A force of 40, N acts on a point B at the end | vedclass

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(C) The moment of a force about a point is given by $	au = rF \sin \phi$,where $r$ is the position vector from the pivot point $A$ to the point of ap

Vedclass · Accepted Answer

$	an 	heta = \frac{1}{2}$

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(C) The moment of a force about a point is given by $	au = rF \sin \phi$,where $r$ is the position vector from the pivot point $A$ to the point of application $B$,$F$ is the magnitude of the force,and $\phi$ is the angle between the position vector $\vec{r}$ and the force vector $\vec{F}$.To maximize the moment for a fixed magnitude of force,the force must be applied perpendicular to the position vector $\vec{r} = \vec{AB}$.Let the coordinates of $A$ be $(0, 4)$ and $B$ be $(2, 0)$. The vector $\vec{AB} = (2 - 0)\hat{i} + (0 - 4)\hat{j} = 2\hat{i} - 4\hat{j}$.The slope of the line $AB$ is $m_{AB} = \frac{0 - 4}{2 - 0} = -2$.For the force to be perpendicular to $AB$,the slope of the force vector $m_F$ must satisfy $m_F \cdot m_{AB} = -1$,so $m_F = -\frac{1}{-2} = \frac{1}{2}$.The force vector makes an angle $	heta$ with the horizontal,so its slope is $	an 	heta$. Thus,$	an 	heta = \frac{1}{2}$.

$A$ force of $40\, N$ acts on a point $B$ at the end of an $L$-shaped object,as shown in the figure. The angle $\theta$ that will produce the maximum moment of the force about point $A$ is given by:

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