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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Laws of Motion
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Two blocks of masses $m_1=5 \;kg$ and $m_2= 6\;kg$ are connected by a light string passing over a light friction less pulley as shown in fig. The mass $m_1$ is at rest on the inclined plane and mass $m_2$ hangs vertically . If the angle of incline $\theta =30^{\circ}$, What is the magnitude and direction of the force of friction on the $5\;kg$ block . Take $g= 10 ms^{-2}$

$\begin{array}{1 1}35\;N\;up\;the\;plane\\35\;N\;down\;the\;plane\\85\;N\;up\;the\;plane\\85\;N\;down\;the\;plane\end{array} $

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Answer : $35\;N\;down\;the\;plane$
Weight of mass $m_2 = 6 \times 10= 60\;N$
The weight of $m_2$ provides the tension . Thus $T= 60\;N$
Opposite this force along the plane is the component $F_1= m_1 g \sin \theta$ of the force $m_1 g$ .
Now $F_1= m_1 g \sin \theta = 5 \times 10 \times \sin 30 ^{\circ}=25 N$$
Since $F_1$ is less that T and is , therefore , insufficient top balance T (see Fig ), the force of friction (f) down the plane is necessary to keep block $m_1$ at rest .
Thus , f must act down the plane.
Since mass $m_1$ is at rest, the net force on $m_1$ along the plane must be zero.
Thus ,
$T- m_1g \sin 30^{\circ}-f=0$
$f= T- m_1 g \sin 30^{\circ}$
$\quad= 60-5 \times 10 \times \sin 30^{\circ}$
$\quad= 60-25=35\;N$


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

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