Question

Vector problem: add 11 N at 58 degrees, 17 N at 158 degrees and 22 N at 258 degrees. How much acceleration would that net force provide on a proton (mp+ + 1.67 x 10^-27 kg)?

Answer 1

A = 1 N at 58 degrees
= 1* cos 58 x+ 1* sin 58 y
= 0.53 x + 0.85 y

B = 17 N at 158 degrees
= 17* cos 158 x+ 17* sin 158 y
= -15.76 x + 6.37 y

C=22 N at 258 degrees
= 22* cos 258 x+ 22* sin 258 y
= -4.57 x -21.52 y

Net F = A + B + C
= 0.53 x + 0.85 y -15.76 x + 6.37 y -4.57 x -21.52 y
= -19.8 x - 14.3 y
Magnitude = sqrt (19.8^2 + 14.3^2) = 24.4 N
angle from x axis = 90 + atan (19.8/14.3) = 90 + 54.2 = 144.2 degree

So net force is 24.4 N at an angle of 144.2 degree

a of proton = F/m = 24.4/ (1.67*10^-27) = 1.46*10^28 m/s^2
Acceleration is 1.46*10^28 m/s^2 at an angle of 144.2 degree