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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Laws of Motion
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Three friends have a bet on the value of the normal force in this free body diagram for an inclined plane problem? Raj says its 500 N Ram says its 250 N and Rahul says that its 433 N. Who is correct?

$\begin{array}{1 1} \text{Rahul} (433 N) \quad \text{Ram} (250 N) \quad \text{Raj} (500 N )\end{array}$

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Raj’s estimate of 500N would be correct only if normal force = force of gravity, which is true only on level surfaces when the normal force is the only up force.
Here the normal force is perpendicular to the surface and equal to the perpendicular component of gravity's pull.
The normal force is equal to the perpendicular component of the weight vector i.e, $500 N mg \times \cos 30 = 433 N$
Ram is incorrect because he has probably incorrectly used the perpendicular component of the weight vector.
The parallel component is $250\; N \;mg \times \sin 30$
answered Aug 26, 2014 by meena.p
edited Sep 2, 2014 by balaji.thirumalai
 

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