Question

Determine the direction of the forces in the DIAGRAM ABOVE. LEAVE your ANSWER IN 3 SIGNIFICANT FIGURE *

Answer 1

Answer:

Direction of net force will be θ = 0.100° in $IV^{th}$ quadrant, with magnitude 4.26 N

Explanation:

How to determine direction of forces in the diagram, they're already given.

What I can do is to find their resultant and determine it's direction.

In positive x-direction, sum of the forces are

10 N + 12cos60° - 14cos30°

⇒ + 3.88 î N

In positive y-direction, sum of forces are

18 N + 14sin30° - 15 N - 12sin60°

⇒ - 0.39 ĵ N

Hence, 3.88 N force is acting on positive x direction, and 0.39 N force is acting on negative y direction.

So, the vector sum of these forces is

Here, tanθ = $\frac{0.39}{3.88}$ ≈ 0.100

⇒ θ ≈ 0.100

R = $\sqrt{(0.39)^{2}+(3.88)^{2}+2*0.39*3.88cos 0.1}$

$R=\sqrt{0.1521+15.05+3}$

R = 4.26 N

And, θ ≈ 0.1 in IVth quadrant.