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 July 9th, 2014, 02:18 PM #1 Senior Member   Joined: Jul 2014 From: Norway Posts: 140 Thanks: 3 Math Focus: geometry How to find the resultant force on the object I am so confused on how you solve these problems. One I have on my summerwork that has a 65 degrees angle then one line is 14 and the other is 9. I am sorry if this is not making any sense ;/ Last edited by girlbadatmath; July 9th, 2014 at 02:29 PM.
 July 10th, 2014, 12:49 PM #3 Senior Member   Joined: Jul 2014 From: Norway Posts: 140 Thanks: 3 Math Focus: geometry Thank you for taking the time to explain this to me! ok so this is what I have concluded. The magnitude is 65degress and the larger velocity is 14 and 9 would be the horizontal line. So I applied the Pythagorean theorem and came up with the resultant force on the object being 16.643. I am confused on how the magnitude has anything o do with getting the answer tho. :/ Oh wait! does 115 degrees have to do with anything? OMG....my brain ...... ****Imagine this picture and on the left side is 65 degress, the arrow closer to the top being 14units and the horizontal arrow being 9. This is what I came up with. http://www.basic-mathematics.com/images/Obtuseangle.gif (click search by image and its the only one that pops up) Sorry this is the only way I can explain what Ive figured out
July 11th, 2014, 02:17 AM   #4
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A great start! You have the wrong answer, but don't worry.

First thing, let's improve your diagram,

i) write down the magnitude of each force (14N or 9N) by each arrow. Remember that your vectors in this problem represent force, not velocity, so you should write the units as N (for Newtons).
ii) draw a set of axes at the point the two forces originate (basically just a vertical line and a horizontal line). Label the axes x and y. Use a ruler! If you get puzzled about the horizontal vector because it overlaps, make the horizontal vector bold by making the arrow a little thicker. You should always draw the axes first.
iii) We know that the 14N force is off to one side, so we can draw a little triangle and put the 65 degree angle in the diagram. This helps a lot with the calculations later.

You should have something that looks like the attached file. I prefer to hand-draw these things but I don't have a scanner to hand

Also, a couple of things to clarify before we move on

i) The magnitude is, in simple terms, how big the vector is. The magnitudes of the forces are 14N and 9N.
ii) The direction is where the arrow is pointing. In our question we are using angles or descriptions (horizontally to the right) to mark the direction.

Every vector must have a magnitude and a direction. If you don't put the direction down, you can lose marks in an exam!

That's step 1 done from my first post. Now let's look at the second step:

Quote:
 2. Determine the horizontal and vertical components of each vector (use trigonometry for this)
I recommend following through each calculation bit by bit on paper as you read this. It will really help, so make sure you have pencil, paper and calculator at the ready and that you have your newly improved diagram in front of you

You see the triangle we drew with the angle 65 degrees? We need to calculate the lengths of the sides of that triangle. That will give us the horizontal and vertical components of the vector. We use trigonometry for this.

Well.... $\displaystyle \sin \theta = \frac{opp.}{hyp.}$ from SOHCAHTOA (if you're not sure what this is, look it up on Google!). In our triangle the opposite is the horizontal side, which we're after, and the hypotenuse is 14N. So, we can rearrange it a bit:

$\displaystyle opp. = hyp. \times \sin\theta$

and we can put the numbers in:

$\displaystyle opp. = 14 \times \sin 65^{\circ} = 14 \times 0.9063 = 12.688$

So the horizontal component of the 14N force is 12.688N to the left. It's important to notice that it goes to the left

You now need to get the vertical component of the 14N force (hint: use cosine...)

The horizontal component of the 9N force is just 9N, because it is horizontal already
The vertical component of the 9N force is 0N, because the arrow is horizontal.

Once you do this step, you're nearly done as this is the annoying bit. Also, the 9N force is a trick question. Let me know what answer you get and, if you're feeling adventurous, have a go at step 3 in my first post.
Attached Images
 mathsforum_1.jpg (5.3 KB, 3 views)

Last edited by Benit13; July 11th, 2014 at 02:30 AM.

 July 11th, 2014, 09:47 AM #5 Senior Member   Joined: Jul 2014 From: Norway Posts: 140 Thanks: 3 Math Focus: geometry Ahhh Thank you so much!!! However I think the 65 degrees is supposed to go to the left of the arrow. I had a tutor session and that was in the answer key but I am still confused on where to go from there
 July 14th, 2014, 01:17 AM #6 Senior Member   Joined: Apr 2014 From: Glasgow Posts: 2,156 Thanks: 731 Math Focus: Physics, mathematical modelling, numerical and computational solutions No worries, just draw the dotted line vertically down from the arrow tip instead of horizontally to make a triangle. Then perform the trigonometry on that one instead. Quick question... could you follow through my other steps? If not I would recommend revising some trigonometry from a maths course/book. You won't be able to do this question without knowing how to use sine, cosine and tangent to get lengths of the sides of triangles.
 July 15th, 2014, 11:58 AM #7 Senior Member   Joined: Jul 2014 From: Norway Posts: 140 Thanks: 3 Math Focus: geometry Thank you! Okay so I got 12.688 of the 14N force even thought I inputted 12 x sin 115degrees . Which is the same you got. Now how do I figure out how to do cosine? I have literally had to teach math to myself my whole life because I struggle with dyscalculia so I am sorry for my apparent struggle. Thank you for being so helpful!!!
July 16th, 2014, 02:06 AM   #8
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No worries. I'm glad you managed to perform the sine of the force to get the horizontal component of the 14N force, but I'm worried that you don't understand why this works. The process of using cosine to get the vertical component is very easy to do once you know how.

Here's what I'm going to do: I'll provide the entire answer here in detail, but I would definitely advise you to get a maths text book and revise some trigonometry, specifically SOHCAHTOA. You should definitely do some exercises in the maths book so you are comfortable with the mathematics. These questions and many others like it will be virtually impossible to do unless you understand trigonometry. The following online resources might be able to help you out too:

SOHCAHTOA:

Finding the length of the side of a triangle:
Link: Finding a Side in a Right Angled Triangle

Here's the answer to the question you posed:

Note: You pointed out that there was a mistake in the original diagram, so I have changed it. Please confirm whether it is the right one. The change in the diagram changes the answer!

Vertical component of 14N force:
$\displaystyle \sin \theta = \frac{opp}{hyp}$
$\displaystyle opp = hyp \times sin\theta = 14\sin65^{\circ} = 12.688N$

Horizontal component of 14N force:
$\displaystyle \cos \theta = \frac{adj}{hyp}$
$\displaystyle adj = hyp \times \cos\theta = 14\cos65^{\circ} = 5.917N$

Vertical component of 9N force = 0 (it is horizontal)

Horizontal component of 9N force = 9N (it is horizontal)

The resultant force has a total vertical component of 12.688 + 0 = 12.688N
The resultant force has a total horizontal component of 9 - 5.917 = 3.083N

Use Pythagoras to get the magnitude of the resultant force:

$\displaystyle hyp^2 = opp^2 + adj^2$
$\displaystyle hyp = \sqrt{12.688^2 + 3.083^2} = 13.057N$

Use tangent to get the direction of the resultant force:

$\displaystyle \tan\theta = \frac{opp}{adj}= \frac{12.688}{3.083} = 4.115$
$\displaystyle \theta = \arctan(4.115) = 76.34^{\circ}$

So the resultant force is 13.057N, 76.34$\displaystyle ^{\circ}$ up from the horizontal.
Attached Images
 mathsforum_1.jpg (4.9 KB, 1 views)

Last edited by Benit13; July 16th, 2014 at 02:07 AM. Reason: Latex bs

 July 16th, 2014, 09:13 AM #9 Senior Member   Joined: Jul 2014 From: Norway Posts: 140 Thanks: 3 Math Focus: geometry Yes Thank you!! That is the correct diagram! And thank you so much for posting the links for SOHCAHTOA! It has already helped me so much!!! Thank you for taking the time out of your life to help me with this, I know people like you don't have to do this and I don't take your help for granted! Thank you once again!!! You are better at explaining things then all the tutors I have had usually they just get frustrated with me because of how long it takes me to grasp concepts. Anyways, Thank you for sticking through this with me!!!! <3 Last edited by girlbadatmath; July 16th, 2014 at 09:16 AM.
 July 17th, 2014, 01:24 AM #10 Senior Member   Joined: Apr 2014 From: Glasgow Posts: 2,156 Thanks: 731 Math Focus: Physics, mathematical modelling, numerical and computational solutions No problem

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