Question

A steel plate of face area 4 cm2 and thickness 0.5 cm is fixed rigidly at the lower surface. A tangential force of 10 N is applied on the upper surface. Find the lateral displacement of the upper surface with respect to the lower surface. Rigidity modulus of steel = 8.4 × 1010 N m−2.

Answer 1

Lateral Displacement is $1.5 \times 10^-^9 m$

Explanation:

Given:  

Face region of steel plate $A = 4\ cm^2 = 4 \times 10^-^4\ m^2$

Thickness of steel plate $d = 0.5\ cm = 0.5 \times 10^-^2 m$

Applied force on the upper surface F = 10 N  

Rigidity modulus of steel = $8.4 \times 10^1^0 N m^-^2$

Let θ be the angular displacement.

Rigidity Modulus $m = \frac {F}{A\theta}$  

$m = \frac {10}{4\times 10^-^4\theta}$

$\theta = \frac {10}{4\times 10^-^4 \times 8.4 \times 10^1^0}$

= $0.297 \times 10^-^6$

∴ Lateral displacement of the upper surface as for the lower surface = $\theta \times d$

⇒ $(0.297) \times 10^-^6\times (0.5) \times 10^-^2$

⇒ $1.5 \times 10^-^9 m$

Henceforth, the necessary Lateral Displacement of the steel plate is $1.5 \times 10^-^9 m$

Answer 2

Answer:Answer is in the attachment attached below...

I KNOW U WILL FORGET TO MARK IT AS BRAINLIEST......