A-200-N-force-inclined-at-40-above-the-horizontal-drag-load-along-the-horizontal-floor-coefficient-of-the-kinetic-friction-between-the-load-is-0-30-and-the-load-experiences-an-acceleration-of-1-2

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Question Number 10914 by Saham last updated on 02/Mar/17

$$\mathrm{A}200\mathrm{N}\mathrm{force}\mathrm{inclined}\mathrm{at}40°\mathrm{above}\mathrm{the}\mathrm{horizontal},\mathrm{drag}\mathrm{load}\mathrm{along}\mathrm{the}$$$$\mathrm{horizontal}\mathrm{floor}.\mathrm{coefficient}\mathrm{of}\mathrm{the}\mathrm{kinetic}\mathrm{friction}\mathrm{between}\mathrm{the}\mathrm{load}\mathrm{is}0.30$$$$\mathrm{and}\mathrm{the}\mathrm{load}\mathrm{experiences}\mathrm{an}\mathrm{acceleration}\mathrm{of}1.2\mathrm{m}/{\mathrm{s}}^2,$$$$\mathrm{What}\mathrm{is}\mathrm{the}\mathrm{mass}\mathrm{of}\mathrm{the}\mathrm{load}.$$

Answered by sandy_suhendra last updated on 02/Mar/17

Commented by sandy_suhendra last updated on 02/Mar/17

$${\mathrm{F}}_{\mathrm{H}}=\mathrm{Fcos}40°=200\times 0.766=153.2\mathrm{N}$$$${\mathrm{F}}_{\mathrm{V}}=\mathrm{Fsin}40°=200\times 0.643=128.6\mathrm{N}$$$$\mathrm{N}=\mathrm{w}-{\mathrm{F}}_{\mathrm{V}}=(10\mathrm{m}-128.6)\mathrm{N}$$$$\mathrm{f}={\mu }_{\mathrm{k}}.\mathrm{N}=0.3\times (10\mathrm{m}-128.6)=(3\mathrm{m}-38.58)\mathrm{N}$$$$$$$$\mathrm{\Sigma }{\mathrm{F}}_{\mathrm{x}}=\mathrm{m}.\mathrm{a}$$$${\mathrm{F}}_{\mathrm{H}}-\mathrm{f}=\mathrm{m}.\mathrm{a}$$$$153.2-(3\mathrm{m}-38.58)=1.2\mathrm{m}$$$$191.78-3\mathrm{m}=1.2\mathrm{m}$$$$4.2\mathrm{m}=191.78$$$$\mathrm{m}=45.66\mathrm{kg}$$$$(\mathrm{I}\mathrm{use}\mathrm{gravity}\mathrm{acceleration}=10\mathrm{m}/{\mathrm{s}}^2)$$

Commented by Saham last updated on 02/Mar/17

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.\mathrm{i}\mathrm{really}\mathrm{appreciate}\mathrm{your}\mathrm{effort}.$$

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