11. Given: -31. 31. 2k, =-1.47+ 2k, a) Compute A (Bx C) b) Compute A (B + C), c)
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11. Given: -31. 31. 2k, =-1.47+ 2k, a) Compute A (Bx C) b) Compute A (B + C), c) Compute AX (BC), 13. Given two forces F1 and F2. The magnitude of F1, F1 = 1 = 10 N (Newtons) and its orientation Ø1- 30° with respect to the positive x-axis. The magnitude of F2, F2 IF2l 15 N and its orientation Ø2 1350 with respect to the positive x-axis. a) Write F1 in component form. b) Write F2 in component form. c) Find the sum : F = Fl t F2. d) Find the magnitude and orientation with respect to the positive x-axis. 14. Given two Electric fields E1 and E2. The magnitude of E1, E1 E 20 X 104 N/C (Newtons/Coulomb) and its orientation 01--45° with respect to the positive x-axis. The magnitude of E2, E2 = E21-30 X 104 N/C and its orientation Ø2 60° with respect to the positive x-axis. a) Write E1 in component form. b) Write E2 in component form. c) Find the sum: E E1+ E2. d) Find the magnitude and orientation with respect to the positive x-axis. 4Explanation / Answer
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13.
Suppose given that Force is F and it makes angle A with +x-axis, then it's components are given by:
Fx = F*cos A
Fy = F*sin A
Using above rule:
F1 = 10 N & angle = 30 deg with +x-axis
F1x = F1*cos A1 = 10*cos 30 deg = 8.66 N
F1y = F1*sin A1 = 10*sin 30 deg = 5 N
F1 = F1x i + F1y j
F1 = (8.66 i + 5 j) N
F2 = 15 N & angle = 135 deg with +x-axis
F2x = F2*cos A2 = 15*cos 135 deg = -10.61 N
F2y = F2*sin A2 = 15*sin 135 deg = 10.61 N
F2 = F2x i + F2y j
F2 = (10.61 i + 10.61 j) N
Now net force will be
Fnet = Fnet)x + Fnet)y
Fnet = (F1x + F2x) i + (F1y + F2y) j
Using above values
Fnet = (8.66 - 10.61) i + (5 + 10.61) j
Fnet = -1.95 i + 15.61 j
Now Magnitude of force will be
|Fnet| = sqrt (Fnet_x^2 + Fnet_y^2)
|Fnet| = sqrt ((-1.95)^2 + 15.61^2)
|Fnet| = 15.73 N
Direction of Fnet will be
Direction = arctan (Fnet_y/Fnet_x)
Direction = arctan (15.61/1.95) = 82.88 deg with -ve x-axis
Direction = 180 - 82.88 = 97.12 deg with positive x-axis
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