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A copper (shear modulus 4.2 x 1010 N/m2) cube, 0.242 m on a side, is subjected to two shearing forces, each of magnitude F = 3.07 x 10 6 N (see the drawing). Find the angle (in degrees), which is one measure of how the shape of the block has been altered by shear deformation.

Respuesta :

Answer:

The angle is 0.00124°.

Explanation:

Given that,

Shear modulus [tex]= 4.2\times10^{10}[/tex]

One side of copper = 0.242 m

Shearing forces [tex]F = 3.07\times10^{6}\ N[/tex]

We calculate the angle

Using formula of shear modulus

[tex]S = \dfrac{F}{A\theta}[/tex]

F = force

A = area

S = shear modulus

Put the value into the formula

[tex]\theta=\dfrac{F}{A\times S}[/tex]

[tex]\theta =\dfrac{3.07\times10^{6}}{0.242^2\times4.2\times10^{10}}[/tex]

[tex]\theta= 0.00124^{\circ}[/tex]

Hence, The angle is 0.00124°.

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