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Transcript
PHYSICS 231 Lecture 8: Forces, forces & examples Next week: Walk-in hour Monday 4-5 pm Tuesday: homework due Wednesday: Review (Prof. Lynch) Friday Exam I am absent Tue-Fri. Remco Zegers PHY 231 1 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 2 Forces seen in the previous lecture • Gravity: Force between massive objects • Normal force: Elasticity force from supporting surface n=-FgL FgL=mgcos Fg//=mgsin Fg=mg PHY 231 3 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 PHY 231 4 QUIZ! (extra credit) In the absence of friction, to keep an object that is traveling with a certain velocity moving with exactly the same velocity over a level floor, one: a) does not have to apply any force on the object b) has to apply a constant force on the object c) has to apply an increasingly strong force on the object PHY 231 5 Jumping! The pelvis has a mass of 30.0 kg. What is its acceleration and direction of acceleration? Decompose all forces in x and y directions Force x (N) y (N) 300 N -122 -274 690 N 236 648 Weight 0 -249 Resultant 114 N 80.3 N 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 PHY 231 6 Tension T The magnitude of the force T acting on the crate, is the same as the tension in the rope. Spring-scale You could measure the tension by inserting a spring-scale... PHY 231 7 question ceiling A block of mass M is hanging from a string. What is the tension in the string? M a) b) c) d) e) T=0 T=Mg with g=9.81 m/s2 T=0 near the ceiling and T=Mg near the block T=M one cannot tell PHY 231 8 Newton’s second law and tension m1 n T Fg No friction. T What is the acceleration of the objects? m2 Fg Object 1: Object 2: F=m1a, so F=m2a, so T=m1a Fg-T=m2a m2g-T=m2a Combine 1&2 (Tension is the same): a=m2g/(m1+m2) PHY 231 9 Problem What is the tension in the string and what will be the acceleration of the two masses? Draw the forces: what is positive & negative??? T T For 3.00 kg mass: F=ma T-9.813.00=3.00a For 5.00 kg mass: F=ma 9.815.00-T=5.00a T=36.8 N a=2.45 m/s2 Fg Fg 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 friction N pull 20 kg Fs Fg A person wants to drag a crate of 20 kg over the floor. If he pulls the crate with a force of 98 N, the crate starts to move. What is s? Answer: Fs= sN=sMg=20x9.8xs=196s Just start to move: Fs=Fpull 98=196s so: s=0.5 After the crate starts moving, the person continues to pull with the same force. Given k=0.4, what is the acceleration of the crate? F=ma F=98-0.4Mg=98-0.4x20x9.8=98-78.4=19.6 N 19.6=20a a=0.98 m/s2 PHY 231 13 20 100 A person is pushing an ice-sledge of 50kg over a frozen lake with a force of 100N to the east. A strong wind is pushing from the south-west and produces a force of20N on the sledge. a) What is the acceleration of the sledge (no friction)? b) if the coefficient of kinetic friction is 0.05, what is the acceleration? a) force Horizontal vertical person 100 0 wind 20cos(45) 20sin(45) sum 100+14.1=114.1 14.1 Total force: (114.12+14.12)=115 N F=ma 115=50a a=2.3 m/s2 b) Ffriction=n=0.05*m*g=0.05*50*9.8=24.5N F=ma 115-24.5=50a a=1.8 m/s2 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. • use in further equations if necessary • Check whether your answer makes sense. PHY 231 15