Question 227: Properties of Matter

Question

A cube has side length 5 cm and modulus of rigidity $10^5 \mathrm{~N} / \mathrm{m}^2$. The displacement produced by a force of 10 N in the upper face of cube is $\_\_\_\_$ mm.

Accepted Answer

For a cube under
shearing force
, modulus of rigidity is given by
$$ \eta = \frac{	ext{shearing stress}}{	ext{shearing strain}} $$
Now,
Side of cube, $l = 5 	ext{ cm} = 0.05 	ext{ m}$
Force applied, $F = 10 	ext{ N}$
Modulus of rigidity, $\eta = 10^5 	ext{ N/m}^2$
Step 1: Find shearing stress
Area of upper face of cube:
$$ A = l^2 = (0.05)^2 = 2.5 	imes 10^{-3} 	ext{ m}^2 $$
So, shearing stress is
$$ \frac{F}{A} = \frac{10}{2.5 	imes 10^{-3}} = 4000 	ext{ N/m}^2 $$
Step 2: Find shearing strain
Let displacement of upper face be $x$.
Shearing strain:
$$ 	ext{strain} = \frac{x}{l} $$
Using
$$ \eta = \frac{F/A}{x/l} $$
So,
$$ 10^5 = \frac{4000}{x/0.05} $$
$$ 10^5 = \frac{4000 	imes 0.05}{x} $$
$$ 10^5 = \frac{200}{x} $$
$$ x = \frac{200}{10^5} = 2 	imes 10^{-3} 	ext{ m} $$
Step 3: Convert into mm
$$ 2 	imes 10^{-3} 	ext{ m} = 2 	ext{ mm} $$
So, the displacement produced is
$$ \boxed{2 	ext{ mm}} $$

Properties of Matter

A cube has side length 5 cm and modulus of rigidity 10^5 \mathrm{~N} / \mathrm{m}^2. The displacement produced by a force of 10 N in the upper face of cube is \_\_\_\_ mm.

Integer / Numeric Answer Type