$\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} $

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 ,

$T- m_1g \sin 30^{\circ}-f=0$

$f= T- m_1 g \sin 30^{\circ}$

$\quad= 60-5 \times 10 \times \sin 30^{\circ}$

$\quad= 60-25=35\;N$

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