In Stewart's calculus book he asks about the magnitude of the torque of $F$ applied on $P$. The formula given in the book is $|\tau|=|\vec{r}\times \vec{F}|=|\vec{r}||\vec{F}|\sin\theta$. So my guess is the only important component in $\vec F$ is the perpendicular one to the axis of rotation, than the torque in this case is $\tau=240\cdot\sin30\cdot2$.
So the other $2m$ (the stick perpendicular to the axis of rotation) doesn't play any import role here?
On
Drop a perpendicular from P onto the line of force vector. Diagonal length is $ 2 \sqrt2 $ meters by Pythagoras thm.
The torque is
$$ 240 \times 2 \sqrt2 \sin ( 45^{\circ}+ 30^{\circ}) \;Nm $$
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Let's write a solution to this without worrying about angles.
Imagine a right handed rectangular coordinate system with origin $(0,0)$ at the joint. Point $P$ is on the negative $x$ axis and $F$ is acting on negative $y$ axis.Let unit vectors along $x,y,z$ axes be $\hat{i},\hat{j},\hat{k}$ respectively.
Let the force be acting on point $Q$.
$P=(-2,0)$ and hence $Q=(0,-2)$
$\vec{PQ}=\vec{r}=\vec{Q}-\vec{P}=(2,-2)=(2\hat{i}-2\hat{j}) m\implies |\vec{r}|=\sqrt{2^2+2^2}=2\sqrt 2 m$
$\vec{F}=240 (-\cos 30 \hat{i}-\sin 30 \hat{j})N$
Torque =$\vec{\tau}=\vec{r}\times\vec{F}=(2\hat{i}-2\hat{j})\times 240 (-\cos 30 \hat{i}-\sin 30 \hat{j})Nm$
Can you take it from here?
Draw $\vec r$ into the diagram. That will make everything much easier to see.
The vector $\vec r$ goes from $P$ to where the force is applied. Pythagoras tells us immediately that $|\vec r| = 2\sqrt2\,m$. Also, the angle between $\vec r$ and $\vec F$ is found to be $45^\circ + 60^\circ = 105^\circ$ (or, more immediately visible, the angle between $-\vec r$ and $\vec F$ is seen to be $45^\circ + 30^\circ = 75^\circ$).