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 | ||
$$\left({a}\right) \\ $$$${R}={resultant}\:{from}\:{F}_{\mathrm{1}} ,{F}_{\mathrm{2}} ,{F}_{\mathrm{3}} \\ $$$${R}_{{x}} =\mathrm{6}+\mathrm{9}×\mathrm{cos}\:\mathrm{45}°=\mathrm{6}+\frac{\mathrm{9}}{\:\sqrt{\mathrm{2}}}=\frac{\mathrm{12}+\mathrm{9}\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$$${R}_{{y}} =\mathrm{12}+\mathrm{9}×\mathrm{sin}\:\mathrm{45}°=\mathrm{12}+\frac{\mathrm{9}}{\:\sqrt{\mathrm{2}}}=\frac{\mathrm{24}+\mathrm{9}\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$$${R}=\sqrt{{R}_{{x}} ^{\mathrm{2}} +{R}_{{y}} ^{\mathrm{2}} }=\frac{\sqrt{\left(\mathrm{12}+\mathrm{9}\sqrt{\mathrm{2}}\right)^{\mathrm{2}} +\left(\mathrm{24}+\mathrm{9}\sqrt{\mathrm{2}}\right)^{\mathrm{2}} }}{\mathrm{2}} \\ $$$$=\mathrm{3}\sqrt{\mathrm{29}+\mathrm{18}\sqrt{\mathrm{2}}}\approx\mathrm{22}.\mathrm{138}\:{units} \\ $$$$\mathrm{tan}\:\theta=\frac{{R}_{{y}} }{{R}_{{x}} }=\frac{\mathrm{24}+\mathrm{9}\sqrt{\mathrm{2}}}{\mathrm{12}+\mathrm{9}\sqrt{\mathrm{2}}}=\mathrm{6}\sqrt{\mathrm{2}}−\mathrm{7} \\ $$$$\Rightarrow\theta=\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{6}\sqrt{\mathrm{2}}−\mathrm{7}\right)\approx\mathrm{56}.\mathrm{05}° \\ $$$$\theta={angle}\:{between}\:{R}\:{and}\:{x}−{axis} \\ $$$$ \\ $$$$\left({b}\right) \\ $$$$\mathrm{tan}\:\alpha=\frac{\mathrm{1}}{\mathrm{2}}\Rightarrow\mathrm{sin}\:\alpha=\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}},\:\mathrm{cos}\:\alpha=\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}} \\ $$$${F}_{\mathrm{4}} \mathrm{cos}\:\alpha+{F}_{\mathrm{5}} \mathrm{sin}\:\alpha+{R}_{{x}} =\mathrm{0} \\ $$$$\left(\mathrm{2}{F}_{\mathrm{4}} +{F}_{\mathrm{5}} \right)\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}+\frac{\mathrm{12}+\mathrm{9}\sqrt{\mathrm{2}}}{\mathrm{2}}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{2}{F}_{\mathrm{4}} +{F}_{\mathrm{5}} =−\frac{\sqrt{\mathrm{5}}\left(\mathrm{12}+\mathrm{9}\sqrt{\mathrm{2}}\right)}{\mathrm{2}}\:\:\:...\left({i}\right) \\ $$$${F}_{\mathrm{4}} \mathrm{sin}\:\alpha+{F}_{\mathrm{5}} \mathrm{cos}\:\alpha+{R}_{{y}} =\mathrm{0} \\ $$$$\left({F}_{\mathrm{4}} +\mathrm{2}{F}_{\mathrm{5}} \right)\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}+\frac{\mathrm{24}+\mathrm{9}\sqrt{\mathrm{2}}}{\mathrm{2}}=\mathrm{0} \\ $$$$\Rightarrow{F}_{\mathrm{4}} +\mathrm{2}{F}_{\mathrm{5}} =−\frac{\sqrt{\mathrm{5}}\left(\mathrm{24}+\mathrm{9}\sqrt{\mathrm{2}}\right)}{\mathrm{2}}\:\:\:...\left({ii}\right) \\ $$$$\mathrm{2}\left({i}\right)−\left({ii}\right): \\ $$$$\mathrm{3}{F}_{\mathrm{4}} =−\mathrm{2}×\frac{\sqrt{\mathrm{5}}\left(\mathrm{12}+\mathrm{9}\sqrt{\mathrm{2}}\right)}{\mathrm{2}}+\frac{\sqrt{\mathrm{5}}\left(\mathrm{24}+\mathrm{9}\sqrt{\mathrm{2}}\right)}{\mathrm{2}} \\ $$$$\Rightarrow{F}_{\mathrm{4}} =−\frac{\mathrm{3}\sqrt{\mathrm{10}}}{\mathrm{2}}\approx−\mathrm{4}.\mathrm{74}\:{units} \\ $$$$\mathrm{2}\left({ii}\right)−\left({i}\right): \\ $$$$\mathrm{3}{F}_{\mathrm{5}} =−\mathrm{2}×\frac{\sqrt{\mathrm{5}}\left(\mathrm{24}+\mathrm{9}\sqrt{\mathrm{2}}\right)}{\mathrm{2}}+\frac{\sqrt{\mathrm{5}}\left(\mathrm{12}+\mathrm{9}\sqrt{\mathrm{2}}\right)}{\mathrm{2}} \\ $$$$\Rightarrow{F}_{\mathrm{5}} =−\frac{\left(\mathrm{12}+\mathrm{3}\sqrt{\mathrm{2}}\right)\sqrt{\mathrm{5}}}{\mathrm{2}}\approx−\mathrm{18}.\mathrm{16}\:{units} \\ $$$${F}_{\mathrm{4}} \:{and}\:{F}_{\mathrm{5}} \:{are}\:{in}\:{opposite}\:{direction} \\ $$$${as}\:{shown}. \\ $$ | ||
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 | ||
$$\mathrm{Sir},\:\mathrm{as}\:\mathrm{i}\:\mathrm{go}\:\mathrm{through}\:\mathrm{the}\:\mathrm{workings},\:\mathrm{i}\:\mathrm{have}\:\mathrm{challenge}\:\mathrm{to}\:\mathrm{underetand} \\ $$$$\mathrm{the}\:\:\left(\mathrm{b}\right)\:\:\mathrm{part}.\:\:\mathrm{How}\:\:\:\:\mathrm{tan}\left(\alpha\right)\:\:=\:\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{and}\:\mathrm{how}\:\mathrm{we}\:\mathrm{relate}\:\:\:\:\:\mathrm{F}_{\mathrm{4}} \mathrm{cos}\left(\alpha\right)\:\:+\:\:\mathrm{F}_{\mathrm{5}} \mathrm{sin}\left(\alpha\right)\:\:+\:\:\mathrm{R}_{\mathrm{x}} \:\:=\:\:\mathrm{0}. \\ $$$$\mathrm{I}\:\mathrm{understand}\:\mathrm{every}\:\mathrm{single}\:\mathrm{steps}\:\mathrm{except}\:\mathrm{those}\:\mathrm{two}\:\mathrm{mention}\:\mathrm{above}. \\ $$ | ||
Commented by mr W last updated on 11/Mar/21 | ||
$${H}\:{is}\:{midpoint}\:{of}\:{BC}\:{as}\:{given}, \\ $$$$\Rightarrow{BH}=\frac{{BC}}{\mathrm{2}}=\frac{{AB}}{\mathrm{2}} \\ $$$$\mathrm{tan}\:\alpha=\frac{{BH}}{{AB}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$ | ||
Commented by mr W last updated on 11/Mar/21 | ||
$${here}\:{R}_{{x}} ={F}_{\mathrm{1}{x}} +{F}_{\mathrm{2}{x}} +{F}_{\mathrm{3}{x}} \:{from}\:{part}\:\left({a}\right) \\ $$$${F}_{\mathrm{4}{x}} ={F}_{\mathrm{4}} \mathrm{cos}\:\alpha \\ $$$${F}_{\mathrm{5}{x}} ={F}_{\mathrm{5}} \mathrm{sin}\:\alpha \\ $$$${in}\:{equilibrium} \\ $$$$\Sigma{F}_{{x}} ={F}_{\mathrm{1}{x}} +{F}_{\mathrm{2}{x}} +{F}_{\mathrm{3}{x}} +{F}_{\mathrm{4}{x}} +{F}_{\mathrm{5}{x}} =\mathrm{0} \\ $$$$\Rightarrow\mathrm{F}_{\mathrm{4}} \mathrm{cos}\left(\alpha\right)\:\:+\:\:\mathrm{F}_{\mathrm{5}} \mathrm{sin}\left(\alpha\right)\:\:+\:\:\mathrm{R}_{\mathrm{x}} \:\:=\:\:\mathrm{0} \\ $$ | ||
Commented by I want to learn more last updated on 11/Mar/21 | ||
$$\mathrm{Sir},\:\mathrm{while}\:\mathrm{checking}\:\mathrm{for}\:\:\mathrm{Ry} \\ $$$$\mathrm{I}\:\mathrm{think}\:\mathrm{it}\:\mathrm{should}\:\mathrm{be}\:\:\:\:\:\mathrm{Ry}\:\:=\:\:\:−\:\:\mathrm{12}\:\:\:−\:\:\:\mathrm{9}\:×\:\mathrm{sin}\left(\mathrm{45}\right)\:\:\:\mathrm{because}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{direction}.\:\mathrm{But}\:\mathrm{you}\:\mathrm{used}\:\:+\:\mathrm{12}\:\:+\:\:\mathrm{9}\:\mathrm{sin}\:\mathrm{45}.\:\:\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{know}\:\mathrm{why}. \\ $$ | ||
Commented by I want to learn more last updated on 11/Mar/21 | ||
Commented by mr W last updated on 11/Mar/21 | ||
$${you}\:{can}\:{see}\:{in}\:{my}\:{diagram}\:{which}\:{x}− \\ $$$${and}\:{y}−{axis}\:{i}\:{used}! \\ $$$${if}\:{you}\:{like},\:{it}\:{is}\:{the}\:{same}\:{as}: \\ $$ | ||
Commented by mr W last updated on 11/Mar/21 | ||
Commented by mr W last updated on 11/Mar/21 | ||
$${now}\:{you}\:{say}\:{R}_{{y}} =\mathrm{12}+\mathrm{9}\:\mathrm{sin}\:\mathrm{45}°\:{is}\:{ok}, \\ $$$${though}\:{actually}\:\:{i}\:{changed}\:{nothing}. \\ $$ | ||
Commented by I want to learn more last updated on 11/Mar/21 | ||
$$\mathrm{Thanks}\:\mathrm{for}\:\mathrm{clarification}\:\mathrm{sir}.\:\mathrm{I}\:\mathrm{appreciate}. \\ $$ | ||