SATHEE: Laws of Motion 3 Question 3

Laws of Motion 3 Question 3

3. A block of mass $10 k g$ is kept on a rough inclined plane as shown in the figure. A force of $3 N$ is applied on the block. The coefficient of static friction between the plane and the block is 0.6 . What should be the minimum value of force $F$ , such that the block does not move downward ? (Take, $g = 10 m s^{- 2}$ )

(2019 Main, 9 Jan I)

(a) $32 N$

(b) $25 N$

(d) $18 N$

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Answer:

Correct Answer: 3. (a)

Solution:

Free body diagram, for the given figure is as follows,

For the block to be in equilibrium i.e., so that it does not move downward, then

$\begin{aligned} Σ f_{x} & = 0 \\ 3 + M g \sin θ - F - f & = 0 \\ or & 3 + M g \sin θ & = F + f \end{aligned}$

As, frictional force, $f = μ R$

$∴ 3 + M g \sin θ = F + μ R$

Similarly,

$Σ f_{y} = 0$

$- M g \cos θ + R = 0$

or $M g \cos θ = R$

Substituting the value of ’ $R$ ’ from Eq. (ii) to Eq. (i), we get

$3 + M g \sin θ = F + μ (M g \cos θ)$

Here, $M = 10 k g, θ = 45^{\circ}, g = 10 m / s^{2}$ and $μ = 0.6$

Substituting these values is Eq. (iii), we get $3 + (10 \times 10 \sin 45^{\circ}) - (0.6 \times 10 \times 10 \cos 45^{\circ}) = F$

$\begin{aligned} \Rightarrow F & = 3 + \frac{100}{\sqrt{2}} - \frac{60}{\sqrt{2}} = 3 + \frac{40}{\sqrt{2}} \\ = 3 + 20 \sqrt{2} = 31.8 N \\ or F & ≃ 32 N \end{aligned}$

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Laws of Motion 3 Question 3

Answer:

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

Laws of Motion 3 Question 3

Answer:

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

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