The uniform rod is bent into the shape of a parabola and has a weight per unit length of \(100 \mathrm{N} / \mathrm{m}\) . Determine the reactions at the fixed support \(A\)
\(
d L=\sqrt{d x^{2}+d y^{2}}
\)
\(
=\sqrt{\left(\frac{d x^{2}}{d y^{2}}\right)+1} \ d y
\)
\(
=\sqrt{\left(\frac{d x}{d y}\right)^{2}+1} \ d y
\)
\(
\left|\begin{array}{c}{y^{2}=x} \\ {2 y d y=d x} \\ {\frac{d x}{d y}=2 y} \\ {\left(\frac{d x}{d y}\right)^{2}=4 y^{2}}\end{array}\right.
\)
\(
L=\int_{0}^{1} \sqrt{1+4 y^{2}} d y
\)
\(
L=\int_{0}^{1} d l=1.48 m
\)
\(
\overline{x}=\frac{\int \tilde{x} d l}{\int d l}
=\frac{\int_{0}^{1} y^{2} \sqrt{1+4 y^{2}} d y}{1.48}=
\frac{0.606}{1.48}=0.41 \mathrm{m}
\)
The uniform rod is bent into the shape of a parabola and has a weight per unit length of \(100 \mathrm{N} / \mathrm{m}\) . Determine the reactions at the fixed support \(A\)
\( d L=\sqrt{d x^{2}+d y^{2}} \)
\( =\sqrt{\left(\frac{d x^{2}}{d y^{2}}\right)+1} \ d y \)
\( =\sqrt{\left(\frac{d x}{d y}\right)^{2}+1} \ d y \)
\( \left|\begin{array}{c}{y^{2}=x} \\ {2 y d y=d x} \\ {\frac{d x}{d y}=2 y} \\ {\left(\frac{d x}{d y}\right)^{2}=4 y^{2}}\end{array}\right. \)
\( L=\int_{0}^{1} \sqrt{1+4 y^{2}} d y \)
\( L=\int_{0}^{1} d l=1.48 m \)
\( \overline{x}=\frac{\int \tilde{x} d l}{\int d l} =\frac{\int_{0}^{1} y^{2} \sqrt{1+4 y^{2}} d y}{1.48}= \frac{0.606}{1.48}=0.41 \mathrm{m} \)
\( +\uparrow \sum F_ y=0 \quad\longrightarrow A_ y-1.48(100) =0 \longrightarrow A_ y=148 N \uparrow \)
\( \sum M_ A=0 \quad\longrightarrow M_ A-(148)(0.41)=M_{A}=60.6 \mathrm{N}.{m} \)
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