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A shear force of 400 N is applied to one face of an aluminum cube with sides of 30 cm. What is the resulting relative displacement? (the shear modulus for aluminum is 2.5 X 1010N/m2).

a) 4.4 X 10-8 m
b) 8.2 X 10-8 m
c) 1.9 X 10-8 m
d) 5.3 X 10-8 m

Respuesta :

Answer:

[tex]\Delta x=5.33\times 10^{-8}m[/tex]

So option (d) will be correct option

Explanation:

We have given shear force F = 400 N

Size of the cube [tex]L_0=30cm=0.3m[/tex]

Shear modulus of aluminium [tex]S=2.5\times 10^{10}N/m^2[/tex]

We have to find the resulting relative displacement

Area of the cube [tex]A=L_0^2=0.3^2=0.09m^2[/tex]

We know that shear force is given by

[tex]F=S\times \frac{\Delta x}{L_0}\times A[/tex]

So [tex]400=2.5\times 10^{10}\times \frac{\Delta x}{0.3}\times 0.09[/tex]

[tex]\Delta x=5.33\times 10^{-8}m[/tex]

So option (d) will be correct option