Transcript on forces
PHYSICS 231 Lecture 9: More on forces 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 General strategy for problems with forces • 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 3 quiz! A homogeneous block of 10 kg is hanging with 2 ropes from a ceiling. The tension in each of the ropes is: a) 0 N b) 49 N c) 98 N d) 196 N e) don’t know PHY 231 4 A 2000 kg sailboat is pushed by the tide of the sea with a force of 3000 N to the East. Because of the wind in its sail it feels a force of 6000 N toward to North-West direction. What is the magnitude and direction of the acceleration? Horizontal Due to tide: 3000 N Due to wind: 6000cos(135)=-4243 Sum: -1243 N N 6000N W 3000N S Vertical 0N 6000sin(135)=+4243 4243 N Magnitude of resulting force: Fsum=[(-1243)2+(4243)2]=4421 N Direction: angle=tan-1(4243/-1243)= 1060 (calc: -730, add 1800) E F=ma so a=F/m=4421/2000=2.21 m/s2 PHY 231 5 T 900 T 1kg A mass of 1 kg is hanging from a rope as shown in the figure. If the angle between the 2 supporting wires is 90 degrees, what is the tension in each rope? TVerL horizontal Vertical TVerR 0 45 0 left rope Tsin(45) Tcos(45) 45 right rope -Tsin(45) Tcos(45) gravity 0 -1*9.81 Sum: 0 2Tcos(45)-9.81 ThorR ThorL Object is stationary, so: 2Tcos(45)-9.81=0 so, T=6.9 N Fg 6 PHY 231 Problem n=-FgL A)If s=1.0, what is the angle for which the block just starts to slide? F =mgsin g// FgL=mgcos B)The block starts moving. Given that k=0.5, what is the acceleration of the block? Fg=mg Fs,k 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(½2-¼2)=a so a=g¼2 PHY 231 7 All the forces come together... n T T If a=3.30 m/s2 (the 12kg block is moving downward), what is the value of k? Fg For the 7 kg block parallel to the slope: T-mgsin-kmgcos=ma For the 12 kg block: Mg-T=Ma Fk Fg Solve for k M ( g a) mg sin ma k 0.25 mg cos PHY 231 8 A lawsuit… A passenger in a bus files a lawsuit claiming that when the driver suddenly pushed the brakes a suitcase came flying toward her from the front of the bus and she got injured. You are in the jury; will you award damages? a) No, this could not have happened b) Yes, this is plausible PHY 231 9 puck on ice After being hit by a hockey-player, a puck is moving over a sheet of ice (frictionless). The forces working on the puck are: a) No force whatsoever b) the normal force only c) the normal force and the gravitational force d) the force that keeps the puck moving e) the normal force, the gravitational force and the force that keeps the puck moving PHY 231 10 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? 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 11