Question

*2-20. if $\phi=45^{\circ}, f_{1}=5 \mathrm{kn}$, and the resultant force is $6 \mathrm{kn}$ directed along the positive $y$ axis, determine the required magnitude of $\mathbf{f}_{2}$ and its direction $\theta$.

Explanation:

Step1: 分解力

根据力的分解原理，将\(F_1\)和\(F_2\)沿\(x\)、\(y\)轴分解。
已知\(F_1 = 5kN\)，\(\phi=45^{\circ}\)，则\(F_{1x}=F_1\sin\phi = 5\sin45^{\circ}\)，\(F_{1y}=F_1\cos\phi = 5\cos45^{\circ}\)。
设\(F_2\)与\(x\)轴夹角为\(\theta\)，则\(F_{2x}=F_2\cos\theta\)，\(F_{2y}=F_2\sin\theta\)。

Step2: 根据合力条件列方程

因为合力\(F_R = 6kN\)沿\(y\)轴，所以\(F_{Rx}=0\)，\(F_{Ry}=6kN\)。
由\(F_{Rx}=0\)可得：\(F_{1x}-F_{2x}=0\)，即\(5\sin45^{\circ}-F_2\cos\theta = 0\) ①。
由\(F_{Ry}=6kN\)可得：\(F_{1y}+F_{2y}=6\)，即\(5\cos45^{\circ}+F_2\sin\theta = 6\) ②。

Step3: 求解方程

由①得\(F_2\cos\theta = 5\sin45^{\circ}=\frac{5\sqrt{2}}{2}\)。
由②得\(F_2\sin\theta = 6 - 5\cos45^{\circ}=6-\frac{5\sqrt{2}}{2}\)。
将\(F_2\cos\theta\)与\(F_2\sin\theta\)两式平方相加：
\(F_2^{2}(\cos^{2}\theta+\sin^{2}\theta)=(\frac{5\sqrt{2}}{2})^{2}+(6 - \frac{5\sqrt{2}}{2})^{2}\)
因为\(\cos^{2}\theta+\sin^{2}\theta = 1\)，所以\(F_2=\sqrt{(\frac{5\sqrt{2}}{2})^{2}+(6 - \frac{5\sqrt{2}}{2})^{2}}\)
\(=\sqrt{\frac{25}{2}+36-30\sqrt{2}+\frac{25}{2}}\)
\(=\sqrt{50 + 36-30\sqrt{2}}\)
\(=\sqrt{86-30\sqrt{2}}\approx 4.53kN\)。
求\(\theta\)：\(\tan\theta=\frac{6 - \frac{5\sqrt{2}}{2}}{\frac{5\sqrt{2}}{2}}=\frac{12 - 5\sqrt{2}}{5\sqrt{2}}\approx\frac{12-7.07}{7.07}=\frac{4.93}{7.07}\approx0.697\)，则\(\theta\approx34.9^{\circ}\)。

Answer:

\(F_2\approx4.53kN\)，\(\theta\approx34.9^{\circ}\)。