Question

The following forces are acting on an object with a mass of 3.2 x 10-6 kg. You may assume there are no other forces present (weight is negligible). Please find the magnitude and direction of the net force on the object, and the magnitude and direction of the object’s acceleration. Show all work and include an appropriate vector diagrams.

F1: 5 N, 30o above the + x axis

F2: 7.2 N, 45o above the – x axis

F3: 3.8 N, 22o below the + x axis

Answer 1

Answer:

$a=2.1125 \times 10^{6}m/s^2$

θ = 65.9°

Explanation:

mass of the body, m = 3.2 x 10^-6 kg

F1 = 5 N , 30° above + X axis

F2 = 7.2 N, 45° above - X axis

F3 = 3.8 N, 22° below + X axis

let F be the net force which makes an angle θ from + X axis.

Now resolve the components of forces along X axis and y axis, we get

$FCos\theta =F_{1}Cos30+F_{2}Cos135+F_{3}Cos(-22))$

FCosθ = 5 Cos30° + 7.2 Cos 135° + 3.8 Cos22°

F Cosθ = 4.33 - 5.09 + 3.52

F Cosθ = 2.76 N    .... (1)

Similarly,

$FSin\theta =F_{1}Sin30+F_{2}Sin135+F_{3}Sin(-22))$

F Sinθ = 5 Sin30° + 7.2 Sin 135° - 3.8 Sin22°

F Sinθ = 2.5 + 5.09 -1.42

F Sinθ = 6.17 N    .... (2)

Squaring both the equations and then add, we get

$F^{2}\left ({Cos^{2}\theta+Sin^{2}\theta }  \right )=\left (2.76  \right )^{2}+\left (6.17  \right )^{2}$

F = 6.76 N

Dividing equation (2) by equation (1), we get

$tan\theta =\frac{6.17}{2.76}$

θ = 65.9°

Now,

Force = mass x acceleration

F = ma

$a=\frac{F}{m}=\frac{6.76}{3.2\times 10^{-6}}=2.1125 \times 10^{6}m/s^2$