0
$\begingroup$

enter image description here

My question is: Is perpendicular distance in the equation $\tau=Fd$ the distance of the yellow or red line?

My belief was that it was the red line, however when I take moments about $O$ I get $$\begin{align*}\tau=4a\sin\theta\, mg&-2a\sin\theta\,4mg=0;\\ 4a\sin\theta\,mg&=8a\sin\theta mg;\\ 4&=8 \end{align*}$$

Which leads me to think my assumption was wrong. I understand the diagram is not accurate but I don't think I can assume $OC$ is perpendicular to the dashed line since the $\theta$ is not necessarily $45^{\mathrm{o}}$.

I have attempted to look elsewhere online but everything I have seen leads me to think the yellow line is the right distance.

edit: $\angle OBC=90^{\mathrm{o}}$

$\endgroup$
3
  • $\begingroup$ It is the yellow distance. $\endgroup$
    – MrYouMath
    Commented May 4, 2016 at 17:25
  • $\begingroup$ What is $4a\sin{\theta}$ supposed to represent? It wouldn't be either the red or the yellow line. $\endgroup$
    – BowlOfRed
    Commented May 5, 2016 at 6:29
  • $\begingroup$ You have taken wrong angle while finding torque due to force exerted on arm BC $\endgroup$
    – ATHARVA
    Commented Feb 24, 2017 at 2:02

4 Answers 4

1
$\begingroup$

enter image description here

Consider the figure above. The torque about $O$ is given by

$$ \vec{T}=\vec{r}\times\vec{F}$$

Its magnitude is

$$||\vec{T}||=||\vec{r}||.||\vec{F}||.\sin{\theta}$$ As you can see in the figure, $||\vec{r}||.\sin{\theta}$ is equal to the length of the green line, which is the perpendicular drawn from $O$ to the line of action of $\vec{F}$ and is called the perpendicular distance.

$\endgroup$
1
$\begingroup$

It is the yellow and green lines. The perpendicular distance is always perpendicular to the direction of the force.

$\endgroup$
1
$\begingroup$

You have used the correct definition of the torque due to a force about a point as the force times the perpendicular distance from the point to the line of action of the force which are you green (length $2a \sin \theta$) and yellow (length $4a \sin \theta$) lines but not realised that there is a net torque about $O$ on the system of $4mg a \sin \theta$ anticlockwise.
That is why you got $4=8$.

$\endgroup$
0
$\begingroup$

The question is rather unclear: if the force under consideration is the weight $mg$ then the lever arm is of course the yellow red, since it must be perpendicular to the force.

$\endgroup$
3
  • $\begingroup$ Thank you for your response. The forces under consideration are the forces acting at point A and point C. So yes the weights 4mg and mg. However I don't understand why there is a contradiction when I use the distance given by the yellow line. $\endgroup$
    – Mike
    Commented May 4, 2016 at 17:45
  • $\begingroup$ In this case the lever arm to be included in the computation of the moment is the yellow one (and the green one for the other force). It follows directly from the definition of the moment as cross product between the position vector (from the point with respect to which you want to compute the torque till the application point of the force) and the force, and in magnitude, the product of the magnitudes of the two vectors times the sine of the angle between them, $\endgroup$
    – Vexx23
    Commented May 8, 2016 at 19:00
  • $\begingroup$ or, in a more explicative way, the product between the magnitude of the force and the length of the lever arm, i.e. the distance between the point used as reference and the action line of the force (exactly the component of the distance vector perpendicular to the force). I hope it may help you (sorry for the double comment but it was too long) $\endgroup$
    – Vexx23
    Commented May 8, 2016 at 19:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.