A copper (shear modulus 4.2 x 1010 N/m2) cube, 0.248 m on a side, is subjected to two shearing forces, each of magnitude F = 3.84 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.
To find the angle of shear deformation, we can use the formula:
θ = (F * L) / (G * A)
Where:
θ = Angle of shear deformation
F = Magnitude of the shearing force (in N)
L = Length of the cube (in m)
G = Shear modulus of the material (in N/m^2)
A = Cross-sectional area of the cube (in m^2)
Given:
F = 3.84 x 10^6 N
L = 0.248 m
G = 4.2 x 10^10 N/m^2
A = L^2 (since it's a cube, area = length * width)
Let's substitute the values into the formula:
θ = (3.84 x 10^6 N * 0.248 m) / (4.2 x 10^10 N/m^2 * 0.248 m^2)
Simplifying:
θ = (9.5232 x 10^5 N·m) / (1.0416 x 10^10 N·m)
θ = 9.1346 x 10^-5
Now, to convert this angle to degrees, we can use the fact that there are 360 degrees in a full circle:
θ_degrees = θ * (180 / π)
θ_degrees = 9.1346 x 10^-5 * (180 / 3.14159)
θ_degrees ≈ 0.0052 degrees
Therefore, the angle of shear deformation is approximately 0.0052 degrees.