# trig

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A 2000 pound car is parked on a street that makes an angle of 12* with the horizontal (think it's a street on a hill). a) find the magnitude of the force required to keep the car from rolling down the hill. b) Find the force perpendicular to the street.

• trig - ,

mass = 2000lbs * 0.454kg/lb = 908kg.

Fc = mg = 908kg * 9.8N/kg = 8898.4N @
12deg.

a. Fp = 8898.4*sin(12) = 1850.4N. = Force parallel to the street = Force required to keep the car from roling down hill.

b. Fv = 8898.4 * cos12 = 8704N = Force
perpendicular to the street.

• Pre-Cal: Vectors - ,

a. Force to prevent car from rolling = v
v=<cos12°,sin12°>
F=projvF=(F•v/|v|^2)v= (F•v)v
<0,2000>•<cos12,sin12>= 0+2000sin12 = 415.82lbs

b. Find the angle of the perpendicular in relation to the force of 2000lbs. To do this, I drew a triangle, one line horizontal, one line being the 12° angle between it and the horizontal, and one being perpendicular to the 12° line(90°). Connect the lines and the third angle should be 78°.
Then you do the same as the last.
Force to prevent car from rolling(perpendicular) = c
c=<cos78°,sin78°>
F=projcF=(F•v/|v|^2)v= (F•v)v
<0,2000>•<cos78,sin78>= 0+2000sin78 = 1956.3lbs

• Yo MUM - ,