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Got more than one force in your life? Use Vectors + Free Body Diagrams solve all your problems Typical Problem: What’s the acceleration on a particle having a charge of 2C, and mass of 100kg if placed in an electric field of 500 N/C pointing 30 degrees northeast? Do pies! (picture, information, equation, solve) 1. Picture picture- label it too, ask yourself forces are acting is there a wire or cable, then there’s___tension__________ does it have mass (of course!), then there’s ___weight______ Is there some surface pushing or holding, then there’s_normal_______ Are there static charges, then there’s ___electric force______________ Are there moving charges or current, then there’s ___magnetic force__________ 2. info: what are we looking for? What do we have? find acceleration, given charge, mass, electric field 3. what equation connects knowns w/unknowns? F=ma 4. Solve ! …. But use vectors if using variables w/direction (force,acceleration,fields) …. Organize! Use charts to keep x, y info separate or different forces …. Reflect? Does your final answer have the right units, seem right size? Solving vector problems…. a. break each variable into x & y components so can seek balance or get net value Electric Force= qE= 2*500=1000 but 30 NE 1000 Weight= mg = 100*9.8= 980 N 980 N 30° b. Socatoa! Electric force X: 1000cos(30)=866N Y: 1000sin(30)= 500N Fex = Fey= Gravity X: 0 Y: =-980 N Wx= Wy= Total x: = 866 N Total y= - 480 N (watch minuses!) c. combine all x & y components to get the net value Total : Fx = 867 + 0= 866 N Fy = 500 – 980 N = -480 N d. draw a picture to fine the resultant (total) and its angle Fx= 866 N Fy=480 total This is always a right triangle since you used x and y components (Total Force)2 = (866) 2 + (480) 2 = 990N Tan < = Fy/Fx = 480/866 < = -1tan (Fy/Fx)= -1tan (480/866)= 29° e. Use the total or resultant to find the final answer a = Fnet / m = 990 N/ 100kg= 9.9 N/kg or 9.9 m/2 (just slightly more than gravity) direction is same as force!!! 29 degrees south of east Sample problem 1: 1. What is the true velocity of a football if it’s launched at a constant speed of 20m/s at 45 degrees above the horizon and the wind is 10 m/s due east? (hint: draw the vectors & use socatoa- find both net speed and direction!) First vector X=20cos(45)=14.1 Y=20sin(45)=14.1 second vector X=10 Y=0 third vector X= Y= total (resultant) X=24.1 Y=14.1 (Total velocity)2 = 24.12 + 14.12 Total velocity = 27.9 m/s Angle = tan-1 (y/x)= 14.1/24.1)= 30.3 north of east 2. Calculate the electric field needed to levitate an electron (assume the electric field points up straight vertically) Q= 1.6 e-19 C, me= 9.1 x 10-31 kg Fe = qE Wt=mg Electric force is equal and opposite weight (no angles, easy vectors!) Fnet = 0 so qE = mg or E= mg/q = 9.1 x 10-31 kg * 9.8/1.6 e-19 C= 5.57e-11 up 3. Two balls of identical charge (q=?) and same mass (2e-4kg) are hung from a string. They stay separated by 30 degrees (top angle) and a distance of .052 m. charge=? (hint: focus on just one ball and do a free body diagram!) electric X= kq2/(.052)2 Y=0 weight X=0 Y=.00196 tension .052 total X=.00113 Y=.00196 Fe= kq*q/d2 = 9e9*q2/(.052)2 Wt= mg= (2e-4kg)*9.8= .00196 T= ? (but we know 1. Ty must balance weight since their the only vertical forces) Ty= Wt =.00196 2. Tx must balance electric force Tx can be found by using trig: tan(30) = Tx/Ty = Tx/.00196 so Tx=.00113 Tx= Fe= kq*q/d2 = 9e9*q2/(.052)2 = .00113 so q= 7.2 e-9 C 30° X=0 Y=0