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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