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aman-walks-600m-at-a-bearing-of-45-0-then-500m-at-a-bearing-90-0-then-300m-at-bearing-of-135-0-then-400m-at-a-bearing-of-225-0-find-the-resultant-displacement-which-the-man-has-made-




Question Number 10475 by ketto last updated on 13/Feb/17
aman walks 600m at a bearing of   45^(0 ) then 500m at a bearing 90^0    then 300m at bearing of 135^(0 )    then 400m at a bearing of 225^0   .find the resultant displacement  which the man has made
$${aman}\:{walks}\:\mathrm{600}{m}\:{at}\:{a}\:{bearing}\:{of}\: \\ $$$$\mathrm{45}^{\mathrm{0}\:} {then}\:\mathrm{500}{m}\:{at}\:{a}\:{bearing}\:\mathrm{90}^{\mathrm{0}} \: \\ $$$${then}\:\mathrm{300}{m}\:{at}\:{bearing}\:{of}\:\mathrm{135}^{\mathrm{0}\:} \: \\ $$$${then}\:\mathrm{400}{m}\:{at}\:{a}\:{bearing}\:{of}\:\mathrm{225}^{\mathrm{0}} \\ $$$$.{find}\:{the}\:{resultant}\:{displacement} \\ $$$${which}\:{the}\:{man}\:{has}\:{made} \\ $$
Answered by sandy_suhendra last updated on 13/Feb/17
ΣF_x =600 cos 45° + 500 cos 90° + 300 cos 135° + 400 cos 225°           =300(√2) + 0 −150(√2) − 200(√2)           =−50(√2) m  ΣF_y  = 600 sin 45° + 500 sin 90° + 300 sin 135° + 400 sin 225°                   = 300(√2) + 500 + 150(√2) −200(√2)            = (250(√2) + 500) m  F_R  = (√((−50(√2))^2 +(250(√2)+500)^2 ))         =856.48 m
$$\Sigma\mathrm{F}_{\mathrm{x}} =\mathrm{600}\:\mathrm{cos}\:\mathrm{45}°\:+\:\mathrm{500}\:\mathrm{cos}\:\mathrm{90}°\:+\:\mathrm{300}\:\mathrm{cos}\:\mathrm{135}°\:+\:\mathrm{400}\:\mathrm{cos}\:\mathrm{225}° \\ $$$$\:\:\:\:\:\:\:\:\:=\mathrm{300}\sqrt{\mathrm{2}}\:+\:\mathrm{0}\:−\mathrm{150}\sqrt{\mathrm{2}}\:−\:\mathrm{200}\sqrt{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:=−\mathrm{50}\sqrt{\mathrm{2}}\:\mathrm{m} \\ $$$$\Sigma\mathrm{F}_{\mathrm{y}} \:=\:\mathrm{600}\:\mathrm{sin}\:\mathrm{45}°\:+\:\mathrm{500}\:\mathrm{sin}\:\mathrm{90}°\:+\:\mathrm{300}\:\mathrm{sin}\:\mathrm{135}°\:+\:\mathrm{400}\:\mathrm{sin}\:\mathrm{225}°\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:=\:\mathrm{300}\sqrt{\mathrm{2}}\:+\:\mathrm{500}\:+\:\mathrm{150}\sqrt{\mathrm{2}}\:−\mathrm{200}\sqrt{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\:\left(\mathrm{250}\sqrt{\mathrm{2}}\:+\:\mathrm{500}\right)\:\mathrm{m} \\ $$$$\mathrm{F}_{\mathrm{R}} \:=\:\sqrt{\left(−\mathrm{50}\sqrt{\mathrm{2}}\right)^{\mathrm{2}} +\left(\mathrm{250}\sqrt{\mathrm{2}}+\mathrm{500}\right)^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:=\mathrm{856}.\mathrm{48}\:\mathrm{m} \\ $$

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