Question

A block of mass 10 kg is placed in a rough inclined plane and theta is 30 the mue in between block and inclined plan is 1 by root 3

Answer 1

$\Huge{\underline{\underline{\mathfrak{Correct \ Question \colon }}}}$

A block of 10 Kg is placed on a rough inclined plane,the coefficient of friction between then is 1/3 and the angle between incline and ground is 30. Find the frictional force acting on the body

$\Huge{\underline{\underline{\mathfrak{Answer \colon}}}}$

From the Question,

• Mass of the block,m = 10 Kg

• Angle of inclination = 30°

• Coefficient of friction = 1/3

Assuming the block is at rest

Since,

The block is said to be at rest,static friction acts on the block

We know that,

$\boxed{ \boxed{ \sf{ {f}_{s} \propto \: N}}} \\ \\ \rightarrow \: \sf{ {f}_{s} = { \mu}_{s}.mg cos \phi \: } \\ \\ \rightarrow \: \sf{ {f}_{s} = \frac{1}{3} \times 10 \times 10 \times cos30} \\ \\ \huge{ \rightarrow \: \boxed{ \boxed{\sf{ {f}_{s} = 16.67 N}}}}$

The frictional force experienced by the block is 16.67 N