Question

All surface are smooth in following figure. Find F such that block remains stationary with respect to wedge​ .please don't spam​

Answer 1

${ \huge \bf { \mid{ \overline{ \underline{Question}}} \mid}}$

All surface are smooth in following figure. Find F such that block remains stationary with respect to wedge?

$\huge{\bold{\underline{\underline{....Answer....}}}}$

$\huge{\bold{\underline{Given:-}}}$

• Let the Normal reaction be "N".
• and, Its Weight be "mg".

$\huge{\bold{\underline{Explanation:-}}}$

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Resolving Normal reaction (N) into Components,

we get :-

Along x - axis = N sinθ

Along y - axis = N cosθ

As the Question specifies that the Block should remain at rest, Therefore, Net force acting on the block should be Zero.

The Net Force Along the Components should also be Zero.

Therefore,

$\large{\boxed{\tt \sum F_y = 0}}$

Now,

$\large{\tt \leadsto N cos \theta - mg = 0}$

$\large{\leadsto {\underline{\underline{\tt N cos \theta = mg}}} \: \tt -------(1)}$

Now, For x - axis,

$\large{\boxed{\tt \sum F_x = 0}}$

Now,

$\large{\tt \leadsto N sin \theta - ma = 0}$

$\large{\leadsto {\underline{\underline{\tt N sin \theta = ma}}} \: \tt -------(2)}$

• Here ma is opposite to the N cosθ value.

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Dividing Equation (1) with (2),

$\large{\boxed{\tt \dfrac{Equation (2)}{Equation (1)}}}$

$\large{\tt \leadsto \dfrac{N sin\theta}{N cos \theta} = \dfrac{ma}{mg}}$

$\large{\tt \leadsto \dfrac{\cancel{N} sin \theta}{\cancel{N} cos \theta} = \dfrac{\cancel{m}a}{\cancel{m}g}}$

$\large{\tt \leadsto \dfrac{ sin \theta}{cos\theta} = \dfrac{a}{g}}$

As we Know sinθ/cosθ = tanθ.

Substituting it,

$\large{\tt \leadsto Tan\theta = \dfrac{a}{g}}$

$\large{\boxed{\tt a = g \: tan \theta}}$

So, the accleration is a = g tanθ.

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From Newton's second law of motion,

$\large{\boxed{\tt F = M_{(Total)}a}}$

$\large{\tt \leadsto F = (M + m) a}$

• M = Weight of Wedge.
• m = Weight of Block.

Substituting the Acceleration value,

$\large{\tt \leadsto F = (M + m) \times g \: tan \theta}$

$\huge{\boxed{\boxed{\tt F = (M + m) g \: tan \theta}}}$

So, the Force Exerted is F = (M + m)g tanθ.

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