Question

Q. A block of mass m is kept on a rough inclined plane of inclination $\theta$. Coefficient of friction between the block and the plane is $\mu$. What horizontal force F should be applied on the block so that it just begins to slide up the plane? (Given $\mu$ < cot$\theta$).

#ANSWER THE QUESTION WITH PROPER EXPLANATION

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Answer 1

Answer:

It is best to draw the Free Body Diagram of the mass m over the inclined plane.

Since the question has mentioned that the block will start to move up, so the friction will be acting downwards.

Refer to the attached photo to understand better .

Since it is a Limiting condition , we can say ;

$\bigstar \: \: F \cos( \theta) = f + mg \sin( \theta)$

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

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

Taking similar terms on one side ;

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

$= > F = \dfrac{mg \{ \sin( \theta) + \mu \cos( \theta) \} }{ \{\cos( \theta) - \mu \sin( \theta) \}}$

Answer 2

Answer :

• $\sf{F \: = \: \dfrac{\big[ mg(\sin \theta \: + \: \mu \cos \theta) \big]}{\cos \theta \: - \: \mu \sin \theta}}$

Explanation :

As we know that from the attached diagram, that the Reaction Force (R) is placed on the opposite side of the mgcosθ.

So,

$\longrightarrow \sf{R \: = \: mg \cos \theta \: + \: F \sin \theta \: \: \: \: \: \: ...(1)}$

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Also, we know that, The Frictional Force is directly proportional to the Reaction force :

$\longrightarrow \sf{f \: \propto \: R} \\ \\ \longrightarrow {\boxed{\sf{f \: = \: \mu R \: \: \: \: \: ...(2)}}}$

Where,

• μ is a constant

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Now, put value of R from (1) in (2)

$\longrightarrow \sf{f \: = \: \mu (mg \cos \theta \: + \: F \sin \theta) \: \: \: \: \: \: ...(3)}$

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Also, from the Diagram

$\longrightarrow \sf{F\cos \theta \: = \: mg \sin \theta \: + \: f} \\ \\ \longrightarrow \sf{F \cos \theta \: = \: (mg \sin \theta) \: + \: \big[ \mu ( mg \cos \theta \: + \: F \sin \theta) \big] } \\ \\ \longrightarrow \sf{F \: = \: \dfrac{\big[ mg(\sin \theta \: + \: \mu \cos \theta) \big]}{\cos \theta \: - \: \mu \sin \theta}} \\ \\ \underline{\underline{\sf{Force \: is\: \dfrac{\big[ mg(\sin \theta \: + \: \mu \cos \theta) \big]}{\cos \theta \: - \: \mu \sin \theta}}}}$