Question

Five forces F1, F2, F3, F4 and F5 are acting on a particle. F1=10N at 30 ; F2 =20N at 90 ; F3 =20N

at 120 ; F4 =50N at 180 ; and F5=60N at 270. All angles are with respect to positive X-axis in anti-clock

direction. Find resultant ​

Answer 1

Explanation:

Given that when all forces are acting on the particle then its acceleration is 4 m/s^2 towards East

So now we can write

\vec F_1 + \vec F_2 + \vec F_3 + \vec F_4 + \vec F_5 = ma[tex] < /p > < p > [tex]\vec F_1 + \vec F_2 + \vec F_3 + \vec F_4 + \vec F_5 = 4*2 = 8\hat i[tex] < /p > < p > Now when force F1 is removed then < /p > < p > [tex]\vec F_2 + \vec F_3 + \vec F_4 + \vec F_5 = ma[tex] < /p > < p > [tex]\vec F_2 + \vec F_3 + \vec F_4 + \vec F_5 = 7*2 = 14 \hat j[tex] < /p > < p > Now subtract above two equations < /p > < p > [tex]\vec F_1 = 8\hat i - 14\hat j[tex] < /p > < p > now acceleration is given by < /p > < p > [tex]a_1 = \frac{F_1}{m}F1+F2+F3+F4+F5=ma[tex]</p><p>[tex]F1+F2+F3+F4+F5=4∗2=8i^[tex]</p><p>NowwhenforceF1isremovedthen</p><p>[tex]F2+F3+F4+F5=ma[tex]</p><p>[tex]F2+F3+F4+F5=7∗2=14j^[tex]</p><p>Nowsubtractabovetwoequations</p><p>[tex]F1=8i^−14j^[tex]</p><p>nowaccelerationisgivenby</p><p>[tex]a1=mF1

a_1 = \frac{8\hat i - 14\hat j}{2}a1=28i^−14j^

a_1 = 4\hat i - 7\hat ja1=4i^−7j^

so above is the acceleration of the mass when only F1 force is acting on it