1. Find the resultant due to the action of three forces if they are F1, 100 N, 60º above the x axis; F2, 200 N, 140º above the x axis; and F3, 250 N, 320º above the x axis.

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:

R = 103.7 N, 31.6° above x-axis

total distance/total time

(85 + 82)/(5+3)

167/8

20 just over