The image of the question is missing, so i have attached it.
Answer:
σ_x ≈ -387.47 psi
σ_y ≈ 454.65 psi
Step-by-step explanation:
From the method of equilibrium and looking at the image of the inclined plane, we can sum moments about the X-Plane and equate to zero and also do the same to the y-plane.
Thus;
At Σf_x = 0 we have;
Δf_x - 400(ΔA × cos 60°)cos 60° + 650(ΔA sin 60°)cos 30° = 0
Δf_x - (400(ΔA × 0.5)0.5) + (650(ΔA × 0.866)0.866) = 0
Δf_x - 100ΔA + 487.47ΔA = 0
Where;
Δf_x is change in force along the x - axis and ΔA is change in area along the X-axis.
Δf_x = 100ΔA - 487.47ΔA
Δf_x = -387.47ΔA
Thus, Δf_x/ΔA = -387.47 psi
As ΔA approaches zero, Δf_x/ΔA which will be the stress along the x-component will be;
σ_x = Δf_x/ΔA ≈ -387.47 psi
Similarly for the y-component;
At Σf_y = 0 we have;
Δf_y - 650(ΔA × sin 60°)sin 30° - 400(ΔA × cos 60°)sin 60° = 0
Δf_y - (650(ΔA × 0.866)0.5) - (400(ΔA × 0.5)0.866) = 0
Δf_y - 281.45ΔA - 173.2ΔA = 0
Δf_y = 281.45ΔA + 173.2ΔA
Δf_y = 454.65ΔA
Δf_y/ΔA = 454.65 psi
Now,
As ΔA approaches zero, Δf_x/ΔA which will be the stress along the y -component will be;
σ_y = Δf_y/ΔA ≈ 454.65 psi