PHYSICS 231 Lecture 9: Revision Remco Zegers Walk-in hour: Thursday 11:30-13:30 Helproom PHY 231

The slides about friction are in lecture 8!! PHY 231

TRIGONOMETRY SOH-CAH-TOA: sin=opposite/hypotenuse cos=adjacent/hypotenuse tan=opposite/adjacent Pythagorean theorem: PHY 231

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

Solving a quadratic equation At2+Bt+c=0 Use the equation: 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

Constant acceleration Constant motion Constant velocity Constant acceleration x(t)=x0 x(t)=x0+v0t+½at2 x(t)=x0+v0t x(m) x(m) x(m) t t t v m/s v m/s v m/s v(t)=0 v(t)=v0+at v(t)=v0 t t t a m/s2 a m/s2 a m/s2 a(t)=0 a(t)=0 a(t)=a0 t t t PHY 231

x v t time PHY 231

Vector operations in equations (xa+b,ya+b) y A+B (xb,yb) B (xa,ya) A x Xa=Acos() Ya=Asin() length/magnitude of A: (Xa2+Ya2) PHY 231

Parabolic motion: decompose x and y directions vx=v0cos vy=v0sin-2g=0 vx=v0cos vy=v0sin-1g vx=v0cos vy=v0sin-3g  V=v0 vx=v0cos vy=v0sin vx=v0cos vy=v0sin-4g vx=v0cos vy=v0sin-5g t=0 t=1 t=2 t=3 t=4 t=5 PHY 231

Parabolic motion X(t)=X0+V0cost Y(t)=Y0+V0sint-1/2gt2  X=X0 Y=Y0 PHY 231

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

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

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

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? Y(t)=y0+v0yt-0.5gt2 =h-0.5gt2 =0 (hits the ground) so: h=0.5gt2 horizontal Vertical t vx x Vy y h X(t)=x0+voxt =0+12.7t =34 (hit the ground) t=34/12.7=2.68 s h=35 m V0=12.7 m/s h? d=34m PHY 231

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 t blue v v0b x 13.3 red v0r Blue ball: yb(t)=y0+v0bt-0.5gt2= =13.3-5.t-0.5gt2 Red ball: yr(t)=yo+vort-0.5gt2= =26.8t-0.5gt2 yb(t)=yr(t) 13.3-5t-0.5gt2=26.8t-0.5gt2 13.3=31.8t so t=0.42s h=13.3m V0=26.8 PHY 231

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? t v 5.09 5.05+ 2.96 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

it feels a force of 6000 N toward to North-West direction. 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? E S W N 6000N 3000N Horizontal Vertical Due to tide: 3000 N 0 N Due to wind: 6000cos(135)=-4243 6000sin(135)=+4243 Sum: -1243 N 4243 N Magnitude of resulting force: Fsum=[(-1243)2+(4243)2]=4421 N Direction: angle=tan-1(4243/-1243)= 1060 (calc: -730, add 1800) F=ma so a=F/m=4421/2000=2.21 m/s2 PHY 231

A mass of 1 kg is hanging from a rope as shown in the figure. 900 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? 450 ThorL TVerL TVerR Fg horizontal Vertical left rope Tsin(45) Tcos(45) right rope -Tsin(45) Tcos(45) gravity 0 -1*9.81 Sum: 0 2Tcos(45)-9.81 Object is stationary, so: 2Tcos(45)-9.81=0 so, T=6.9 N ThorR PHY 231