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Question: An external force of 10N acts normally on a square area each of side 50cm. The stress produced in eq...

An external force of 10N acts normally on a square area each of side 50cm. The stress produced in equilibrium state is –

& \text{A) 10 N}{{\text{m}}^{-2}} \\\ & \text{B) 20 N}{{\text{m}}^{-2}} \\\ & \text{C) 40 N}{{\text{m}}^{-2}} \\\ & \text{D) 50 N}{{\text{m}}^{-2}} \\\ \end{aligned}$$
Explanation

Solution

The stress produced is the force applied per unit area of the applied surface. The normal force acting on a surface can cause the surface to deform along the direction of the force. This deformation is characterised as the stress applied on the surface.

Complete answer:
The stress applied is the force applied per unit area normally on a surface where the force is applied. It is the pressure in terms of the deformation caused in a material due to a normal force. The stress applied on a surface can produce a strain depending on the angle of force acting on the surface. A normal force is completely utilised to deform, so the entire stress is converted to strain on the surface.
The stress applied on a surface can be given by –
Stress=ForceUnit AreaStress=\dfrac{\text{Force}}{\text{Unit Area}}
The force is said to be acting normal to the surface of the given square tile. Therefore, we needn’t consider the components along other directions.
The area of the square tile of side 50cm is –

& \text{Area of the square surface, }A={{a}^{2}} \\\ & \Rightarrow \text{ }A=0.5\times 0.5 \\\ & \Rightarrow \text{ }A=0.25{{m}^{2}} \\\ \end{aligned}$$ The stress applied to the surface can be calculated from the formula as – $$\begin{aligned} & Stress,S=\dfrac{Force}{Area} \\\ & S=\dfrac{F}{A} \\\ & \Rightarrow \text{ }S=\dfrac{10N}{0.25{{m}^{-2}}} \\\ & \Rightarrow \text{ }S=40N{{m}^{-2}} \\\ \end{aligned}$$ The stress applied on the surface is found to be $$40N{{m}^{-2}}$$ . **So, the correct answer is “Option C”.** **Note:** The unit of stress is not given in Pascals. Even though the physical quantity which is derived from dividing the force by the unit area is the same as pressure, the use of Pascal is limited to the fluid mechanics than that of solids and surfaces which undergo deformation.