Express F as a vector in terms of the unit vectors i and j

[hagdt37r]

You can find the question on the left and the working out done on the right (by my tutor) but there are parts of this I don't understand:

1) i understand that[math]\sqrt{5^2+12^2}[/math] gives you 13 and to calculate the magnitude of a line across the x and y axis is Fcos theta and Fsin theta
2) what i don't understand is:
- why does sinTheta = 5/13 and cosTheta=12/13, i'm not sure how you get a resultant angle from this? In most problems using these functions I usually just input the angle that it given, but obviously there is no angle given and i'm left to calculate it myself but how does -6.5(12/13) and -6.5(5/13) result in the angle being calculated im confused

If anybody could explain the process step by step it would be helpful

 

[Otis]

Hi. I don't see the answer in the requested form (i.e., expressed in terms of the given unit vectors). However, there is an easier way to find the magnitudes 2.5 and 6.

Use similar triangles. With the given right triangle (legs 5 and 12), we see that the hypotenuse is 13. We want the right triangle whose hypotenuse is half as long (6.5). Therefore, we shrink the given triangle by 50%. In other words, all three sides become half as long.

5/2 = 2.5

12/2 = 6



[imath]\;[/imath]

 

[Harry_the_cat]

View attachment 32069

You can find the question on the left and the working out done on the right (by my tutor) but there are parts of this I don't understand:

1) i understand that[math]\sqrt{5^2+12^2}[/math] gives you 13 and to calculate the magnitude of a line across the x and y axis is Fcos theta and Fsin theta
2) what i don't understand is:
- why does sinTheta = 5/13 and cosTheta=12/13, i'm not sure how you get a resultant angle from this? In most problems using these functions I usually just input the angle that it given, but obviously there is no angle given and i'm left to calculate it myself but how does -6.5(12/13) and -6.5(5/13) result in the angle being calculated im confused

If anybody could explain the process step by step it would be helpful

Why does sin Theta = 5/13 ? Look at the triangle and go back to basic trigonometry.
sin Theta = opp/hyp.

 

[The Highlander]

View attachment 32069

You can find the question on the left and the working out done on the right (by my tutor) but there are parts of this I don't understand:

1) i understand that[math]\sqrt{5^2+12^2}[/math] gives you 13 and to calculate the magnitude of a line across the x and y axis is Fcos theta and Fsin theta
2) what i don't understand is:
- why does sinTheta = 5/13 and cosTheta=12/13, i'm not sure how you get a resultant angle from this? In most problems using these functions I usually just input the angle that it given, but obviously there is no angle given and i'm left to calculate it myself but how does -6.5(12/13) and -6.5(5/13) result in the angle being calculated im confused

We would like to help you but the best person to provide the help you need in this case is, clearly, you, yourself!

I have taught both Maths & Physics for over thirty years and one of the perennial problems I have faced is pupils adopting a mechanistic approach instead of acquiring a thorough understanding of what is being taught! This appears to be the approach that you have adopted too and that is why you need to go back and revise the basics of "Right-Angled Triangle Trigonometry".

You say: "I usually just input the angle that it (is?) given, but obviously there is no angle given and i'm left to calculate it myself" but this indicates that you have learned that you can get the horizontal (Fcosθ) and vertical (Fsinθ) components of a force vector (F) by using F and the angle, θ, of its orientation. You are then expecting to be given both F & θ to simply substitute into the expressions you have learned but are now struggling when, as in this instance, θ itself has not been explicitly provided for you. That it why it is no use just learning a couple of little 'formulas' ('Fsinθ' & 'Fcosθ'), rather you need to understand how these expressions are first derived (from the basic "Right-Angled Triangle Trigonometry").

I suggest you study the diagram I have provided (below) for you (it would do no harm to copy it into your jotters/workbooks and refer back to it regularly) and ensure that you understand the relationships that all the elements in it have with each other, eg:-
[math]cos(θ) =\left(\frac{x}{h}\right) \Rightarrow x=hcos(θ)[/math] You need to understand how x, y, h & θ all relate to each other via both Pythagoras' Theorem and the Trigonometric Ratios (as they were defined for you).

Once you have a thorough understanding of these relationships you will then know the answer to your question ("why does sinTheta = 5/13 and cosTheta=12/13"). Furthermore, in this particular question, you have no need whatsoever to know (or calculate) the "angle" involved. The expression you that you are familiar with ("Fcos theta") does not involve multiplying F by the angle, it involves multiplying F by the Cosine of the angle and that (as defined) is a Ratio (one number divided by another). That is why "cosTheta=12/13"!

One final comment:-
Should you wish to seek further help in this forum it is best to supply the original question, exactly as it was provided to you. The picture you posted appears to have extracts from the original question accompanied by "working" that has been added "(by my tutor)" so we are left to guess what information was originally provided for you and what your 'tutor' has added into the mix.

You have, unfortunately, added to the confusion by stating: "You can find the question on the left and the working out done on the right (by my tutor)" whereas the added workings appear to be on the LEFT! (This conclusion is supported by the fact that there is a (small) error on the bottom line where it should say "F_y" (not "F_x") so that would appear to be part of your tutor's additions). This means we are now left entirely unsure as to what parts of your picture came from the original question and what has been added by you/your tutor, eg: the red triangle; was that added or part of the original question and did the question specifically state that the hypotenuse of that triangle was parallel to the vector representing the 6.5 kN Force?

Please give us all the information pertaining to your problem(s) including original questions (in their entirety and unamended in any way) plus, any and all subsequent workings/attempts at solution that you (or anyone else) have done. That way we are best placed to provide you with the assistance you may need.

Please come back and let us know that you have done the revision
suggested, understand all the relationships identified in my diagram and now fully comprehend what is going on in your picture. ?

 

[LCKurtz]

I am not one of the moderators here, but I do have a suggestion. I find the use of multiple size large fonts and bold, italic, and underlining to be very distracting. The above post is over the top in this regard. If I were the poster to whom that post was directed I would likely have additional negative reaction to it.

 

[AvgStudent]

I am not one of the moderators here, but I do have a suggestion. I find the use of multiple size large fonts and bold, italic, and underlining to be very distracting. The above post is over the top in this regard. If I were the poster to whom that post was directed I would likely have additional negative reaction to it.

I'm more concerned with the length. I would most likely skip the post as it's too long lol. I think people here are mostly looking for pointers, not a lesson. But that's just my opinion.

 

[The Highlander]

I am not one of the moderators here, but I do have a suggestion. I find the use of multiple size large fonts and bold, italic, and underlining to be very distracting. The above post is over the top in this regard. If I were the poster to whom that post was directed I would likely have additional negative reaction to it.

I'm more concerned with the length. I would most likely skip the post as it's too long lol. I think people here are mostly looking for pointers, not a lesson. But that's just my opinion.

My "lesson" wasn't aimed at either of you but thank you for your comments.
(Noted & filed.)