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Question Number 27128 by NECx last updated on 02/Jan/18

A body resting on a rough horizontal plane require a pull of 18N inclined at 30° to the plane first to move it.It was found that a push of 22N inclined at 30° to the plane just moved the body. Determine the weight and coefficient of friction.

$${A}\:{body}\:{resting}\:{on}\:{a}\:{rough} \\ $$$${horizontal}\:{plane}\:{require}\:{a}\:{pull}\:{of} \\ $$$$\mathrm{18}{N}\:{inclined}\:{at}\:\mathrm{30}°\:{to}\:{the}\:{plane} \\ $$$${first}\:{to}\:{move}\:{it}.{It}\:{was}\:{found} \\ $$$${that}\:{a}\:{push}\:{of}\:\mathrm{22}{N}\:{inclined}\:{at}\:\mathrm{30}° \\ $$$${to}\:{the}\:{plane}\:{just}\:{moved}\:{the}\:{body}. \\ $$$${Determine}\:{the}\:{weight}\:{and}\: \\ $$$${coefficient}\:{of}\:{friction}. \\ $$

Answered by mrW1 last updated on 02/Jan/18

Pull: F_1 cos α_1 =μ(mg−F_1 sin α_1 ) ...(i) Push: F_2 cos α_2 =μ(mg+F_2 sin α_2 ) ...(ii) (i)/(ii): ⇒((F_1 cos α_1 )/(F_2 cos α_2 ))=((mg−F_1 sin α_1 )/(mg+F_2 sin α_2 )) ⇒((F_1 cos α_1 )/(F_2 cos α_2 ))=((mg+F_2 sin α_2 −F_1 sin α_1 −F_2 sin α_2 )/(mg+F_2 sin α_2 )) ⇒((F_1 cos α_1 )/(F_2 cos α_2 ))=1−((F_1 sin α_1 +F_2 sin α_2 )/(mg+F_2 sin α_2 )) ⇒1−((F_1 cos α_1 )/(F_2 cos α_2 ))=((F_1 sin α_1 +F_2 sin α_2 )/(mg+F_2 sin α_2 )) ⇒mg=((F_1 sin α_1 +F_2 sin α_2 )/(1−((F_1 cos α_1 )/(F_2 cos α_2 ))))−F_2 sin α_2 ⇒mg=F_2 sin α_2 (((1+((F_1 sin α_1 )/(F_2 sin α_2 )))/(1−((F_1 cos α_1 )/(F_2 cos α_2 ))))−1) α_1 =α_2 =30° ⇒mg=22×(1/2)(((1+((18)/(22)))/(1−((18)/(22))))−1)=11×9=99 N (ii)−(i): F_2 cos α_2 −F_1 cos α_1 =μ(F_2 sin α_2 +F_1 sin α_1 ) ⇒μ=((F_2 cos α_2 −F_1 cos α_1 )/(F_2 sin α_2 +F_1 sin α_1 )) ⇒μ=((22−18)/(22+18))×((cos 30°)/(sin 30°))=((√3)/(10))=0.173

$${Pull}:\:{F}_{\mathrm{1}} \mathrm{cos}\:\alpha_{\mathrm{1}} =\mu\left({mg}−{F}_{\mathrm{1}} \mathrm{sin}\:\alpha_{\mathrm{1}} \right)\:\:\:...\left({i}\right) \\ $$$${Push}:\:{F}_{\mathrm{2}} \mathrm{cos}\:\alpha_{\mathrm{2}} =\mu\left({mg}+{F}_{\mathrm{2}} \mathrm{sin}\:\alpha_{\mathrm{2}} \right)\:\:\:\:...\left({ii}\right) \\ $$$$\left({i}\right)/\left({ii}\right): \\ $$$$\Rightarrow\frac{{F}_{\mathrm{1}} \mathrm{cos}\:\alpha_{\mathrm{1}} }{{F}_{\mathrm{2}} \mathrm{cos}\:\alpha_{\mathrm{2}} }=\frac{{mg}−{F}_{\mathrm{1}} \mathrm{sin}\:\alpha_{\mathrm{1}} }{{mg}+{F}_{\mathrm{2}} \mathrm{sin}\:\alpha_{\mathrm{2}} } \\ $$$$\Rightarrow\frac{{F}_{\mathrm{1}} \mathrm{cos}\:\alpha_{\mathrm{1}} }{{F}_{\mathrm{2}} \mathrm{cos}\:\alpha_{\mathrm{2}} }=\frac{{mg}+{F}_{\mathrm{2}} \mathrm{sin}\:\alpha_{\mathrm{2}} −{F}_{\mathrm{1}} \mathrm{sin}\:\alpha_{\mathrm{1}} −{F}_{\mathrm{2}} \mathrm{sin}\:\alpha_{\mathrm{2}} }{{mg}+{F}_{\mathrm{2}} \mathrm{sin}\:\alpha_{\mathrm{2}} } \\ $$$$\Rightarrow\frac{{F}_{\mathrm{1}} \mathrm{cos}\:\alpha_{\mathrm{1}} }{{F}_{\mathrm{2}} \mathrm{cos}\:\alpha_{\mathrm{2}} }=\mathrm{1}−\frac{{F}_{\mathrm{1}} \mathrm{sin}\:\alpha_{\mathrm{1}} +{F}_{\mathrm{2}} \mathrm{sin}\:\alpha_{\mathrm{2}} }{{mg}+{F}_{\mathrm{2}} \mathrm{sin}\:\alpha_{\mathrm{2}} } \\ $$$$\Rightarrow\mathrm{1}−\frac{{F}_{\mathrm{1}} \mathrm{cos}\:\alpha_{\mathrm{1}} }{{F}_{\mathrm{2}} \mathrm{cos}\:\alpha_{\mathrm{2}} }=\frac{{F}_{\mathrm{1}} \mathrm{sin}\:\alpha_{\mathrm{1}} +{F}_{\mathrm{2}} \mathrm{sin}\:\alpha_{\mathrm{2}} }{{mg}+{F}_{\mathrm{2}} \mathrm{sin}\:\alpha_{\mathrm{2}} } \\ $$$$\Rightarrow{mg}=\frac{{F}_{\mathrm{1}} \mathrm{sin}\:\alpha_{\mathrm{1}} +{F}_{\mathrm{2}} \mathrm{sin}\:\alpha_{\mathrm{2}} }{\mathrm{1}−\frac{{F}_{\mathrm{1}} \mathrm{cos}\:\alpha_{\mathrm{1}} }{{F}_{\mathrm{2}} \mathrm{cos}\:\alpha_{\mathrm{2}} }}−{F}_{\mathrm{2}} \mathrm{sin}\:\alpha_{\mathrm{2}} \\ $$$$\Rightarrow{mg}={F}_{\mathrm{2}} \mathrm{sin}\:\alpha_{\mathrm{2}} \left(\frac{\mathrm{1}+\frac{{F}_{\mathrm{1}} \mathrm{sin}\:\alpha_{\mathrm{1}} }{{F}_{\mathrm{2}} \mathrm{sin}\:\alpha_{\mathrm{2}} }}{\mathrm{1}−\frac{{F}_{\mathrm{1}} \mathrm{cos}\:\alpha_{\mathrm{1}} }{{F}_{\mathrm{2}} \mathrm{cos}\:\alpha_{\mathrm{2}} }}−\mathrm{1}\right) \\ $$$$\alpha_{\mathrm{1}} =\alpha_{\mathrm{2}} =\mathrm{30}° \\ $$$$\Rightarrow{mg}=\mathrm{22}×\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}+\frac{\mathrm{18}}{\mathrm{22}}}{\mathrm{1}−\frac{\mathrm{18}}{\mathrm{22}}}−\mathrm{1}\right)=\mathrm{11}×\mathrm{9}=\mathrm{99}\:{N} \\ $$$$ \\ $$$$\left({ii}\right)−\left({i}\right): \\ $$$${F}_{\mathrm{2}} \mathrm{cos}\:\alpha_{\mathrm{2}} −{F}_{\mathrm{1}} \mathrm{cos}\:\alpha_{\mathrm{1}} =\mu\left({F}_{\mathrm{2}} \mathrm{sin}\:\alpha_{\mathrm{2}} +{F}_{\mathrm{1}} \mathrm{sin}\:\alpha_{\mathrm{1}} \right) \\ $$$$\Rightarrow\mu=\frac{{F}_{\mathrm{2}} \mathrm{cos}\:\alpha_{\mathrm{2}} −{F}_{\mathrm{1}} \mathrm{cos}\:\alpha_{\mathrm{1}} }{{F}_{\mathrm{2}} \mathrm{sin}\:\alpha_{\mathrm{2}} +{F}_{\mathrm{1}} \mathrm{sin}\:\alpha_{\mathrm{1}} } \\ $$$$\Rightarrow\mu=\frac{\mathrm{22}−\mathrm{18}}{\mathrm{22}+\mathrm{18}}×\frac{\mathrm{cos}\:\mathrm{30}°}{\mathrm{sin}\:\mathrm{30}°}=\frac{\sqrt{\mathrm{3}}}{\mathrm{10}}=\mathrm{0}.\mathrm{173} \\ $$

Commented by NECx last updated on 02/Jan/18

please can you post a diagram for better understanding. This is because I never knew that there was a difference between the reaction of forces when in a push and pull.

$${please}\:{can}\:{you}\:{post}\:{a}\:{diagram}\:{for} \\ $$$${better}\:{understanding}.\:{This}\:{is} \\ $$$${because}\:{I}\:{never}\:{knew}\:{that}\:{there} \\ $$$${was}\:{a}\:{difference}\:{between}\:{the}\:{reaction} \\ $$$${of}\:{forces}\:{when}\:{in}\:{a}\:{push}\:{and}\:{pull}. \\ $$

Commented by mrW1 last updated on 02/Jan/18

Commented by NECx last updated on 02/Jan/18

please the diagram you displayed is not showing please can you post it as a question.

$${please}\:{the}\:{diagram}\:{you}\:{displayed} \\ $$$${is}\:{not}\:{showing}\:{please}\:{can}\:{you}\:{post} \\ $$$${it}\:{as}\:{a}\:{question}. \\ $$

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