In a two-dimensional stress analysis, the state of stress at a point is shown in the figure. The values of length $PQ$, $QR$, and $RP$ are $4$, $3$, and $5$ units, respectively. The principal stresses are (round off to one decimal place)
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For triangular stress elements, use unit normal/tangent on the inclined face and multiply by its length to get resultant force; then apply $\sum F_x=\sum F_y=0$ to recover the Cartesian stresses.
Step 1: Geometry and unit vectors.
$PQ=4$, $QR=3$, $RP=5$ (a $3$–$4$–$5$ triangle).
For the slanted face $RP$, outward unit normal $\mathbf{n}=(4/5,\,3/5)$ and unit tangent $\mathbf{t}=(-3/5,\,4/5)$.
Step 2: Tractions on $RP$.
Given normal stress $\sigma_n=120$ MPa and shear stress $\tau=70$ MPa on $RP$.
Resultant force on $RP$ (unit thickness):
\[
\mathbf{F}_{RP}= (\sigma_n\,\mathbf{n}+\tau\,\mathbf{t})\times \text{area}
= (120\,\mathbf{n}+70\,\mathbf{t})\times 5.
\]
Compute components:
$120\times5\,\mathbf{n}=120(4,3)=(480,\,360)$, $70\times5\,\mathbf{t}=70(-3,4)=(-210,\,280)$.
Hence $\mathbf{F}_{RP}=(480-210,\,360+280)=(270,\,640)$.
Step 3: Forces on the other faces.
On $PQ$ (length $4$): normal stress $\sigma_x$ acts $\rightarrow$ force $\mathbf{F}_{PQ}=(-\sigma_x\cdot4,\,0)$.
On $QR$ (length $3$): normal stress $\sigma_y$ acts $\rightarrow$ force $\mathbf{F}_{QR}=(0,\,-\sigma_y\cdot3)$.
Step 4: Force equilibrium of the wedge.
$\mathbf{F}_{PQ}+\mathbf{F}_{QR}+\mathbf{F}_{RP}=\mathbf{0}$.
Thus, $-4\sigma_x+270=0 \Rightarrow \sigma_x=270/4=67.5\ \text{MPa}$.
And $-3\sigma_y+640=0 \Rightarrow \sigma_y=640/3=213.3\ \text{MPa}$.
\[
\boxed{\sigma_x=67.5\ \text{MPa}, \sigma_y=213.3\ \text{MPa}}
\]
