Question

ANSWER THIS QUESTION WITH PROPER EXPLANATION

#No Spamming Plz​

Answer 1

Answer:

The basic concept is to draw FBD of each blocks , put the forces and use Translational Equilibrium equation .

I will be putting the angle of wedge as θ , mass as m and frictional coefficient as μ.

Since the body is in Translational Equilibrium :

$mg \sin( \theta) = f + ma \cos( \theta)$

$= > mg \sin( \theta) = \mu N + ma \cos( \theta)$

$= > mg \sin( \theta) = \mu \{mg \cos( \theta) + ma \sin( \theta) \} + ma \cos( \theta)$

Cancelling mass on both sides :

$= > \cancel mg \sin( \theta) = \mu \{ \cancel mg \cos( \theta) + \cancel ma \sin( \theta) \} + \cancel ma \cos( \theta)$

Putting all the available values :

$= > \dfrac{g}{ \sqrt{2} } = 0.5 \bigg \{ \dfrac{g}{ \sqrt{2} } + \dfrac{a}{ \sqrt{2} } \bigg \} + \dfrac{a}{ \sqrt{2} }$

$= > \dfrac{g}{ \sqrt{2} } = \dfrac{1}{2} \bigg \{ \dfrac{g}{ \sqrt{2} } + \dfrac{a}{ \sqrt{2} } \bigg \} + \dfrac{a}{ \sqrt{2} }$

$= > \dfrac{g}{ \sqrt{2} } - \dfrac{g}{2 \sqrt{2} } = \dfrac{a}{2 \sqrt{2} } + \dfrac{a}{ \sqrt{2} }$

$= > \dfrac{g}{2 \sqrt{2} } = \dfrac{3a}{2 \sqrt{2} }$

$= > a = \dfrac{g}{3}$

So final answer :

$\boxed{ \red{ \huge{ \bold{a = \dfrac{g}{3} }}}}$

Answer 2

Solution :

⏭ Given:

✏ mass of block = m = 1kg

✏ mass of wedge = M = 10kg

✏ co-efficient of friction = 0.5

⏭ To Find:

✏ The minimum horizontal acceleration to the wedge such that block will not slide down on the wedge.

⏭ Concept:

✏ In this kind of questions, first we have to draw FBD of block...Please see the attachment for better understanding...

✏ This question is completely based on concept of 'Force equilibrium'.

⏭ Calculation:

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• Equilibrium in y-axis

✏ N = mgsin45° + macos45°

✏ N = m(gsin45° + acos45°)

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• Equilibrium in x-axis

$\circ\sf\: mg\sin45\degree = ma\cos45\degree + \mu\times N \\ \\ \circ\sf \: mg\sin45\degree = ma\cos45\degree + \mu(mg\cos45\degree+masin45\degree)\\ \\ \circ\sf \: g\sin45\degree= a\cos45\degree + \mu(g\sin45\degree + a\sin45\degree)\\ \\ \circ\sf \: a(\cos45\degree+\mu\sin45\degree) = g(\sin45\degree - \mu\cos45\degree)\\ \\ \circ\sf \: a[\dfrac{1}{\sqrt{2}}+ \dfrac{0.5}{\sqrt{2}}] = g[\dfrac{1}{\sqrt{2}}-\dfrac{0.5}{\sqrt{2}}]\\ \\ \circ\sf \: a(\dfrac{1.5}{\sqrt{2}})= g(\dfrac{0.5}{\sqrt{2}})\\ \\ \circ\sf \: a= \dfrac{0.5g}{1.5}\\ \\ \circ\: \boxed{\sf{\pink{\large{a= \dfrac{g}{3}}}}}$

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✒ Option-C)