problem. force t is applied at point c on the semi - circle shown below. determine the moment due to force t…

Explanation:

Step1: Recall moment formula

The moment of a force $\vec{F}$ about a point $P$ is given by $\vec{M}=\vec{r}\times\vec{F}$, and its magnitude $M = rF\sin\theta$, where $\vec{r}$ is the position - vector from point $P$ to the point of application of the force, $F$ is the magnitude of the force, and $\theta$ is the angle between $\vec{r}$ and $\vec{F}$.

Step2: Calculate moment about point O

The position vector $\vec{r}_{OC}=r\hat{i}$ (assuming $O$ as the origin). The force $\vec{T}$ can be written in component form as $\vec{T}=T\cos\alpha\hat{i}+T\sin\alpha\hat{j}$. The moment $\vec{M}O=\vec{r}{OC}\times\vec{T}$. Using the cross - product formula $\vec{a}\times\vec{b}=(a_y b_z - a_z b_y)\hat{i}+(a_z b_x - a_x b_z)\hat{j}+(a_x b_y - a_y b_x)\hat{k}$, here $a_x = r,a_y = 0,a_z = 0,b_x=T\cos\alpha,b_y=T\sin\alpha,b_z = 0$. So, $M_O=rT\sin\alpha$.

Step3: Calculate moment about point A

The position vector $\vec{r}_{AC}=(r + r\cos\theta)\hat{i}+r\sin\theta\hat{j}$. $\vec{M}A=\vec{r}{AC}\times\vec{T}$. [ \begin{align*} \vec{M}_A&=\begin{vmatrix} \hat{i}&\hat{j}&\hat{k}\ r(1 + \cos\theta)&r\sin\theta&0\ T\cos\alpha&T\sin\alpha&0 \end{vmatrix}\ &=rT((1 + \cos\theta)\sin\alpha-\sin\theta\cos\alpha) \end{align*} ]

Step4: Calculate moment about point B

The position vector $\vec{r}_{BC}=(r\cos\theta)\hat{i}+(r + r\sin\theta)\hat{j}$. $\vec{M}B=\vec{r}{BC}\times\vec{T}$. [ \begin{align*} \vec{M}_B&=\begin{vmatrix} \hat{i}&\hat{j}&\hat{k}\ r\cos\theta&r(1+\sin\theta)&0\ T\cos\alpha&T\sin\alpha&0 \end{vmatrix}\ &=rT(\cos\theta\sin\alpha-(1 + \sin\theta)\cos\alpha) \end{align*} ]

Step5: Calculate moment about point C

The position vector $\vec{r}_{CC}=\vec{0}$. So, $M_C = 0$ (since $\vec{M}=\vec{r}\times\vec{F}$ and $\vec{r}=\vec{0}$).

Answer:

a) $M_O=rT\sin\alpha$ b) $M_A=rT((1 + \cos\theta)\sin\alpha-\sin\theta\cos\alpha)$ c) $M_B=rT(\cos\theta\sin\alpha-(1 + \sin\theta)\cos\alpha)$ d) $M_C = 0$