Posted by Dani on .
Two forces are applied to a car in an effort to move it, as shown in the following figure, where F1 = 402 N and F2 = 364 N. (Assume up and to the right as positive directions.)(F1 is 10degrees and F2 is 30degrees)
(a) What is the resultant of these two forces?
(b) If the car has a mass of 3,000 kg, what acceleration does it have? Ignore friction.
F1 = 402N @ 10 Deg.
F2 = 364N @ 30 Deg.
a. X = hor. = 402cos10 + 364cos30,
X = 395.9 + 315.2 = 711.1N.
Y = ver. = 402sin10 + 364sin30,
Y = 69.8 + 182 = 251.8N.
tanA = Y/X = 251.8 / 711.1 = 0.3541,
A = 19.5 Deg. = Direction.
M = X/cosA = 711.1 / cos19.5 = 754.4N =
b. F = ma,
a = F/m = 754.4 / 3000 = 0.25m/s^2
remember if the degrees is negative then subtract it from 360
F1 @ -10 deg
F2 @ 30 deg
F1 is actually @ 350 deg
F2 is still 30 deg
You have to remember that there's no force being created on the x axis, the answer below must've been wrong because of that.