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Question Number 134704 by I want to learn more last updated on 06/Mar/21

Answered by mr W last updated on 07/Mar/21

Commented by mr W last updated on 07/Mar/21

$$\left(a\right)$$$$R=resultantfrom{F}_1,F_2,F_3$$$$R_x=6+9\times \cos 45°=6+\frac{9}{\sqrt{2}}=\frac{12+9\sqrt{2}}{2}$$$$R_y=12+9\times \sin 45°=12+\frac{9}{\sqrt{2}}=\frac{24+9\sqrt{2}}{2}$$$$R=\sqrt{R_x^2+R_y^2}=\frac{\sqrt{{(12+9\sqrt{2})}^2+{(24+9\sqrt{2})}^2}}{2}$$$$=3\sqrt{29+18\sqrt{2}}\approx 22.138units$$$$\tan \theta =\frac{R_y}{R_x}=\frac{24+9\sqrt{2}}{12+9\sqrt{2}}=6\sqrt{2}-7$$$$\Rightarrow \theta ={\tan }^{-1}(6\sqrt{2}-7)\approx 56.05°$$$$\theta =anglebetweenRandx-axis$$$$$$$$\left(b\right)$$$$\tan \alpha =\frac{1}{2}\Rightarrow \sin \alpha =\frac{1}{\sqrt{5}},\cos \alpha =\frac{2}{\sqrt{5}}$$$$F_4\cos \alpha +F_5\sin \alpha +R_x=0$$$$(2F_4+F_5)\frac{1}{\sqrt{5}}+\frac{12+9\sqrt{2}}{2}=0$$$$\Rightarrow 2F_4+F_5=-\frac{\sqrt{5}(12+9\sqrt{2})}{2}...\left(i\right)$$$$F_4\sin \alpha +F_5\cos \alpha +R_y=0$$$$(F_4+2F_5)\frac{1}{\sqrt{5}}+\frac{24+9\sqrt{2}}{2}=0$$$$\Rightarrow F_4+2F_5=-\frac{\sqrt{5}(24+9\sqrt{2})}{2}...\left(ii\right)$$$$2\left(i\right)-\left(ii\right):$$$$3F_4=-2\times \frac{\sqrt{5}(12+9\sqrt{2})}{2}+\frac{\sqrt{5}(24+9\sqrt{2})}{2}$$$$\Rightarrow F_4=-\frac{3\sqrt{10}}{2}\approx -4.74units$$$$2\left(ii\right)-\left(i\right):$$$$3F_5=-2\times \frac{\sqrt{5}(24+9\sqrt{2})}{2}+\frac{\sqrt{5}(12+9\sqrt{2})}{2}$$$$\Rightarrow F_5=-\frac{(12+3\sqrt{2})\sqrt{5}}{2}\approx -18.16units$$$$F_{4}and{F}_{5}areinoppositedirection$$$$asshown.$$

Commented by I want to learn more last updated on 07/Mar/21

$$\mathrm{Wow},\mathrm{thanks}\mathrm{sir},\mathrm{i}\mathrm{really}\mathrm{appreciate}.\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.$$

Commented by I want to learn more last updated on 11/Mar/21

$$\mathrm{Thanks}\mathrm{sir},\mathrm{i}\mathrm{understand}\mathrm{very}\mathrm{well}\mathrm{now}.$$

Commented by I want to learn more last updated on 11/Mar/21

$$\mathrm{Sir},\mathrm{as}\mathrm{i}\mathrm{go}\mathrm{through}\mathrm{the}\mathrm{workings},\mathrm{i}\mathrm{have}\mathrm{challenge}\mathrm{to}\mathrm{underetand}$$$$\mathrm{the}\left(\mathrm{b}\right)\mathrm{part}.\mathrm{How}\tan \left(\alpha \right)=\frac{1}{2}$$$$\mathrm{and}\mathrm{how}\mathrm{we}\mathrm{relate}{\mathrm{F}}_4\cos \left(\alpha \right)+{\mathrm{F}}_5\sin \left(\alpha \right)+{\mathrm{R}}_{\mathrm{x}}=0.$$$$\mathrm{I}\mathrm{understand}\mathrm{every}\mathrm{single}\mathrm{steps}\mathrm{except}\mathrm{those}\mathrm{two}\mathrm{mention}\mathrm{above}.$$

Commented by mr W last updated on 11/Mar/21

$$HismidpointofBCasgiven,$$$$\Rightarrow BH=\frac{BC}{2}=\frac{AB}{2}$$$$\tan \alpha =\frac{BH}{AB}=\frac{1}{2}$$

Commented by mr W last updated on 11/Mar/21

$$here{R}_x=F_{1x}+F_{2x}+F_{3x}frompart\left(a\right)$$$$F_{4x}=F_4\cos \alpha$$$$F_{5x}=F_5\sin \alpha$$$$inequilibrium$$$$\mathrm{\Sigma }{F}_x=F_{1x}+F_{2x}+F_{3x}+F_{4x}+F_{5x}=0$$$$\Rightarrow {\mathrm{F}}_4\cos \left(\alpha \right)+{\mathrm{F}}_5\sin \left(\alpha \right)+{\mathrm{R}}_{\mathrm{x}}=0$$

Commented by I want to learn more last updated on 11/Mar/21

$$\mathrm{Sir},\mathrm{while}\mathrm{checking}\mathrm{for}\mathrm{Ry}$$$$\mathrm{I}\mathrm{think}\mathrm{it}\mathrm{should}\mathrm{be}\mathrm{Ry}=-12-9\times \sin \left(45\right)\mathrm{because}\mathrm{of}\mathrm{the}$$$$\mathrm{direction}.\mathrm{But}\mathrm{you}\mathrm{used}+12+9\sin 45.\mathrm{I}{\mathrm{don}}^{\prime }\mathrm{t}\mathrm{know}\mathrm{why}.$$

Commented by I want to learn more last updated on 11/Mar/21

Commented by mr W last updated on 11/Mar/21

$$youcanseeinmydiagramwhichx-$$$$andy-axisiused!$$$$ifyoulike,itisthesameas:$$

Commented by mr W last updated on 11/Mar/21

Commented by mr W last updated on 11/Mar/21

$$nowyousay{R}_y=12+9\sin 45°isok,$$$$thoughactuallyichangednothing.$$

Commented by I want to learn more last updated on 11/Mar/21

$$\mathrm{Thanks}\mathrm{for}\mathrm{clarification}\mathrm{sir}.\mathrm{I}\mathrm{appreciate}.$$

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