Question

A block of mass Toke is at rest on an inclined plane which is making angle 30 with the
horizontal The co-eficient of friction between the block and plane is 0.5 then in the frictional
for acting between the surface​

Answer 1

$\huge\underline\mathfrak\red{Answer}$

I WILL SOLVE THE QUESTION ACC. TO PICTURE NOT ACC. TO DATA GIVEN BECAUSE BOTH ARE DIFFERENT

$mass = 10kg \\ \gamma = 1.5 \\ \alpha = 30 \\ \\ normal \: force = mg \cos \alpha \\ \\ normal \: force = 10 \times 10 \times \cos30 \\ \\ normal \: force = 10 \times 10 \times \frac{ \sqrt{3} }{2} \\ \\ normal \: force = \frac{170}{2} \\ \\ normal \: force = 85 \: newton \\ \\ force \: of \: friction = \gamma \times normal \: force \\ \\ friction \: force(f) = 1.5 \times 85 = 127.5 \\ \\ mg \sin \alpha = force \: along \: wedge \\ \\ eq(1) \\ mg \sin \alpha - f = ma \\ \\ where \: a \: is \: the \: acceleration \: of \: body \\ \\ 10 \times 10 \times \sin30 - 127.5 = ma \\ \\ 100 \times \frac{1}{2} - 127.5 = 10a \\ \\ 50 - 127.5 = 10a \\ \\ - 77.5 = 10a \\ \\ - 7.75 = a \\ \\ now \: i \: done \: all \: the \: calculation \\ just \: to \: conclude \: that \: the \: body \: does \: \\ not \: move \: as \: friction \: force \: is \: more \\ than \: gravitation \: component \: along \: wedge$