4chan archive /sci/ (index)
2013-09-01 02:07 6002931 Anonymous (423425235.png 348x285 40kB)
hey /sci/, is here anybody who can solve and explain me this problem. Started learning statics, can solve really basic problem, but get stuck on this one

6 min later 6002937 Anonymous
>>6002931 bump

20 min later 6002947 Anonymous
sum of forces in x direction = 0 sum of forces in y direction = 0

23 min later 6002950 Anonymous
as in, -3cos60 - 2cos30 +Ry = 0 2sin30 - 3sin30 + Rx = 0 R=rt(Rx^2 + Ry2)

1 hours later 6002984 Anonymous
>>6002950 and theta = arctan (R_y/R_x)

1 hours later 6003004 Anonymous
>>6002947 Dear god, you're retarded. Not every damn problem is just sticking numbers into he same damn formula(s). This is why most maths education sucks. Not ONCE in the question does it say that the system is in equilibrium. Fuck Jesus, it just asks for the resultant of two vectors. Christ.

1 hours later 6003008 Anonymous
>>6002931 Seeing as the guy who asked before is retarded, I'll tell you how to do it. You have the two vectors in polar form. Convert to Cartesian form. Sum components. Find magnitude and direction. I'm not going to tell you how to do each little step. Textbooks are your friend.

1 hours later 6003012 Anonymous
>>6003004 I agree with your angriness but don't sage...it's not gonna help. >>6002947 should be : sum of the two forces in x direction = resultant in x direction sum of the two forces in y direction = resultant in y direction So you need to do projections. Be careful with the signs. >>6002950 becomes R_x = -3cos60 - 2cos30 (units : degrees and kN) R_y = 2sin30 - 3sin60 R=sqrt[ (R_x)^2+(R_y)^2 ] >>6002984 remains correct.

1 hours later 6003013 Anonymous
>>6003008 oh, ok. I approve this one, but I already replied.

2 hours later 6003032 Anonymous
>>6002950 okay, i have solved correctly R, Fx and Fy, but when i try to solve the angle I get 20degrees and it should be 208

2 hours later 6003033 Anonymous
>>6003032 this is what i do arctan(1.59/3.23)= 26.21 degrees

2 hours later 6003038 Anonymous
>>6002931 every one of these problems, you need to immediately choose an axis-set (frame of reference) and break up all external forces into their projections along these axis, then add them up (as well as look for moments of inertia). every. single. problem. its so mundane, you would write a program in your calculator to do this automatically for you, if you were an A-tier student.

2 hours later 6003040 Anonymous
>>6002947 >not immediately recognizing that the resultant force has a net projection in the negative x direction... moron.

2 hours later 6003043 Anonymous
>>6003032 use >>6003008 or >>6003012 because >>6002950 was wrong. Try to use radians instead of degrees, it's more convenient. Last thing, if R_x or R_y are negatives, then draw a picture for understanding what is Arctan(R_y/R_x). You may have to "adjust" with + or - Pi to get your theta.

2 hours later 6003080 Anonymous
>>6003043 this is what i did fx= -3cos60i-2cos30i kN fy=-3sin60j+2sin30j kN fx=-1.5i-1.73i=-3.23i kN fy=-2.59j+1j=1.59 kN R=sqrt(-3.23^2+(-1.59^2))=3.60 which is the right answer according to the book degree=tan^-1(1.59/3.23)= 26.21degrees, but according to the book it should be somthing like 206degrees

2 hours later 6003087 Anonymous
>>6003080 since fx<0 and fy<0 theta = [180 + Arctan()] =180+26,21 ~206° :) Draw a picture with just the vector R and its projections on y-axis and x-axis to unnderstand the why of "+180".

2 hours later 6003091 Anonymous
>>6003087 oh my god, thank you so much.

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