Math, asked by xjahid, 9 months ago

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

Answered by amitnrw
14

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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Answered by xXitzSweetMelodyXx
13

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

xXitzSweetMelodyXx

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