Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 11

# 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. - Physics

Short Note

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

#### Solution

Given:
Face area of steel plate = 4 cm2 = 4 × 10−4 m2
Thickness of steel plate d = 0.5 cm = 0.5 × 10−2 m
Applied force on the upper surface F   = 10 N
Rigidity modulus of steel = 8.4 × 1010 N m−2
Let θ be the angular displacement.
Rigidity modulus $\text{ m } = \frac{F}{A\theta}$

$\Rightarrow m = \left( \frac{10}{4 \times {10}^{- 4} \theta} \right)$

$\Rightarrow \theta = \frac{10}{4 \times {10}^{- 4} \times 8 . 4 \times {10}^{10}}$

$= 0 . 297 \times {10}^{- 6}$

∴ Lateral displacement of the upper surface with respect to the lower surface = θ × d
⇒ (0.297) × 10−6 × (0.5) × 10−2
⇒ 1.5 × 10−9 m

Hence, the required lateral displacement of the steel plate is 1.5 × 10−9 m.

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#### APPEARS IN

HC Verma Class 11, 12 Concepts of Physics 1
Chapter 14 Some Mechanical Properties of Matter
Q 15 | Page 301