Question

Example 2.3. A triangle ABC has its side AB = 40 mm along positive
x-axis and side BC = 30 mm along positive y-axis. Three forces of 40 N, 50
N and 30 N act along the sides AB, BC and CA respectively. Determine
magnitude of the resultant of such a system of forces.​

Answer 1

Answer:

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Answer 2

Given,

ΔABC

AB = 40 mm (along positive x-axis)

BC = 30 mm (along positive y-axis)

F1 = 40N, F2= 50N and F3 = 30N acting along sides AB, BC and CA of the ΔABC respectively.

To find,

The magnitude of the resultant force F.

Solution,

Using Pythagora's theorem for ΔABC we can say,

$AC^{2} = AB^{2} + BC^{2}$

putting the values we get,

AC = 50 mm

Let ∠BAC =θ

sinθ = 30/50 = 0.6

cosθ = 40/50 = 0.8

Resolving the forces along x-axis we get,

$F_{x}$ = 40 - 30cosθ = 16 N

Resolving the forces alomg y-axis we get,

$F_{y}$ = 50 - 30sinθ = 32 N

Magnitude of the resultant force,

$F = \sqrt{F_{x}^2 + F_{y}^2}$

⇒ F = $\sqrt{16^2 + 32^2}$

⇒ F = 35.8 N