Expert: Sherry Wallin - 5/9/2010

Question
QUESTION: I am stuck !
Can u help ?
Calculate the total sum of voltages, Vtot, of the following voltages
V1=10sin(theta+120')
V2=4sin(theta+30')
V3=8sin(theta-30')
V4=6sin(theta-130')

ANSWER: sin(x+y) = sin x cos y + cos x sin y and
sin(x-y) = sin x cos y - cos x sin y and with that said it is just a matter of rewriting each of your voltages:

V1 = 10[sin(theta)*cos(120') + cos(theta)*sin(120')]
= 10[sin(theta)*(-1/2) + cos(theta)*(sqrt(3)/2)]
= -5*sin(theta) + 5sqrt(3)*cos(theta)

V2 = 4[sin(theta)*cos(30') + cos(theta)*sin(30')
= 4[sin(theta)*sqrt(3)/2 + cos(theta)*(1/2)]
= 2*sqrt(3)*sin(theta) + 2*cos(theta)

V3 = 8[sin(theta)*cos(30') - cos(theta)*sin(30')]
= 8[sin(theta)*sqrt(3)/2 - cos(theta)*(1/2)]
= 4*sqrt(3)*sin(theta) - 4*cos(theta)

V4 = 6[sin(theta)*cos(130') - cos(theta)*sin(130')]
= 6[sin(theta)*(-.643) - cos(theta)*.766]
= 3.858*sin(theta) - 4.596*cos(theta)

V1 + V2 + V3 + V4 = -5*sin(theta) + 5sqrt(3)*cos(theta)+2*sqrt(3)*sin(theta) + 2*cos(theta)+
4*sqrt(3)*sin(theta) - 4*cos(theta) + 3.858*sin(theta) - 4.596*cos(theta)

Now you can combine like terms:

first combine your sines:

-5*sin(theta) + 2*sqrt(3)*sin(theta) + 4*sqrt(3)*sin(theta) + 3.858*sin(theta)
= sin(theta)(-5 + 2*sqrt(3) + 4*sqrt(3) + 3.858) = sin(theta)(-1.142 + 6*sqrt(3))

and finally combine your cosines:

5*sqrt(3)*cos(theta) + 2*cos(theta) - 4*cos(theta) - 4.596*cos(theta)
= cos(theta)(5*sqrt(3) + 2 - 4 - 4.596) = cos(theta)(-6.596 +5*sqrt(3)) and

all together, V1 + V2 + V3 + V4 = sin(theta)(-1.142 + 6*sqrt(3)) + cos(theta)(-6.596 +5*sqrt(3))

Now for any value of theta you can calculate the total voltage

Math Prof



---------- FOLLOW-UP ----------

QUESTION: Explain why a 40 tonne truck, travelling on a dry concrete road, has a shorter stopping distance than a train of equivalent size and speed?
Co-effecients for steel/steel dry,  us 0.7 , uf 0.6
Co-effecient for rubber/concrete dry, us1.0 , uf 0.8

How do i answer this ????????????????

Answer
Neil~
    Notice the coefficients of kinetic and static friction are both less for the steel on dry steel as opposed to concrete on dry concrete respectively so that there is less friction applied to the train tracks and therefore less force in helping the train to stop. A body in motion would stay in motion if there weren't any forces working upon it. Both the truck and the train have the same gravitational force working on it so the only difference is that the truck has higher coefficients of friction on dry concrete and thus more opposing force and a quicker stop as a result.

Math Prof