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Download 2 The slides about friction are in lecture 8!! 3 TRIGONOMETRY
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PHYSICS 231 Lecture 9: Revision Remco Zegers Walk-in hour: Thursday 11:30-13:30 Helproom PHY 231 1 The slides about friction are in lecture 8!! PHY 231 2 TRIGONOMETRY SOH-CAH-TOA: sin=opposite/hypotenuse cos=adjacent/hypotenuse tan=opposite/adjacent Pythagorean theorem: c a b 2 PHY 231 2 2 3 Units • Convert all units in your problem to be in the SI system • When adding/subtracting two quantities check whether their units are the same. • If you are unsure about an equation that you want to use, perform the dimensional analysis and make sure that each part of the equation that is set equal/subtracted/added have the same dimensions • When 2 quantities are multiplied, their units do not have to be the same. The result will have as unit the multiplied units of the quantities being multiplied. • Sin, cos and tan of angles are dimensionless PHY 231 4 Solving a quadratic equation At2+Bt+c=0 • Use the equation: b b 4ac t 2a 2 • CHECK THE ANSWER • OR just calculate it! t2+(B/A)t+C/A=0 and use (t+d)2=t2+2dt+d2 [t+(B/2A)]2-B2/4A2+C/A=0 t=-B/2A(B2/4A2-C/A) PHY 231 5 Constant motion x(t)=x0 x(m) v m/s t v(t)=0 Constant velocity x(m) x(t)=x0+v0t t v m/s v(t)=v0+at t a m/s2 a(t)=0 t t v m/s v(t)=v0 t a m/s2 Constant acceleration x(t)=x0+v0t+½at2 x(m) a m/s2 a(t)=0 PHY 231 t t a(t)=a0 t 6 x v t time PHY 231 7 Vector operations in equations X a b X a X b X a X b Ya b Ya Yb Ya Yb X a b X a X b X a X b Ya b Ya Yb Ya Yb y (xb,yb) B (xa+b,ya+b) A+B A (xa,ya) x Xa=Acos() Ya=Asin() length/magnitude of A: (Xa2+Ya2) PHY 231 8 Parabolic motion: decompose x and y directions vx=v0cos vy=v0sin-2g=0 V=v0 vx=v0cos vy=v0sin vx=v0cos vy=v0sin-1g vx=v0cos vy=v0sin-3g vx=v0cos vy=v0sin-4g vx=v0cos vy=v0sin-5g t=0 t=1 t=2 PHY 231 t=3 t=4 t=5 9 Parabolic motion X(t)=X0+V0cost Y(t)=Y0+V0sint-1/2gt2 X=X0 Y=Y0 t=0 t=1 t=2 PHY 231 t=3 t=4 t=5 10 2D motion • When trying to understand the motion of an object in 2D decompose the motion into vertical and horizontal components. • Be sure of your coordinate system; is the motion of the object you want to study relative to another object? • Write down the equations of motion for each direction separately. • If you cannot understand the problem, draw motion diagrams for each of the directions separately. • Make sure you understand which quantity is unknown, and plug in the equation of motions the quantities that you know (givens). Then solve the equations. PHY 231 11 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 12 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 13 A snowball is launched horizontally from the top of a building at v=12.7 m/s. If it lands 34 m away from the bottom, how high was the building? horizontal Vertical V0=12.7 m/s h x y vx t h? t Vy X(t)=x0+voxt =0+12.7t 2 Y(t)=y +v t-0.5gt 0 0y =34 (hit the 2 =h-0.5gt ground) =0 (hits the ground) t=34/12.7=2.68 s so: h=0.5gt2 d=34m h=35 m PHY 231 14 A red ball is thrown upward with a velocity of 26.8 m/s. A blue ball is dropped from a 13.3 m high building with initial downward velocity of 5.00 m/s. At what time will the balls be at the same height. Vo=-5.00 blue Blue ball: red 13.3 yb(t)=y0+v0bt-0.5gt2= x v0r =13.3-5.t-0.5gt2 x Red ball: 2= y (t)=y +v t-0.5gt r o or t t h=13.3m v0b =26.8t-0.5gt2 v v V0=26.8 yb(t)=yr(t) 13.3-5t-0.5gt2=26.8t-0.5gt2 13.3=31.8t so t=0.42s PHY 231 15 A car starts at rest and travels for 5.09 s with uniform acceleration of +1.48 m/s2. The driver then brakes, causing uniform acceleration of -1.91 m/s2. If the brakes are applied for 2.96 s how fast is the car going after that? v 5.09 5.05+ 2.96 t In first period: v(t)=v0+at=0+1.48·5.09=7.53 m/s In 2nd period: v(t)=v0+at=7.53-1.91·2.96=1.88 m/s PHY 231 16 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 17 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 18 PHY 231