help guys

- anonymous

help guys

- chestercat

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- anonymous

statics

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- anonymous

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- triciaal

@jim_thompson5910 please help I don't remember this stuff

- anonymous

help me please :(

- triciaal

@nincompoop will you please help?

- anonymous

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- rajat97

i think i can help you but what is that 2N/m written in the blocks??

- anonymous

@rajat97 those are distributed loads. you multiply the force by the length of the distribution so you can treat it as a single force. Like the load between A and B is 2 N/m distributed over 3 m, so it's 6 N acting in the middle 1.5 m from either end.
I don't see a way to do this without setting up a long series of equations

- rajat97

Thanks a lot @peachpi

- anonymous

- anonymous

help me guys

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- anonymous

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- anonymous

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- anonymous

i cant solve this stuff

- anonymous

help me please

- rajat97

i have the answer i don't know it is correct or wrong
normal reaction at K=4.571N
normal reaction at L=3N
normal reaction at M=9.5715N
normal reaction at N=8.1905N
please let me know if any of my answers are correct
if they are, i would be able to help you
I used only two concepts - newton's third law and rotational equilibrium

- anonymous

can you show it ?

- anonymous

- anonymous

@rajat97 ur answer is correct , how did you get that :((( , please help me with that :((

- mathmate

There is no system of equations to solve, you only have to resolve the reactions of each of the 7 beams, from top to bottom, and convert them to loads.
To find the reaction of the supports, you take moments about the other support, and solve for the reaction.
For example, for beam AB, length = 3 m, load = 2 N/m, so total load = 3 m * 2 N/m = 6 N. By symmetry, each support supports a load of 6/2=3 N, thus the load on support B is equivalent to 3 N.
|dw:1436705214582:dw|
Note that for calculation of reactions (external loads only), the uniform load can be replaced by an equivalent load equal to the total load (udl*length) applied at the middle of the uniform load. Remember that this should not be done when you are calculating the internal forces of the beam.
Now to tackle beam DC, you will make a load diagram, take moments about D and C, and check that the sum of reactions equal to the total load.

- rajat97

yeah i'll upload an image
please wait for a moment

- rajat97

mathmate is correct
i needed one equation
and then needed to resolve the forces and the torque things

- mathmate

|dw:1436705928662:dw|

- anonymous

i dont get it

- anonymous

- anonymous

how did you get the 3 ?

- rajat97

oh just choose point k as your ref. point and then balance torque on the stick KL (the torque due to 5N thing and the opposite torque due to normal reaction at point L)

- mathmate

R(C)=(3*1+5*3+8*6)/8=33/4 N
R(D)=(3*7+5*5+8*2)/8=31/4 N
Check: R(C)+R(D)=(31+33)/4=16N = (3+5+8) N checks.
Either you use symmetry, total load=2*3=6N, each reaction gets 3N.
or
you take moments about, say, A, with a total load of (2*3)=6 N in the middle of a span of 2 m, then M about A = 6*1 - R(B)*2= 0, or
solve for R(B)=6/2=3 N.

- mathmate

Correction: the span was 3 m, so
you take moments about, say, A, with a total load of (2*3)=6 N in the middle of a span of 3 m, then M about A = 6*1.5 - R(B)*3= 0, or
solve for R(B)=9/3=3 N.

- anonymous

@mathmate can you rewrite it , i dont understand sorry

- rajat97

i don't have a good camera currently so i cannot upload the image of my work and it may take too long to draw the things

- anonymous

ijust dont get it

- mathmate

@jacalneaila
Do you have a text book that shows you how to find the reactions of a simply supported beam? You need to start from there! Otherwise
If you are doing an engineering course, try this
http://bendingmomentdiagram.com/tutorials/calculating-reactions-at-supports/
and if you are doing a physics course, try this:
https://www.youtube.com/watch?v=5-QZ0JKamPU

- rajat97

first of all you have to find the normal reaction at each end
we can start from the top
we'll do it for span AB
we choose the reference point as A and balance torque on the span w.r.t. point A
6N down at a distance of 1.5m from point A will give a clockwise torque of 6 x 1.5 = 9Nm
assume the normal force at point B due to the span below it to be Nb. The torque due to the normal force will be anticlockwise which will balance the torque due to the distributed mass.
so we get 9Nm=Nb x 3 which gives me Nb=3N

- rajat97

now, the span AB will apply the third law pair of the normal force Nb on the span DC in the opposite direction but of same magnitude
and the force due to the distributed load will be 8N at a distance of 6m from end D
now we balance torque on the span DC

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- anonymous

their answer is different now

- anonymous

K = 4.76
K = 4.76
L = 3
M= 5.265
N = 6.755

- rajat97

okay no problem i'll try it once more

- rajat97

i'll get it to you in sometime

- rajat97

okay i got it i just missed out some normal reactions

- rajat97

do you live in india??

- rajat97

- rajat97

i gotta go but 'll be right back after sometime

- anonymous

thank you :)

- anonymous

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- mathmate

Is that a new problem?

- rajat97

Thanks for the medal @mathmate

- mathmate

You're welcome! You deserve the medal with all the work and efforts you put in! :)

- anonymous

another problem

- anonymous

- anonymous

- anonymous

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- rajat97

i'm on it

- rajat97

by internal forces do you mean tensions in the strings??

- rajat97

- anonymous

yes

- anonymous

tension in the strings

- anonymous

@rajat97 i have a class tommorow morning , i just want to sleep hehe , but can help me with that ? please , just my reviewr for my exam :)))

- rajat97

okay we need to draw a so called free body diagram for each knot and balance the forces on the knot as it is at equilibrium
and sure i'll help ya

- anonymous

i need this , because this coming week i have a prelim exam in STATICS

- rajat97

for the lowest knot: i.e. point B
let the tension in the string AB be T1 and that in string BC be T2
then by breaking the tensions in horizontal and vertical components, we get,
the vertical component will balance the load of 50N and the horizontal components of the tensions will balance each other
i'll post a drawing and it'll be more clear

- rajat97

|dw:1436975835661:dw|

- anonymous

@rajat97 i will back tommorrow after school :) i promise :)

- rajat97

yeah sure until then i'll solve it for you
and i'll have my dinner now

- anonymous

@rajat97 bye , and thank you :))

- rajat97

no need for thanks, chill!

- rajat97

so i got tension in string AB=40.5827N
and tension in string BC=36.3855N

- rajat97

Thanks for the medal @peachpi

- rajat97

Now, we write Newtons law equation for the point C i.e. knot C
assume the tension in the string DC to be T3 and that in string CG be T4 so, just look at the FBD of the point C|dw:1437011797701:dw|
so from the figure, we get, T3sin(5deg)+T4sin(20deg) = 50N+T2sin(15deg)(i forgot to write the component of T2 acting on the point C in the downward direction hope you understand)
and also T3cos(5deg) = T4cos(20deg)
from these equations, we get T3=132.1129N
and T4=140.0688N
please let me know if my answers are correct
hope this helps you
now i think that you should solve the remaining part of the question because i'm tired of typing(lol):p
just joking
if you have any problem in the remaining part of the question, you may feel free to ask anything.
well i'm not sure if my answers are the correct ones but i tried my level best

- anonymous

@rajat97 , im home :)) thanks

- anonymous

@rajat97 i will try this thing :))

- rajat97

Please, let me know if my answers are correct or not:)

- mathmate

Here's my input:
|dw:1437221537394:dw|

- mathmate

|dw:1437222001918:dw|

- mathmate

Please note that direction of forces (in tension) are indicated at each joint to permit correct setup of the joint equilibrium equations. It is imperative that a diagram is drawn and CHECKED before proceeding to solve for the forces.
All forces must be indicated on the diagram to ensure consistent use of symbols.
Now proceed joint by joint, from bottom up.
All forces are in tension, so we can put in the direction of forces at each joint to help set up equations.
I will not make the solution more busy by solving systems of 2 equations at each joint. I will set up the equations, and give the results.
Now proceeding to solve for forces T1 to T8 in that order, and CHECK results of external forces (attachments to wall and ceiling) for accuracy of work.
Joint B:
T1cos30=T2cos15
T1sin30+T2sin15=P
=>
T1=1.366P
T2=1.225P
Joint C:
T3cos5+T2cos15=T4cos20
T3sin5+T4sin20=T2sin15+P
=>
T3=1.971P
T4=3.348P
Joint D:
T5cos10=T3cos5+T6cos60
T5sin10+T6sin60=T3sin5+P
=>
T5=2.433P
T6=0.865P
Joint E:
T7cos45+T4cos20=T8cos25
T7sin45+T8sin25=T4sin20+P
=>
T7=0.654P
T8=3.982P
CHECK: (Very important to provide correct answer when solving statics problems)
Equilibrium of forces in the vertical direction:
T1sin30+T5sin10+T6sin60+T7sin45+T8sin25=4.000P
Since the total applied vertical forces equals 4P, vertical equilibrium is satisfied.
Equilibrium of forces in the horizontal direction:
T1cos30+T5cos10-T6cos60+T7cos45-T8cos25=0.000P
Since there is no applied horizontal force, horizontal equilibrium is established.
Summary of results (and conclusion)
P=50N
T1 = 1.3660P
T2 = 1.2247P
T3 = 1.9709P
T4 = 3.3484P
T5 = 2.4330P
T6 = 0.8652P
T7 = 0.6539P
T8 = 3.9819P

- mathmate

@rajat97 you can check your answer against the above.

- anonymous

i try :) , if i have a the same answer :)) @mathmate thanks

- anonymous

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- mathmate

Congrats, your numbers are all good!
If you look at the answers I gave, and substitute P=50N, you will get exactly the same numbers as you have!
Good job! :)

- anonymous

- anonymous

- mathmate

Yes?

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