Split up each force into horizontal and vertical components.
• 300 N at N30°E :
(300 N) (cos(30°) i + sin(30°) j)
• 400 N at N60°E :
(400 N) (cos(60°) i + sin(60°) j)
• 500 N at N80°E :
(500 N) (cos(80°) i + sin(80°) j)
The resultant force is the sum of these forces,
∑ F = (300 cos(30°) + 400 cos(60°) + 500 cos(80°)) i
… … … + (300 sin(30°) + 400 sin(60°) + 500 sin(80°)) j N
∑ F ≈ (546.632 i + 988.814 j) N
so ∑ F has a magnitude of approximately 1129.85 N and points in the direction of approximately N61.0655°E.
Answer:
Step-by-step explanation:
so you will move constant to the right side and change the sign. 72x=223-7
then you will subtract the numbers 72x=223-7 and that will give you 72=216
then you will divide both sides by 72 and that will give you
- x=3 .
I think it would be (F,E)
You can isolate r to one side of the equation by dividing by adding three, dividing by s, multiplying by a and you get:
r=a(3+k)/s
