help me please :(
i think i can help you but what is that 2N/m written in the blocks??
@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
i cant solve this stuff
help me please
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
can you show it ?
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.
yeah i'll upload an image please wait for a moment
mathmate is correct i needed one equation and then needed to resolve the forces and the torque things
i dont get it
how did you get the 3 ?
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)
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.
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.
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
ijust dont get it
@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
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
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
their answer is different now
K = 4.76 K = 4.76 L = 3 M= 5.265 N = 6.755
okay no problem i'll try it once more
i'll get it to you in sometime
okay i got it i just missed out some normal reactions
do you live in india??
i gotta go but 'll be right back after sometime
thank you :)
Is that a new problem?
You're welcome! You deserve the medal with all the work and efforts you put in! :)
i'm on it
by internal forces do you mean tensions in the strings??
tension in the strings
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
i need this , because this coming week i have a prelim exam in STATICS
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
yeah sure until then i'll solve it for you and i'll have my dinner now
no need for thanks, chill!
so i got tension in string AB=40.5827N and tension in string BC=36.3855N
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
Please, let me know if my answers are correct or not:)
Here's my input: |dw:1437221537394:dw|
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
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! :)