The head of a pin is a square of side 10 mm. Find the pressure on it due to a force of 20 N. Ans. (2x10⁵ Nm⁻²)


Given:  

A side of Pin of square head = L = 10 mm = 1ｘ10 ｘ10⁻³ m =  1 ｘ10⁻²  m
Area of the head of Pin = A = L ｘ L = 1 ｘ10⁻²  m ｘ1 ｘ10⁻²  m = 1 ｘ10⁻⁴  m²
Force = F =  20 N

To Find:

Pressure = P =?  

Solution:   

As Pressure P is defined as force per unit area. So, 

P = `\frac {F}{A}`

 by putting values

P = `\frac {20 N}{1 ｘ10⁻⁴ m²}`

P = 20 ｘ10⁴ N m⁻²

or

P = 2 ｘ10⁵ N m⁻²

thus, the pressure on the pin's head is 2 ｘ10⁵ N m⁻²

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