastic surface

  22. 500 m A smoothly varnished surface. 50 m

  23. Polished carbon steel surfaces

  24. Since even the smoothest of surfaces are microscopically rough, friction results from the sliding up and over of craggy surfaces, and even the chipping and breaking of jagged peaks. There are TWO TYPES of friction. Static Friction Acts to prevent objects from starting to slide Forces can range from zero to an upper limit Sliding Friction Acts to stop objects that are already sliding Forces of sliding friction have a fixed value that depends on the particular surfaces involved.

  25. Frictional forces increase when you: force the sliding surfaces together more tightly (increase an object’s weight). The peak static force is always greater than sliding force Surface features interpenetrate more deeply when stationary objects settle. Friction force drops when sliding begins Cold welds are broken and moving objects ride across the craggy surfaces higher.

  26. f W The force of friction, f, is directly proportional to the total force (usually W for objects sliding horizontally) that presses the sliding surfaces together: We write: f =W where  is known as the “coefficient of friction”

  27. Typical coefficients of friction maximum Materialstaticsliding Rubber on dry concrete 0.90 0.80 Steel against steel 0.74 0.57 Glass across glass 0.94 0.40 Wood on wood 0.58 0.40 Wood on leather 0.50 0.40 Copper on steel 0.53 0.36 Rubber on wet concrete 0.30 0.25 Steel on ice 0.10 0.06 Waxed skis on snow 0.10 0.05 Steel across teflon 0.04 0.04 Synovial joints (hip, elbow) 0.01 0.01 What happens when objects slide to rest? Where does the lost kinetic energy go? It generates heat, an additional form of energy.

  28. Rotation Velocity Wheels can circumvent friction by using the fact that objects can roll without sliding

  29. If friction prevents slipping at this point, the foot planted at bottom stays stationary as the entire assembly tips forward, rotating about its axis.

  30. Notice while the planted foot stays put, the axle moves forward at half the speed that the top edge of our wheel does! Remember:pathlength out a distance r from the center of a rotation: s = r  and the tangential speed at that point: v = r  2v v v = 0

  31. Each time this tethered ball comes around, a wack of the paddle gives it a boost of speed speed v . r m But this v is directly related to an angular velocity,  (in radians/sec) v = r For an individual mass m rotating in an orbit of radius r rotational kinetic energy

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