Forces act at a point. The magnitude of is 9.00 N, and its direction is 60.0o above the x-axis in the second quadrant. The magnitude of is 6.00 N, and its direction is 53.1o below the x-axis in the third quadrant. (a) What are the x- and y-components of the resultant force? (b) What is the magnitude of the resultant force?

Solution 6E Step 1: According to the question, the vectors lie in this manner on the coordinate system. 0 The total angle between these two forces is, 60 + 53.1 = 113.1 . The resultant vector’s magnitude we can find by, R = F 12+ F 22+ 2F 1 2os = + 6 + 2 × 9 × 6 × cos 113.1 0 = 81 + 36 42.37 = 72.648 = 8.523N STEP 2: As we got the resultant, it’s not a big deal to find the angle it makes with the horizontal axis. By the parallelogram law of vector addition, F + F = R 1 2 2 2 F =2R F = 1 R + F 1 2 × R × F1× cos 2 2 2 6 = 8.523 + 9 2 × 8.523 × 9 × cos 36 = 153.64 153.414 cos cos = (36 153.64)/( 153.414) = 0.766 = cos (0.766) = 40 .0 This is the angle between the resultant vector and the vector F . 1 0 So, the angle AOB is given as 60 . 0 The angle AOR we got just now as 40 . 0 The required angle ROB is 60 40 = 20 . CONCLUSION: So, as we know the horizontal component will be, x comp = R cos = 8.523 × cos 20 = 8.009 The vertical component will be, y comp = R sin = 8.523 × sin 20 = 2.915 And the magnitude of the resultant vector is 8.523.