Asked by Allie
"Only two forces act on an object (mass = 3.90 kg), as in the drawing. (F = 76.0 N.) Find the magnitude and direction (relative to the x axis) of the acceleration of the object. " (The angle is 45 degrees and the horizontal force is 40N)
Answers
Answered by
Henry
F1 = 76 N @ 45 Deg.
F2 = 40 N. @ 0 Deg.
X = 76*cos45+40*cos(0) = 93.74 N.
Y = 76*sin45 = 53.74 N.
tanA = Y/X = 53.74 / 93.74 = 0.57329.
A = 29.825 Deg.
Fn = X/cosA = 93.74 / cos29.825 = 108.05 N. @ 29.825 Deg.
Fn = ma.
a = Fn/m = 108.05 / 3.90 = 27.71 m/s^2
@ 29.825 Deg.
F2 = 40 N. @ 0 Deg.
X = 76*cos45+40*cos(0) = 93.74 N.
Y = 76*sin45 = 53.74 N.
tanA = Y/X = 53.74 / 93.74 = 0.57329.
A = 29.825 Deg.
Fn = X/cosA = 93.74 / cos29.825 = 108.05 N. @ 29.825 Deg.
Fn = ma.
a = Fn/m = 108.05 / 3.90 = 27.71 m/s^2
@ 29.825 Deg.
Answered by
Anonymous
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