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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Laws of Motion
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A uniform sphere of radius R and mass M is held at rest on an inclined plane of angle $\theta$ by a horizontal string , as shown in the figure. Let $R= 20 \;cm,M= 3\;kg,$ and $ \theta= 30^{\circ}$. What is the normal force exerted on the sphere by the inclined plane ?

$\begin{array}{1 1}3.76\;N\\36\;N\\29.4\;N \\46\;N\end{array} $

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Answer : $29.4\;N$
Step 1:
Free Body Diagram :
Step 2:
Apply $\sum \overrightarrow{T}=0$ about an axis through the center of the sphere :
Apply $\sum F_x=0$ to the sphere
Substitute for f and solve for T :
Substitute numerical values and evaluate T .
$fR -TR =0 => T =f$
$f+T \cos \theta- Mg \sin \theta=0$
$T= \large\frac{Mg \sin \theta}{1+\cos \theta}$
$ T= \large\frac{(3 \;kg)(9.81 \; m/s^2) \sin 30^{\circ}}{1+\cos 30^{\circ}}$$=7.89\;N$
Step 3:
Apply $\sum F_y =0$ to the sphere :
Solve for $F_n$
Substitute numerical values and evaluate $F_n$ :
$F_n-T \sin \theta - Mg \cos \theta =0$
$F_n =T \sin \theta+Mg \cos \theta$
$F_n= (7.89 \;N) \sin 30^{\circ}+ (3 \;kg)(9.81\;m/s^2) \cos 30^{\circ}$
$\qquad= 29.4 \;N$


answered Aug 28, 2014 by meena.p
edited Aug 29, 2014 by balaji.thirumalai

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