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The fig shows a horizontal force \underset{F}{\rightarrow} acting on a block of mass 'M' on an inclined plane. What is the normal reaction of the block?

  • Option 1)

    mg \sin \theta + F \cos \theta

  • Option 2)

    mg \sin \theta - F \cos \theta

  • Option 3)

    mg \cos \theta - F \sin \theta

  • Option 4)

    mg \cos \theta + F \sin \theta

 

Answers (1)

As we learnt in

Minimum Force to move a body downward along the surface of Inclined Plane -

  

Use condition of equilibrium.

R+P sin \alpha=W cos \lambda

F= P cos \alpha + W sin \lambda

Use  F=\mu R

P=\frac{W sin(\theta-\lambda)}{cos(\alpha-\theta)}

- wherein

P = Pulling force 

\alpha= Angle of P with horizontal 

\lambda= Inclined angle

R = Reaction

W = Weight

 N = mg\cos \theta +F\sin \theta


Option 1)

mg \sin \theta + F \cos \theta

This solution is incorrect.

Option 2)

mg \sin \theta - F \cos \theta

This solution is incorrect.

Option 3)

mg \cos \theta - F \sin \theta

This solution is incorrect.

Option 4)

mg \cos \theta + F \sin \theta

This solution is correct.

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