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Net force (Diagrams incl) Please check my solutions 
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#1
Mar212, 05:28 PM

P: 209

1. The problem statement, all variables and given/known data
Calculate the net force acting on each object indicated in the following diagrams. Show your work. The diagrams are in attachment, you dont have to download them to view them :) 2. Relevant equations c^2=a^2+b^22abCosC c^2=√a^2+b^2 SOH CAH TOA 3. The attempt at a solution 29a) since we are given a 90 degree angle and the rest is 35 degrees, we can add 90+35 =125° since we now have 2 sides of a triangle and the middle angle, we can use cosine to find the net force. c^2=a^2+b^22abCosC =(38)^2+(22)^22(38)(22)Cos125 =1444+4841672Cos125 =19281672Cos125 =1928959 c^2=√969 c=31.12 N Fnet=31.12 N v This 29b) is the important one, because this is where im not sure which solution is the right one. 29b) Since one side of a line has 180 degrees, and we are given 45 degrees, one can assume 18045 degrees would give you the angle on the left side of Z. I attempted 29b) in 2 different ways, im not sure which one is the better way and gives the correct solution. In method 1: i deal only with the downward vector (10 N) and the north west vector (17N) and since 18045=135 we have the angle in the middle, which is 135 degrees. so using this information i use the cosine law to find the net force. c^2=a^2+b^22abCosC =(17)^2+(10)^22(17)(10)cos135 =289+100340cos135 =389240.4 =√148.6 Fnet=12.19 N Method 2: This method, looks at the entire diagram, and assumes its one big right angle triangle. using the smaller right triangle, we find side x. Sin45=opposite/hypotenuse sin45=x/75 sin45(17)=x 12.02N = x (other side we use the given, which is 8 Newtons + 10 Newtons = 18 Newtons) now that we have every side besides the hypotenuse we use Pythagorean to find the hypotnuse which will give us the net force. c=√a^2+b^2 =(18)^2+(12)^2 =√324+144 =√468 c=21.63N Fnet = 21.63 N 29c) here we add the 2 given angles. resulting in the middle angle, which is 32+24=56 ° since we have the 2 sides and the middle angle, once again we use cosine law, to find net force c^2=a^2+b^22abCosC =(15)^2 + (12)^2 2(15)(12)Cos56 =225+144360cos56 =√369201.3 =√167.7 =12.94 N Fnet= 12.94 N 


#2
Mar212, 06:21 PM

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What you are doing is vector subtraction, not vector addition.



#3
Mar212, 10:26 PM

P: 209

um..what?



#4
Mar312, 12:23 AM

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Net force (Diagrams incl) Please check my solutions
I appears that the numerical results you have for the magnitudes of the resultant forces is correct. If you also need the direction of the resultants, your figures will lead you to the wrong answer in each part. 


#5
Mar312, 08:33 AM

P: 209

oh i have the directions, i just didnt post them because there were LOTS of calculations involved, and then id have posted a wall of text, i wanted to minimize the amount of work required by the good people of the forums to check my solutions, in any case 29b) i have 2 possible solutions, which one do you think works?



#6
Mar412, 10:03 AM

P: 209

anyone? for 29b) ?



#7
Mar412, 10:41 AM

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#8
Mar412, 11:08 AM

P: 209




#9
Mar412, 02:26 PM

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To mathematically do the equivalent of the graphical method, find the individual components of all the vectors (EW components and NS) components, corresponding to X and Y components on a Cartesian XY plot) and add the like components. Can you write the components of each vector? Make a vertical list of the xcomponents and another of the ycomponents. Add the columns to find the components of your resultant. 


#10
Mar512, 07:42 AM

P: 209

Anyhow, is there any way you can show me how to do this, in plain english please? :) 


#11
Mar512, 09:09 AM

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The resultant vector is the 'result' of adding A and B. The component of the resultant along the Xaxis is equal to the sum of the Xcomponents of A and B. Similarly, the Yaxis component of the resultant is equal to the sum of the Ycomponents of A and B. Use whatever trigonometry methods you know to extract the individual components. ##R_x = A_x + B_x## ##R_y = A_y + B_y## ##R = \sqrt{R_x^2 + R_x^2}## ##\theta_R = atan \left(\frac{R_y}{R_x}\right)~~~## note: beware of quadrant placement when dealing with negative valued components! 


#12
Mar612, 09:05 AM

P: 209

hm... thanks a lot! So what your saying is each point on the resultant is the sum of a point on vector A and B so the x coordinate of a point on resultant is the sum of an x coordinate on vector A and B? and likewise for each y coordinate on the resultant? Also if possible can you give me an example of subtracting vectors? Similarly to how you showed me how to add vectors in the above post :) Or maybe to subtract vectors would be essentially the same thing as adding except, in negative axis? (x and y) 


#13
Mar612, 10:49 AM

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Vectors are quantities with magnitude and direction. Vectors are not line segments, although they are represented that way as a handy graphical notation which facilitates thinking about the math. 


#14
Mar812, 11:15 AM

P: 209

hm.. ok i think i finally figured out this vector addition/subtraction thing
(yay google) i did the diagram again, and got it done, my question is to find resultant force(mathematically) ill have to use cosine, since i cant use Pythagorean. My question is, the original 45 degree angle, its between the resultant vector and the 17 N vector right? So i should be able to use it in calculations via cosine to find resultant? 


#15
Mar812, 11:55 AM

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The procedure for summing vectors is to take each of the vectors individually and establish their x and y components. This is done using trigonometry and any angles you've been given that establish the vector's direction, and hence it's projections along the x and y axes of the coordinate system. The x components for each vector are summed to given the x component of the resultant. The y components for each vector are summed to given the y component of the resultant. These components define the result vector. 


#16
Mar912, 10:20 AM

P: 209

for the very first one (29a) since i know now that vectors are always added head/tail method, is the diagram i did incorrect? shouldnt it be 38 N vector, with the 22N vectors tail on the 38 N vectors head?



#17
Mar912, 10:32 AM

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