Answer:
R = 103.7 N, 31.6° above x-axis
Explanation:
First we find the x components of all the forces:
F1x = F1 Cos 60°
F1x = (100 N)(Cos 60°)
F1x = 50 N
F2x = F2 Cos 140°
F2x = (200 N)(Cos 140°)
F2x = -153.2 N
F3x = F3 Cos 320°
F3x = (250 N)(Cos 320°)
F3x = 191.5 N
So, the x component of resultant will be the sum of the x component of each force:
Rx = F1x + F2x + F3x
Rx = 50 N - 153.2 N + 191.5 N
Rx = 88.3 N
Now we find the y components of all the forces:
F1y = F1 Sin 60°
F1y = (100 N)(Sin 60°)
F1y = 86.6 N
F2y = F2 Sin 140°
F2y = (200 N)(Sin 140°)
F2y = 128.5 N
F3y = F3 Sin 320°
F3y = (250 N)(Sin 320°)
F3y = -160.7 N
So, the y component of resultant will be the sum of the y component of each force:
Ry = F1y + F2y + F3y
Ry = 86.6 N + 128.5 N - 160.7 N
Ry = 54.4 N
Hence, the magnitude of resultant force will be:
|R| = √(Rx² + Ry²)
|R| = √[(88.3 N)² + (54.4 N)²]
|R| = √10756.25 N²
|R| = 103.7 N
And the direction θ will be:
θ = tan⁻¹(Ry/Rx)
θ = tan⁻¹(54.4/88.3)
θ = 31.6° above x-axis
Hence, the resultant vector will be:
<u>R = 103.7 N, 31.6° above x-axis</u>
