Question

A 150 N picture hangs from a nail in the wall. The maximum shear stress the nail can sustain is 1.00 × 10^6 N/m2. The smallest diameter the nail can be is?

Answer 1

Given:

Force $F = 150$ N

Shear stress $= 1 \times 10^{6} \frac{N}{m^{2} }$

To Find:

The smallest diameter of nail,

From the formula shear stress,

  Shear stress $= \frac{F}{A}$

Where $A =$ area of nail = $\pi r^{2}$

First find radius of nail,

   $A = \frac{150}{1 \times 10^{6} }$

   $r^{2} = 47.77 \times 10^{-6}$

     $r = \sqrt{47.77 \times 10^{-6} }$

     $r = 6.91 \times 10^{-3}$ m

For finding the diameter of diameter,

    $D = 2r$

    $D = 13.82 \times 10^{-3}$ m

Therefore, the diameter of nail is $13.82 \times 10^{-3}$ m

Answer 2

Given :

Weight of the picture = 150 N

Maximum shear stress = 1×10⁶ N/m²

To find :

The smallest diameter of the nail

Solution :

• Area of cross section required =  weight / shear stress

                                                   = 150 / 1×10⁶

• Area of cross section  = 150×10⁻⁶ m² = 150mm²

• Diameter = √(A×4)/π

                =√150×4/π)

                =13.8166 mm

The diameter Of the nail is 13.816mm