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PHYSICS 231 Lecture 8: Forces, forces & examples

PHYSICS 231 Lecture 8: Forces, forces & examples. Remco Zegers Walk-in hour: Monday 11:30-13:30 Helproom. Newton’s Laws.

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PHYSICS 231 Lecture 8: Forces, forces & examples

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  1. PHYSICS 231Lecture 8: Forces, forces & examples Remco Zegers Walk-in hour: Monday 11:30-13:30 Helproom PHY 231

  2. Newton’s Laws • First Law: If the net force exerted on an object is zero the object continues in its original state of motion; if it was at rest, it remains at rest. If it was moving with a certain velocity, it will keep on moving with the same velocity. • Second Law: The acceleration of an object is proportional to the net force acting on it, and inversely proportional to its mass: F=ma • If two objects interact, the force exerted by the first object on the second is equal but opposite in direction to the force exerted by the second object on the first: F12=-F21 PHY 231

  3. n=-FgL Fg//=mgsin FgL=mgcos  Fg=mg Forces seen in the previous lecture • Gravity: Force between massive objects • Normal force: Elasticity force from supporting surface  PHY 231

  4. Gravity, mass and weights. • Weight=mass times gravitational acceleration • Fg(N)=M(kg) g(m/s2) • Newton’s law of universal gravitation: • Fgravitation=Gm1m2/r2 • G=6.67·10-11 Nm2/kg2 • For objects on the surface of the earth: • m1=mearth=fixed • r=“radius” of earth=fixed • The earth is not a point object relative to m2 PHY 231

  5. Measuring mass and weight. Given that gearth=9.81 m/s2, gsun=274 m/s2, gmoon=1.67 m/s2, what is the mass of a person on the sun and moon if his mass on earth is 70 kg? And what is his weight on each of the three surfaces? • The mass is the same on each of the surfaces • On Earth: w=686.7 N • On the Moon: w=116.7 N • On the Sun: 19180 N PHY 231

  6. Decompose all forces in x and y directions Total Force: F=(1142+80.32)=139 N Direction: =tan-1(Fy/Fx)=35.2o Acceleration: a=F/m=139/30.0=4.65 m/s2 Jumping! The pelvis has a mass of 30.0 kg. What is its acceleration? PHY 231

  7. T Spring-scale You could measure the tension by inserting a spring-scale... Tension The magnitude of the force T acting on the crate, is the same as the tension in the rope. PHY 231

  8. n T T Fg Fg Object 1: F=m1a, so T=m1a Object 2: F=m2a, so Fg-T=m2a m2g-T=m2a Combine 1&2 (Tension is the same): a=m2g/(m1+m2) Newton’s second law and tension m1 No friction. What is the acceleration of the objects? m2 PHY 231

  9. Draw the forces: what is positive & negative??? For 3.00 kg mass: F=ma T-9.813.00=3.00a T T For 5.00 kg mass: F=ma 9.815.00-T=5.00a T=36.8 N a=2.45 m/s2 Fg Fg Problem What is the tension in the string and what will be the acceleration of the two masses? PHY 231

  10. Friction • Friction are the forces acting on an object due to interaction • with the surroundings (air-friction, ground-friction etc). • Two variants: • Static Friction: as long as an external force (F) trying to make an object move is smaller than fs,max, the static friction fs equals F but is pointing in the opposite direction: no movement! • fs,max=sn s=coefficient of static friction • Kinetic Friction: After F has surpassed fs,max, the object starts moving but there is still friction. However, the friction will be less than fs,max! • fk=kn k=coefficient of kinetic friction PHY 231

  11. PHY 231

  12. Fs,k n=-FgL Fg//=mgsin FgL=mgcos   Fg=mg Problem A)If s=1.0, what is the angle  for which the block just starts to slide? B)The block starts moving. Given that k=0.5, what is the acceleration of the block? A) Parallel direction: mgsin-sn=0 (F=ma) Perpendicular direction: mgcos-n=0 so n=mgcos Combine: mgsin-smgcos=0 s=sin/cos=tan=1 so =45o B) Parallel direction: mgsin(45o)- smgcos(45o)=ma (F=ma) g(½2-¼2)=a so a=g¼2 PHY 231

  13. n T T Fk Fg Fg All the forces come together... If a=3.30 m/s2 (the 12kg block is moving downward), what is the value of k? For the 7 kg block parallel to the slope: T-mgsin-kmgcos=ma For the 12 kg block: Mg-T=Ma Solve for k PHY 231

  14. General strategy • If not given, make a drawing of the problem. • Put all the relevant forces in the drawing, object by object. • Think about the axis • Think about the signs • Decompose the forces in direction parallel to the motion and perpendicular to it. • Write down Newton’s first law for forces in the parallel direction and perpendicular direction. • Solve for the unknowns. • Check whether your answer makes sense. PHY 231

  15. Next Lecture: Revision and go through an exam May the Force be with you! PHY 231

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