Question

Two forces are applied to a car in an effort to move it, as shown in the figure below. (Let F1 = 445 N and F2 = 353 N. Assume up and to the right are in the positive directions.) An illustration shows an overhead view of a car as it approaches a fork in the road. The car is oriented such that it is directed toward the top of the illustration. A force F1 extends from the nose of the car and points along the left-side forked road at an angle of 10° measured counterclockwise from the car's initial direction. A force F2 extends from the nose of the car and points along the right-side forked road at an angle of 30° measured clockwise from the car's initial direction. (a) What is the resultant vector of these two forces? magnitude N direction ° to the right of the forward direction (b) If the car has a mass of 3,000 kg, what acceleration does it have? Ignore friction. m/s2

Answer 1

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 * cos1 + F2 * cos2
Where 1 = 10 and 2 = 30 degrees
Fx = 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/m
Where m is the mass of the car.
a = 750.5 / 3000
= 0.250 m/s²