4. Given that the three sides of $\triangle A B C$ are exactly three consecutive positive integers, and its perimeter and area are $p_{1}$ and $S_{1}$, respectively. If the three sides of $\triangle A B C$ are each increased by 10, the new $\triangle A^{\prime} B^{\prime} C^{\prime}$ has a perimeter and area of $p_{2}$ and $S_{2}$, respectively. When $p_{1} p_{2}=S_{1} S_{2}$, the sine of the smallest angle of $\triangle A B C$ is $\qquad$.
4. $\frac{3}{5}$.

Let the three sides of $\triangle A B C$ be $a-1, a, a+1$, and the three sides of $\triangle A^{\prime} B^{\prime} C^{\prime}$ be $b-1, b, b+1$, where $b-a=10$.
From the triangle area formula, we have
$$
\begin{array}{l}
S_{1} S_{2}=\sqrt{\frac{3}{2} a \cdot \frac{1}{2} a\left(\frac{1}{4} a^{2}-1\right) \cdot \frac{3}{2} b \cdot \frac{1}{2} b\left(\frac{1}{4} b^{2}-1\right)}, \\
p_{1} p_{2}=9 a b . \\
\text { Then } \frac{3}{2} a \cdot \frac{1}{2} a\left(\frac{1}{4} a^{2}-1\right) \cdot \frac{3}{2} b \cdot \frac{1}{2} b\left(\frac{1}{4} b^{2}-1\right) \\
=81 a^{2} b^{2} .
\end{array}
$$

Simplifying, we get $a^{2} b^{2}-4\left(a^{2}+b^{2}\right)=16 \times 143$.
$$
\begin{array}{l}
\text { Also, } 4\left(a^{2}+b^{2}\right)=4(a-b)^{2}+8 a b \\
=400+8 a b,
\end{array}
$$

Then $(a b)^{2}-8 a b-16 \times 168=0$.
Solving, we get $a b=56, a b=-48$ (discard).
By $b-a=10$, solving the system of equations, we get
$$
a=4, b=14 \text {. }
$$

Therefore, the three sides of $\triangle A B C$ are $3,4,5$.
Thus, the sine value of the smallest angle of $\triangle A B C$ is
$$
\sin A=\frac{3}{5} \text {. }
$$