Question #95863
Herbert and Randy is playing a friendly game of Serengeti Squash ball (which is basically football but with catchier name) with his antelope friends Thomas, Fred and Darius. On one play Herbert is tackled simultaneously by all three of his long-horned friends, experiencing the following forces:
Thomas: 600N, E 25 degrees S
Fred: 550N, W 40 degrees S
Darius: 610 N, W 37 degrees N

Find the net force on Herbert.
1
Expert's answer
2019-10-07T10:42:12-0400

Let us take west -east line to be x-axis then

Resultant of force applied by Fred and Darius is given by

F1=5502+6102+2×550×610cos77°     =908.6 NF_1=\sqrt{550^2+610^2+2\times550\times610\cos77\degree}\\ \ \ \ \ \ =908.6\ N

Angle which the resultant will make is θ=tan1(550sin77°610+550cos77°)=36.14° towards 550 Nforce wrt. 610 N force\theta=\tan^{-1}(\frac{550\sin77\degree}{610+550\cos77\degree})=36.14\degree\ towards\ 550\ N force\ wrt. \ 610\ N\ force

Now angle between this resultant force and force applied by Thomas = 155.86°155.86\degree

Final resultant force = 908.62+6002+2×908.6×600cos155.86°\sqrt{908.6^2+600^2+2\times908.6\times600\cos155.86\degree}{}

332.41 Nresultant angle=tan1908.6sin155.86°600+908.6cos155.86°332.41\ N\\resultant\ angle=\tan^{-1}\frac{908.6\sin155.86\degree}{600+908.6\cos155.86\degree} =180°tan1(1.62167)=121.66°180\degree -tan^{-1}(1.62167)=121.66\degree

122.66° clockwise towards 908.6 N force wrt. 600N force122.66\degree\ clockwise\ towards\ 908.6\ N\ force\ wrt.\ 600\N\ force or we can say that 32.34° south of west32.34\degree\ south \ of\ west



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