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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
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 m/s^2 , What is the mass of the load.
$$\mathrm{A}\:\mathrm{200}\:\mathrm{N}\:\mathrm{force}\:\mathrm{inclined}\:\mathrm{at}\:\mathrm{40}°\:\mathrm{above}\:\mathrm{the}\:\mathrm{horizontal}\:,\:\mathrm{drag}\:\mathrm{load}\:\mathrm{along}\:\mathrm{the} \\ $$$$\mathrm{horizontal}\:\mathrm{floor}.\:\mathrm{coefficient}\:\mathrm{of}\:\mathrm{the}\:\mathrm{kinetic}\:\mathrm{friction}\:\mathrm{between}\:\mathrm{the}\:\mathrm{load}\:\mathrm{is}\:\mathrm{0}.\mathrm{30}\: \\ $$$$\mathrm{and}\:\mathrm{the}\:\mathrm{load}\:\mathrm{experiences}\:\mathrm{an}\:\mathrm{acceleration}\:\mathrm{of}\:\mathrm{1}.\mathrm{2}\:\mathrm{m}/\mathrm{s}^{\mathrm{2}} , \\ $$$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{mass}\:\mathrm{of}\:\mathrm{the}\:\mathrm{load}. \\ $$
Answered by sandy_suhendra last updated on 02/Mar/17
Commented by sandy_suhendra last updated on 02/Mar/17
F_H =Fcos40°=200×0.766=153.2 N F_V =Fsin40°=200×0.643=128.6 N N=w−F_V =(10m−128.6) N f=μ_k .N=0.3×(10m−128.6)=(3m−38.58) N ΣF_x =m.a F_H −f=m.a 153.2−(3m−38.58)=1.2m 191.78−3m=1.2m 4.2m=191.78 m=45.66 kg (I use gravity acceleration=10 m/s^2 )
$$\mathrm{F}_{\mathrm{H}} =\mathrm{Fcos40}°=\mathrm{200}×\mathrm{0}.\mathrm{766}=\mathrm{153}.\mathrm{2}\:\mathrm{N} \\ $$$$\mathrm{F}_{\mathrm{V}} =\mathrm{Fsin40}°=\mathrm{200}×\mathrm{0}.\mathrm{643}=\mathrm{128}.\mathrm{6}\:\mathrm{N} \\ $$$$\mathrm{N}=\mathrm{w}−\mathrm{F}_{\mathrm{V}} =\left(\mathrm{10m}−\mathrm{128}.\mathrm{6}\right)\:\mathrm{N} \\ $$$$\mathrm{f}=\mu_{\mathrm{k}} .\mathrm{N}=\mathrm{0}.\mathrm{3}×\left(\mathrm{10m}−\mathrm{128}.\mathrm{6}\right)=\left(\mathrm{3m}−\mathrm{38}.\mathrm{58}\right)\:\mathrm{N}\:\:\:\:\: \\ $$$$ \\ $$$$\Sigma\mathrm{F}_{\mathrm{x}} =\mathrm{m}.\mathrm{a} \\ $$$$\mathrm{F}_{\mathrm{H}} −\mathrm{f}=\mathrm{m}.\mathrm{a} \\ $$$$\mathrm{153}.\mathrm{2}−\left(\mathrm{3m}−\mathrm{38}.\mathrm{58}\right)=\mathrm{1}.\mathrm{2m} \\ $$$$\mathrm{191}.\mathrm{78}−\mathrm{3m}=\mathrm{1}.\mathrm{2m} \\ $$$$\mathrm{4}.\mathrm{2m}=\mathrm{191}.\mathrm{78} \\ $$$$\mathrm{m}=\mathrm{45}.\mathrm{66}\:\mathrm{kg} \\ $$$$\left(\mathrm{I}\:\mathrm{use}\:\mathrm{gravity}\:\mathrm{acceleration}=\mathrm{10}\:\mathrm{m}/\mathrm{s}^{\mathrm{2}} \right) \\ $$
Commented by Saham last updated on 02/Mar/17
God bless you sir. i really appreciate your effort.
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{i}\:\mathrm{really}\:\mathrm{appreciate}\:\mathrm{your}\:\mathrm{effort}. \\ $$

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