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A square lead slab of side 50 cm and thickness 10 cm is subject to a shearing force (on its narrow face) of 9.0 × 104 N. The lower edge is riveted to the floor. How much will the upper edge be displaced?

A square lead slab of side 50 cm and thickness 10 cm is subject to a shearing force (on its narrow face) of 9.0 × 104 N. The lower edge is riveted to the floor. How much will the upper edge be displaced?

Grade:11

1 Answers

Ishaan
14 Points
5 years ago
Area of the face on which force is applied:
A = 50 x 10 
= 500 cm²
= 500 x 10 ^ -4 m²
Length 
L = 0.5 m 
Rigidity: 
n = 5.6 x 10⁹ Pa
Force:
F = 9 x 10⁴ N
Expression for Rigidity modulus:
n = FL/AΔL
F is force of shearing
L is side
A is the area of the face on which force is applied
Displacement  of upper edge is 
ΔL = FL/An
= (9x10^4) (0.5)/ (500 x 10^ - 4) ( 5.6 x 10^9)
 = 0.00016 m 
= 0.16 mm 

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