Two forces are applied to a ca
Two forces are applied to a car in an effort to move it, asshown in the figure below. (Let F1 = 445 N and F2 = 353 N. Assumeup and to the right are in the positive directions.) Anillustration shows an overhead view of a car as it approaches afork in the road. The car is oriented such that it is directedtoward the top of the illustration. A force F1 extends from thenose of the car and points along the left-side forked road at anangle of 10° measured counterclockwise from the car’s initialdirection. A force F2 extends from the nose of the car and pointsalong the right-side forked road at an angle of 30° measuredclockwise from the car’s initial direction. (a) What is theresultant vector of these two forces? magnitude N direction ° tothe right of the forward direction (b) If the car has a mass of3,000 kg, what acceleration does it have? Ignore friction. m/s2
Answer:
a)Consider the net force as F,F = Fx 
+ Fy 
Where is along the direction of car and is towards the right of the direction of the car.
Fx = F1 * cos
1+ F2 * cos2Where 1 = 10 and 2 = 30 degreesFx = 445 * cos(10) + 353 * cos(10)= 743.95 N
Fy = F2 * sin2- F1 * sin1= 353 * sin(30) – 445 * sin(10)= 99.23 N
Net force, F = SQRT[Fx2 + Fy2]= SQRT[(743.95)2 + (99.23)2]= 750.5 N
tan= Fx/Fy= tan-1(Fx/Fy)= tan-1(743.95 / 99.23)= 82.4 degrees
b)Acceleration, a = F/mWhere m is the mass of the car.a = 750.5 / 3000= 0.250 m/s2
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