3
$\begingroup$

In my engineering textbook, I didnt understand the example given to explain the magnitude of the moment. It asks to calculate the magnitude of the moment about the base point O of the 600-N force in five different ways. Now, I dont understand the third way it calculated the moment. What I don't understand is why d(1) = 4 + 2 tan(40) and so the moment is 460(d(1)).

It used the principle of transmissibliity to move the vector force to point B and so we dont need one of the component as it is parallel with the distance. And to calculate the distance we have 4 + 2 tan(40). I understand upto 4 but I dont understand why we add 2tan(40). How is 2 vectored with tan(40)?

Use picture I have uploaded.

| improve this question | |
$\endgroup$
  • $\begingroup$ You would be much better off if you started to use cross product for the moment. $\endgroup$ – joojaa Nov 14 '18 at 6:13
  • $\begingroup$ The exercise is apparently to work out the moment in 5 different ways, only three of which are shown here - perhaps cross product was used in one of the other two? If not, a nice sixth method for OP to read up on! $\endgroup$ – Jonathan R Swift Nov 14 '18 at 9:57
3
$\begingroup$

This is a trigonometry question.

Point B is a certain distance upwards from Point A. How much? You know the distance that it is to the left from Point A ($2\text{m}$) and the angle that it is from point A ($40°$), so using standard trigonometry $\tan(\theta)=\frac{opposite}{adjacent}$, you can calculate the vertical distance by rearranging to give: $opposite=adjacent\tan(\theta)$, or in this specific case, $x=2\tan(40)$, such that $d_1 = 4+2\tan(40)$.

| improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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