5. A 15-nC point charge is at the origin in free space. Calculate V₁ if point P, is located at
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The image of the load applied to the polystyrene is missing, so i have attached it.
Answer:
a_new = 2.00302 in
b_new = 2.00552
Explanation:
From the image attached, we can see that the load of 500 lb/in is applied in the x-direction while the load of 350 lb/in acts in the y-direction.
Now, formula for stress is;
Stress(σ) = Force/Area
We are not given force and area but the load and plate thickness.
Thus, stress = load/thickness
We are given;
Load in x - direction = 500 lb/in.
Load in y - direction = 350 lb/in.
Thickness; t = 0.25 in
Thus;
σ_x = 500/0.25
σ_x = 2000 ksi
σ_y = 350/0.25
σ_y = 1400 ksi
From Hooke's law for 2 dimensions, strain is given by the formula;
ε_x = (1/E)(σ_x - vσ_y)
ε_y = (1/E)(σ_y - vσ_x)
We are given v_p = 0.25 and Ep = 597 × 10³ psi
Thus;
ε_x = (1/(597 × 10^(3)))(2000 - (0.25 × 1400)
ε_x = 0.00276
ε_y = (1/(597 × 10^(3)))(1400 - (0.25 × 2000)
ε_y = 0.00151
From elongation formula, we know that;
Startin is: ε = ΔL/L
Thus; ΔL = Lε
We are given a = 2 and b = 2
Thus;
ΔL_x = 2 × 0.00276
ΔL_x = 0.00552
ΔL_y = 2 × 0.00151
ΔL_y = 0.00302
New dimensions are;
a_new = 2 + 0.00302
a_new = 2.00302 in
b_new = 2 + 0.00552
b_new = 2.00552
