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  • As shown in the figure a block of mass 10 kg lying on a horizontal surface is pulled by a force F acting at an angle <img alt="30^{\circ}C" src="https://entrancecorner.oncodecogs.com/gif.latex?30%5

As shown in the figure a block of mass 10 kg lying on a horizontal surface is pulled by a force F acting at an angle 30^{\circ}C, with horizontal. For \mu_{\mathrm{s}}=0.25 the block will just start to move for the value of \left[\text { Given } \mathrm{g}=10 \mathrm{~ms}^{-2}\right. \text { ] }

 

Option: 1

20 \mathrm{~N}


Option: 2

33.3 \mathrm{~N}


Option: 3

25.2 \mathrm{~N}


Option: 4

35.7 \mathrm{~N}


Answers (1)

best_answer

\text { Givne }\: \mathrm{m}=10 \mathrm{~kg}, \mu_{\mathrm{s}}=0.25, \theta=30^{\circ}, \mathrm{g}=10 \mathrm{~m} / \mathrm{sec} \text {. }

F.B.D. (Free Body Diagram)

F \cos 30^{\circ}=\mathrm{f} ___________(i)

F \sin 30^{\circ}+\mathrm{N}=\mathrm{mg} \Rightarrow \mathrm{N}=\mathrm{Mg}-\mathrm{F} \sin 30^{\circ} _____(ii)

From equation (1)

\begin{aligned} & F \sin 30^{\circ}=\mu_s N \\ & F \cos 30^{\circ}=\mu_s\left(\mathrm{mg}-\mathrm{F} \sin 30^{\circ}\right) \\ & F \cos 30^{\circ}=\mu_s \mathrm{mg}-\mu_s \mathrm{~F} \sin 30^{\circ} \\ & F\left(\cos 30^{\circ}+\mu_s \sin 30^{\circ}\right)=\mu_{\mathrm{s}} \mathrm{mg} \\ & \mathrm{F}=\frac{\mu_{\mathrm{s}} \mathrm{mg}}{\cos 30^{\circ}+\mu_{\mathrm{s}} \sin 30^{\circ}}=\frac{0.25 \times 10 \times 10}{\sqrt{3} / 2 \times 0.25 \times 1 / 2} \\ & \mathrm{~F}=\frac{25}{\sqrt{3} / 2+\frac{0.25}{2}}=\frac{50}{1.73+0.25}=\frac{50}{1.98}=25.2 \mathrm{~N} \end{aligned}

 

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Rishabh

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