Question

A block of weight 15 N slides on a
horizontal table, the coefficient of sliding
friction = 0.4, the area of the block in
contact with table is 0.05m- What is the
shear stress?​

Answer 1

• A=0.05m^2
• μ=0.4
• N=mg=15N
• F=μN=0.4×15=6N

Stress, S= F/A

S= 6/0.05

☑️☑️ S=120N/m^2☑️☑️

Answer 2

Given,

• Weight of the block $\sf{=15\ N}$

• Coefficient of friction, $\sf{\mu=0.4}$

• Area of block in contact with table, $\sf{A=0.05\ m^2}$

The magnitude of the reaction acting on the block due to the table is equal to the weight of the block.

• $\sf{R=15\ N}$

Thus the frictional force acting on the block is,

$\longrightarrow\sf{f=\mu\,R}$

$\longrightarrow\sf{f=0.4\times15\ N}$

$\longrightarrow\sf{f=6\ N}$

This frictional force is the shearing force acting on the block due to the table.

Hence the shear stress of the block is,

$\longrightarrow\sf{Shear\ Stress=\dfrac{Shearing\ Force}{Contact\ Area}}$

$\longrightarrow\sf{Shear\ Stress=\dfrac{6}{0.05}}$

$\longrightarrow\underline{\underline{\sf{Shear\ Stress=120\ Pa}}}$