problem 3.5 - forces p1, p2, p3, each of magnitude p, act on the edge of a cube of side \a\. determine the…

Explanation:

Step1: Define position vectors

Let the origin be at point A. Position vectors of relevant points: $\vec{r}_C=(0,0,a)$, $\vec{r}B=(a,a,0)$. The unit - vector along the diagonal CB is $\hat{u}{CB}=\frac{\vec{r}_B - \vec{r}_C}{\vert\vec{r}_B - \vec{r}_C\vert}=\frac{(a,a, - a)}{\sqrt{a^{2}+a^{2}+(-a)^{2}}}=\frac{1}{\sqrt{3}}(1,1, - 1)$.

Step2: For force $\vec{P}_1$

Let $\vec{P}1 = P\hat{i}$. Position vector of the point of application of $\vec{P}1$ (say point H) with respect to A is $\vec{r}H=(a,0,a)$. The moment of $\vec{P}1$ about point C is $\vec{M}{1C}=\vec{r}{HC}\times\vec{P}1$, where $\vec{r}{HC}=( - a,0,0)$. Then $\vec{M}{1C}=( - a,0,0)\times(P,0,0)=(0,0,0)$. The moment of $\vec{P}1$ about the diagonal CB is $M{1}=\hat{u}{CB}\cdot\vec{M}_{1C}=0$.

Step3: For force $\vec{P}_2$

Let $\vec{P}2 = P\hat{j}$. Position vector of the point of application of $\vec{P}2$ (say point G) with respect to A is $\vec{r}G=(0,a,a)$. $\vec{r}{GC}=(0, - a,0)$. $\vec{M}{2C}=\vec{r}{GC}\times\vec{P}2=(0, - a,0)\times(0,P,0)=(0,0,0)$. The moment of $\vec{P}2$ about the diagonal CB is $M{2}=\hat{u}{CB}\cdot\vec{M}_{2C}=0$.

Step4: For force $\vec{P}_3$

Let $\vec{P}3 = P\hat{i}$. Position vector of the point of application of $\vec{P}3$ (point F) with respect to A is $\vec{r}F=(a,a,0)$. $\vec{r}{FC}=( - a, - a,a)$. $\vec{M}{3C}=\vec{r}{FC}\times\vec{P}3=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\-a&-a&a\P&0&0\end{vmatrix}=(0,Pa,Pa)$. The moment of $\vec{P}3$ about the diagonal CB is $M{3}=\hat{u}{CB}\cdot\vec{M}_{3C}=\frac{1}{\sqrt{3}}(1,1, - 1)\cdot(0,Pa,Pa)=\frac{Pa - Pa}{\sqrt{3}} = 0$.

Answer:

The moment about diagonal CB due to $\vec{P}_1$ is 0, due to $\vec{P}_2$ is 0, and due to $\vec{P}_3$ is 0.