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Question Number 62587 by Jmasanja last updated on 23/Jun/19

three forces having equal magnitude  s of 10N,20N and 30N make angles   of 30°,120° and 210° respectively with  the positive direction of the x axis.  By scale drawing find the magnitude  and the direction of the resultant   force

$${three}\:{forces}\:{having}\:{equal}\:{magnitude} \\ $$$${s}\:{of}\:\mathrm{10}{N},\mathrm{20}{N}\:{and}\:\mathrm{30}{N}\:{make}\:{angles}\: \\ $$$${of}\:\mathrm{30}°,\mathrm{120}°\:{and}\:\mathrm{210}°\:{respectively}\:{with} \\ $$$${the}\:{positive}\:{direction}\:{of}\:{the}\:{x}\:{axis}. \\ $$$${By}\:{scale}\:{drawing}\:{find}\:{the}\:{magnitude} \\ $$$${and}\:{the}\:{direction}\:{of}\:{the}\:{resultant}\: \\ $$$${force} \\ $$

Commented by MJS last updated on 23/Jun/19

if they have equal magnitudes how can these forces  be 10, 20, 30?

$$\mathrm{if}\:\mathrm{they}\:\mathrm{have}\:\mathrm{equal}\:\mathrm{magnitudes}\:\mathrm{how}\:\mathrm{can}\:\mathrm{these}\:\mathrm{forces} \\ $$$$\mathrm{be}\:\mathrm{10},\:\mathrm{20},\:\mathrm{30}? \\ $$

Answered by MJS last updated on 23/Jun/19

F_1 =f_1  (((cos 30°)),((sin 30°)) )=f_1  ((((√3)/2)),((1/2)) )  F_2 =f_2  (((cos 120°)),((sin 120°)) )=f_2  (((−(1/2))),(((√3)/2)) )  F_3 =f_3  (((cos 210°)),((sin 210°)) )=f_3  (((−((√3)/2))),((−(1/2))) )  F_1 +F_2 +F_3 =(1/2) ((((√3)f_1 −f_2 −(√3)f_3 )),((f_1 +(√3)f_2 −f_3 )) )=       [if f_1 =f_2 =f_3 =f]  =(f/2) ((1),((√3)) )

$${F}_{\mathrm{1}} ={f}_{\mathrm{1}} \begin{pmatrix}{\mathrm{cos}\:\mathrm{30}°}\\{\mathrm{sin}\:\mathrm{30}°}\end{pmatrix}={f}_{\mathrm{1}} \begin{pmatrix}{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}\\{\frac{\mathrm{1}}{\mathrm{2}}}\end{pmatrix} \\ $$$${F}_{\mathrm{2}} ={f}_{\mathrm{2}} \begin{pmatrix}{\mathrm{cos}\:\mathrm{120}°}\\{\mathrm{sin}\:\mathrm{120}°}\end{pmatrix}={f}_{\mathrm{2}} \begin{pmatrix}{−\frac{\mathrm{1}}{\mathrm{2}}}\\{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}\end{pmatrix} \\ $$$${F}_{\mathrm{3}} ={f}_{\mathrm{3}} \begin{pmatrix}{\mathrm{cos}\:\mathrm{210}°}\\{\mathrm{sin}\:\mathrm{210}°}\end{pmatrix}={f}_{\mathrm{3}} \begin{pmatrix}{−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}\\{−\frac{\mathrm{1}}{\mathrm{2}}}\end{pmatrix} \\ $$$${F}_{\mathrm{1}} +{F}_{\mathrm{2}} +{F}_{\mathrm{3}} =\frac{\mathrm{1}}{\mathrm{2}}\begin{pmatrix}{\sqrt{\mathrm{3}}{f}_{\mathrm{1}} −{f}_{\mathrm{2}} −\sqrt{\mathrm{3}}{f}_{\mathrm{3}} }\\{{f}_{\mathrm{1}} +\sqrt{\mathrm{3}}{f}_{\mathrm{2}} −{f}_{\mathrm{3}} }\end{pmatrix}= \\ $$$$\:\:\:\:\:\left[\mathrm{if}\:{f}_{\mathrm{1}} ={f}_{\mathrm{2}} ={f}_{\mathrm{3}} ={f}\right] \\ $$$$=\frac{{f}}{\mathrm{2}}\begin{pmatrix}{\mathrm{1}}\\{\sqrt{\mathrm{3}}}\end{pmatrix} \\ $$

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