Question

Two mass m1 and m2 are connected by a light string over a smooth pulley
If the system is at rest find
1 ) Angle of Inclination
2 ) Force excreted by inclined plane on the mass m1

NOTE:

Values are to be calculated in terms of masses and constants​

Answer 1

$\huge\mathfrak\blue{Answer}$

Given:

• We have been given two masses m1 and m2 connected by a light string over a smooth pulley
• The system is at rest

To Find:

• We have to find the angle of Inclination
• Force exerted by inclined plane on the black of mass m1

Concept Used:

• One Vector can be divided into two or more smaller vectors . This process is known as Resolution of vectors
• A body is said to be in rest when the net force acting on it is zero

Solution:

On analyzing the given data the following figure can be made

[Refer to the Attachment]

In the Figure

θ = Angle of Inclination

N = Force exerted by the inclined plane

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For Block A of Mass m1

$\boxed{\sf{Weight = m_1 \times g}}$

On Resolving the components of weight

Weight can be resolved in its sine and cosine component

[As shown in the Figure]

Since the system is at rest

Then net force of Block A will be zero

The Tension T in the string will balance the mgsinθ component of Block

$\implies \boxed{\sf{T = m_1 g \: sin θ }}$ --------( 1 )

And

$\implies \boxed{\sf{N = m_1 g \: cos θ }}$ --------------( 2 )

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For Block B of Mass m2

Since the black B is also at rest

Hence the net force on block will be zero

The Tension in the string will balance the weight of Block

$\implies \boxed{\sf{T = m_2 g }}$ -------- ( 3 )

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Equating Equation ( 1 ) and ( 3 ) we get

$\implies \sf{ m_1 g \: sin θ = m_2 g }$

$\implies \sf{sin θ =\dfrac{m_2 g}{m_1 g}}$

$\implies \sf{sin θ =\dfrac{m_2}{m_1 }}$

$\implies \boxed{\sf{θ = sin^{-1} \left ( \dfrac{m_2}{m_1 } \right ) }}$

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Force excreted by the plane = Normal reaction

Using Equation ( 2 ) we get

$\implies \sf{ N = m_1 g cos θ}$

$\implies \sf{ N = m_1 g cos \left [ sin^{-1} \left ( \dfrac{m_2}{m_1 } \right ) \right ] }$

Using Inverse Trigonometric Rule

$\boxed{\sf{sin^{-1} x = cos^{-1} \sqrt{1 - x^2} }}$

Using Identity in the Above Equation

$\implies \sf{ N = m_1 g cos \left [ cos^{-1} \sqrt{1 - {\left ( \dfrac{m_2}{m_1 } \right )}^2} \right ] }$

$\implies \boxed{\sf{ N = m_1 g \sqrt{1 - {\left ( \dfrac{m_2}{m_1 } \right )}^2}} }$

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$\huge\underline{\sf{\red{A}\orange{n}\green{s} \pink{w}\blue{e} \purple{r}}}$

$\large\boxed{\sf{\red{Angle \: of \: Inclination = sin^{-1} \left ( \dfrac{m_2}{m_1 } \right )}}}$

$\boxed{\sf{\red{ N = m_1 g \sqrt{1 - {\left ( \dfrac{m_2}{m_1 } \right )}^2}}} }$

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$\purple {\large\underline{\underline{\sf{Extra \: Information:}}}}$

• Normal Reaction : Normal reaction is the force exerted by an object to counter a weight or a force exerted by other body which is in contact with it
• Tension : Tension can be defined as a pulling force inhibited by strings or an inextesible rod

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