Finding the resultant of a force (using trig)

[Inertia_Squared]

Hi there, I'm a senior HS student doing engineering in Australia, and I've been having some problems with this one question, I have solved and proven it multiple times but I am never able to get the right answer (that matches the answer sheet), the answers seem a little dodgy (I might explain this at the end but context of the question is required first), so I'd like to know if I'm doing something wrong, or if the answers are wrong.

The question is "In the diagram shown, determine the magnitude of the resultant" (hypotenuse in this specific case) "using the cosine rule and the angle it makes to the horizontal using the sine rule if the angle between the two forces is 60 degrees." In a nutshell, it wants me to find the length of the resultant, and the angle between both the smaller and larger of the forces from the resultant.



I recently re-did the question in a neater format so it was readable, and re-did it only to get the same answer I got last time, which is still different to the question. Because of this, I only proved the resultant and not the angles, as I didn't want to waste my time if I was doing something wrong, and to my knowledge, you will always have to find the resultant first in order to find the angles.

Here is my cleaned up working (dashed line should end up being the resultant if I'm not mistaken):


I always get 28.618N as my resultant, however, the answer states that it is 23N (answers are rounded), with the two adjacent angles being 33* and 26* (this is the dodgy part as the angles should still add up to 60* if rounded correctly but it is close enough, so it still seems legit). Could someone confirm whether or not I have done this correctly and if I have done something wrong, what am I missing?

Thanks for taking the time to read this, I appreciate it!

 

[Harry_the_cat]

I also get 26.618N. Either the answer is wrong, or you have transcribed it incorrectly.

(BTW there is no "hypotenuse in this specific case" as there is no right angled triangle involved.)

 

[Dr.Peterson]

I agree with 28.618, which I got directly applying the cosine rule to triangle ADC. Something is definitely wrong. I get 26.996 degrees for the angle you marked in red.

Can we see the original problem as an image? And, of course, make sure the 23 is really given as the answer to this problem!

 

[Inertia_Squared]

I agree with 28.618, which I got directly applying the cosine rule to triangle ADC. Something is definitely wrong. I get 26.996 degrees for the angle you marked in red.

Can we see the original problem as an image? And, of course, make sure the 23 is really given as the answer to this problem!

Sure thing, I can confirm the answer, as I double-checked my teacher about what answers corresponding to what before coming here (although I didn't have time to ask him to re-do the question himself). I don't think I missed anything from the original question but I'll post it just in case...

(sorry about the eraser marks, needed to erase my old working out for the original to be visible , also please ignore the faded red right-angles, It's irrelevant to the question, I just sometimes draw right angles in random places for fun lol, idk why, but I digress.)

 

[Inertia_Squared]

(BTW there is no "hypotenuse in this specific case" as there is no right angled triangle involved.)

Wow I didn't know that was a thing, thanks!

 

[Deleted member 4993]

Sure thing, I can confirm the answer, as I double-checked my teacher about what answers corresponding to what before coming here (although I didn't have time to ask him to re-do the question himself). I don't think I missed anything from the original question but I'll post it just in case...

(sorry about the eraser marks, needed to erase my old working out for the original to be visible , also please ignore the faded red right-angles, It's irrelevant to the question, I just sometimes draw right angles in random places for fun lol, idk why, but I digress.)

View attachment 16745

The problem statement is incorrect.



The diagram does not (nor the verbiage) define the DIRECTIONS of the applied forces. Without direction the "vector" (force) is not defined - hence cannot be added.

In your work (for response #1) you have shown arrow-heads - hence the directions of the forces are defined. If those directions are correct - you have the correct answer.



On the other hand, if one of those arrow-heads are pointing another way - the answer would be different.

 

[Steven G]

The problem statement is incorrect.

View attachment 16751

The diagram does not (nor the verbiage) define the DIRECTIONS of the applied forces. Without direction the "vector" (force) is not defined - hence cannot be added.

In your work (for response #1) you have shown arrow-heads - hence the directions of the forces are defined. If those directions are correct - you have the correct answer.

View attachment 16752

On the other hand, if one of those arrow-heads are pointing another way - the answer would be different.

Subhotosh,
I have a question for you. How can the angle between the two vectors be 60 if one of the vectors goes in the opposite direction from what the OP showed in their diagram? I always thought that the angle between two vectors always had their tales at the vertex. Is that actually wrong?
Steve

 

[Deleted member 4993]

Subhotosh,
I have a question for you. How can the angle between the two vectors be 60 if one of the vectors goes in the opposite direction from what the OP showed in their diagram? I always thought that the angle between two vectors always had their tales at the vertex. Is that actually wrong?
Steve

In engineering, we take special care to draw arrow-heads. If we "add" vectors, using "polygon" construction - we draw the vectors head-to-tail. This is preferred method in engineering (head-to-tail), while adding vectors graphically - because multiple vectors can be added and resultant can be found.

In mathematics, on the other hand, things can be "imagined" to have tails without head, introducing unnecessary "complex" and "irrational" behavior.

 

[Inertia_Squared]

The problem statement is incorrect.

View attachment 16751

The diagram does not (nor the verbiage) define the DIRECTIONS of the applied forces. Without direction the "vector" (force) is not defined - hence cannot be added.

In your work (for response #1) you have shown arrow-heads - hence the directions of the forces are defined. If those directions are correct - you have the correct answer.

View attachment 16752

On the other hand, if one of those arrow-heads are pointing another way - the answer would be different.

I did notice this before posting here and attempted to clarify this with my teacher, they told me that the direction was moving in the manner I represented graphically (provided I didn't assume that's what he meant, he's a rather flamboyant character that tends to speak in half-riddles, but he's an excellent teacher), I recently asked them if they could re-do the question themselves to see if they get the same answer or how/why they don't. Hopefully I'll find out sometime next week and when I find out I'll be sure to keep this thread posted!