Find the magnitude and direction of the resultant of the concurrent forces of 8 N, 12 N, 15N and 20 N making angles of 30°, 70°, 120°.25 and 155° respectively with a fixed-line. [Ans for the check. 39.5 N; 111.7°] ?
Answers
Given : concurrent forces of 8 N, 12 N, 15N and 20 N making angles of 30°, 70°, 120° and 155° respectively with a fixed-line
To find : magnitude and direction
Solution:
Let say Fixed line is x axis
8 N 30°
=> x Component = 8Cos30° = 8√3/2 = 6.9 N
& y Component = 8Sin30° = 8(1/2) = 4 N
12 N 70°
=> x Component = 12Cos70° = 4.1 N
& y Component = 12Sin70° = 11.3 N
15 N 120°
=> x Component = 15Cos120° = -7.5 N
& y Component = 15Sin120° = 13 N
20 N 155°
=> x Component = 20Cos155° = -18.1 N
& y Component = 20Sin155° = 8.5 N
x Component = 6.9 + 4.1 - 7.5 -18.1 = -14.6N
y Component = 4 + 11.3 + 13 + 8.5 = 36.8 N
magnitude = √(-14.6)² + (36.8)² = 39.5 N
Direction = Tan⁻¹(-14.6/36.8) = 111.7 ° ( as lies in 2nd Quadrant x -ve & y + ve)
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Step-by-step explanation:
Choose coordinate system whose X axis is along the 10 Newton force and which is such that the components of the 15 N force are both positive.
The resultant force then has X component 10 + 15 Cos[45 degrees] = 20.6 Newton and Y component 15 Sin[45 Degree]= 10.6 Newto