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Question Number 111282 by Aina Samuel Temidayo last updated on 03/Sep/20

A particle of mass 500g is placed on a plane inclined at an angle 30° to the horizontal. What force is (i) acting parallel to the plane (ii) acting horizontally is required to hold the particle at rest (g=10m/s^2 )

$$\mathrm{A}\:\mathrm{particle}\:\mathrm{of}\:\mathrm{mass}\:\mathrm{500g}\:\mathrm{is}\:\mathrm{placed}\:\mathrm{on}\:\mathrm{a}\: \\ $$$$\mathrm{plane}\:\mathrm{inclined}\:\mathrm{at}\:\mathrm{an}\:\mathrm{angle}\:\mathrm{30}° \\ $$$$\mathrm{to}\:\mathrm{the}\:\mathrm{horizontal}.\:\mathrm{What}\:\mathrm{force}\:\mathrm{is} \\ $$$$\left(\mathrm{i}\right)\:\mathrm{acting}\:\mathrm{parallel}\:\mathrm{to}\:\mathrm{the}\:\mathrm{plane} \\ $$$$\left(\mathrm{ii}\right)\:\mathrm{acting}\:\mathrm{horizontally}\:\mathrm{is}\:\mathrm{required}\:\mathrm{to} \\ $$$$\mathrm{hold}\:\mathrm{the}\:\mathrm{particle}\:\mathrm{at}\:\mathrm{rest}\:\left(\mathrm{g}=\mathrm{10m}/\mathrm{s}^{\mathrm{2}} \right) \\ $$

Answered by ajfour last updated on 03/Sep/20

Commented by Aina Samuel Temidayo last updated on 03/Sep/20

Solution please?

$$\mathrm{Solution}\:\mathrm{please}? \\ $$

Commented by ajfour last updated on 03/Sep/20

(i) F=mgsin 30° = (1/2)×10×(1/2) = 2.5N (ii) F= Nsin 30° Ncos 30° = mg = (1/2)×10 = 5 ⇒ F=(5/(cos 30°))×sin 30° = (5/( (√3))) newtons.

$$\left({i}\right)\:\:\:{F}={mg}\mathrm{sin}\:\mathrm{30}° \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{10}×\frac{\mathrm{1}}{\mathrm{2}}\:=\:\mathrm{2}.\mathrm{5}{N} \\ $$$$\left({ii}\right)\:\:\:{F}=\:{N}\mathrm{sin}\:\mathrm{30}° \\ $$$$\:\:\:\:\:\:\:\:\:\:{N}\mathrm{cos}\:\mathrm{30}°\:=\:{mg}\:=\:\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{10}\:=\:\mathrm{5} \\ $$$$\Rightarrow\:\:{F}=\frac{\mathrm{5}}{\mathrm{cos}\:\mathrm{30}°}×\mathrm{sin}\:\mathrm{30}°\:=\:\frac{\mathrm{5}}{\:\sqrt{\mathrm{3}}}\:{newtons}. \\ $$

Commented by Aina Samuel Temidayo last updated on 03/Sep/20

Kindly explain no. 2 in details please. Thanks.

$$\mathrm{Kindly}\:\mathrm{explain}\:\mathrm{no}.\:\mathrm{2}\:\mathrm{in}\:\mathrm{details}\:\mathrm{please}. \\ $$$$\mathrm{Thanks}. \\ $$

Commented by ajfour last updated on 03/Sep/20

force balance along vertical and horizontal gives: Ncos 30°=mg Nsin 30°=F dividing (F/(mg)) = tan 30° = (1/( (√3))) ⇒ F=((mg)/( (√3))) = (((1/2)×10)/( (√3))) = (5/( (√3))) = ((5(√3))/3) .

$${force}\:{balance}\:{along}\:{vertical}\:{and} \\ $$$${horizontal}\:{gives}: \\ $$$${N}\mathrm{cos}\:\mathrm{30}°={mg} \\ $$$${N}\mathrm{sin}\:\mathrm{30}°={F} \\ $$$${dividing} \\ $$$$\frac{{F}}{{mg}}\:=\:\mathrm{tan}\:\mathrm{30}°\:=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}} \\ $$$$\Rightarrow\:\:{F}=\frac{{mg}}{\:\sqrt{\mathrm{3}}}\:=\:\frac{\left(\mathrm{1}/\mathrm{2}\right)×\mathrm{10}}{\:\sqrt{\mathrm{3}}}\:=\:\frac{\mathrm{5}}{\:\sqrt{\mathrm{3}}}\:=\:\frac{\mathrm{5}\sqrt{\mathrm{3}}}{\mathrm{3}}\:. \\ $$

Commented by Aina Samuel Temidayo last updated on 03/Sep/20

I don′t really understand the diagram for (ii)

$$\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{really}\:\mathrm{understand}\:\mathrm{the}\:\mathrm{diagram} \\ $$$$\mathrm{for}\:\left(\mathrm{ii}\right) \\ $$

Commented by Aina Samuel Temidayo last updated on 03/Sep/20

Is that the correct position for F?

$$\mathrm{Is}\:\mathrm{that}\:\mathrm{the}\:\mathrm{correct}\:\mathrm{position}\:\mathrm{for}\:\mathrm{F}? \\ $$

Commented by ajfour last updated on 03/Sep/20

yes it′ll be like i′ve drawn.

$${yes}\:{it}'{ll}\:{be}\:{like}\:{i}'{ve}\:{drawn}. \\ $$

Commented by Aina Samuel Temidayo last updated on 03/Sep/20

Why is that so?

$$\mathrm{Why}\:\mathrm{is}\:\mathrm{that}\:\mathrm{so}? \\ $$

Commented by Aina Samuel Temidayo last updated on 03/Sep/20

Is it horizontal to the plane like that?

$$\mathrm{Is}\:\mathrm{it}\:\mathrm{horizontal}\:\mathrm{to}\:\mathrm{the}\:\mathrm{plane}\:\mathrm{like}\:\mathrm{that}? \\ $$

Commented by ajfour last updated on 03/Sep/20

yes the force that is horizontal and needed to hold the particle against the frictionless incline will be directed suchwise and F=mgtan θ. horizontal means parallel to flat ground, you should not say horizontal to plane (inclined)!

$${yes}\:{the}\:{force}\:{that}\:{is}\:{horizontal} \\ $$$${and}\:{needed}\:{to}\:{hold}\:{the}\:{particle} \\ $$$${against}\:{the}\:{frictionless}\:{incline} \\ $$$${will}\:{be}\:{directed}\:{suchwise}\:{and} \\ $$$${F}={mg}\mathrm{tan}\:\theta. \\ $$$${horizontal}\:{means}\:{parallel}\:{to}\:{flat} \\ $$$${ground},\:{you}\:{should}\:{not}\:{say}\:{horizontal} \\ $$$${to}\:{plane}\:\left({inclined}\right)! \\ $$

Commented by Aina Samuel Temidayo last updated on 03/Sep/20

What about the dotted line opposite to it?

$$\mathrm{What}\:\mathrm{about}\:\mathrm{the}\:\mathrm{dotted}\:\mathrm{line}\:\mathrm{opposite} \\ $$$$\mathrm{to}\:\mathrm{it}? \\ $$

Commented by ajfour last updated on 03/Sep/20

the component of Normal reaction along that dotted line is Nsin 30° which the horizontal forcw F is balancing.

$${the}\:{component}\:{of}\:{Normal}\:{reaction} \\ $$$${along}\:{that}\:{dotted}\:{line}\:{is}\:{N}\mathrm{sin}\:\mathrm{30}° \\ $$$${which}\:{the}\:{horizontal}\:{forcw}\:{F}\:{is} \\ $$$${balancing}. \\ $$

Commented by Aina Samuel Temidayo last updated on 03/Sep/20

So which angle are we using?

$$\mathrm{So}\:\mathrm{which}\:\mathrm{angle}\:\mathrm{are}\:\mathrm{we}\:\mathrm{using}? \\ $$

Commented by ajfour last updated on 03/Sep/20

Commented by ajfour last updated on 03/Sep/20

Ncos (90°−30°)=F ⇒ Nsin 30°=F

$${N}\mathrm{cos}\:\left(\mathrm{90}°−\mathrm{30}°\right)={F} \\ $$$$\Rightarrow\:{N}\mathrm{sin}\:\mathrm{30}°={F} \\ $$

Commented by Aina Samuel Temidayo last updated on 03/Sep/20

How did you get that 30° close to N?

$$\mathrm{How}\:\mathrm{did}\:\mathrm{you}\:\mathrm{get}\:\mathrm{that}\:\mathrm{30}°\:\mathrm{close}\:\mathrm{to}\:\mathrm{N}? \\ $$

Commented by Aina Samuel Temidayo last updated on 03/Sep/20

I think I understand now. Thanks.

$$\mathrm{I}\:\mathrm{think}\:\mathrm{I}\:\mathrm{understand}\:\mathrm{now}.\:\mathrm{Thanks}. \\ $$

Commented by ajfour last updated on 03/Sep/20

Commented by Aina Samuel Temidayo last updated on 03/Sep/20

Yes. Thanks for your time. I really appreciate your efforts.

$$\mathrm{Yes}.\:\mathrm{Thanks}\:\mathrm{for}\:\mathrm{your}\:\mathrm{time}.\:\mathrm{I}\:\mathrm{really} \\ $$$$\mathrm{appreciate}\:\mathrm{your}\:\mathrm{efforts}. \\ $$

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