# Mechanics question

1. Jan 23, 2006

### KoGs

I am very bad at physics. Anyways there is question 4 and question 6 I am having problems with. I kind of glanced over 6 and knew right away I couldn't do it. But I figured I might as well do my assignment in order, so I am looking at #4 first. Ok this is what I have done so far (which isn't much). I assigned teach beam a tension: t1, t2,......., t7. And I labelled every single vertex: A, B, ......, E. Now I know the sume of the forces on every vertex must add up to 0. And well the algebra is there as you can see. But that doesn't seem to work for me. Because that would make T1 = 0, which doesn't make a whole lot of sense. So I am thinking I am missing some force(s). Normal force perhaps?

Thanks for any help.

Here is the assignment question: http://members.shaw.ca/KingofGods/Phys24421.jpg

And here is my attempt: http://members.shaw.ca/KingofGods/Phys24422.jpg

2. Jan 23, 2006

### Pyrrhus

First off you need to find the normal forces exerted by the ground to the nodes in contact with it, then you start by solving throught each node or by sections.

3. Jan 23, 2006

### KoGs

But the beams are massless, so how would you calculate normal force? Ok let's start at the bottom, vertx C in my diagram.

So we have a horizontal tension T6. This must be balanced out by the horizontal tension of T2 since we have no other horizontal forces, is this correct?
So T6 = T2cos60.

Now for the vertical force.
T6 applies no vertical force to vertex C. So we must have the normal force balancing the verticle force of T2.
or N = T2sin60.
but N = mg
and m = 0 doesn't it?
That's gonna make T2 = 0, and everything else = 0.

4. Jan 23, 2006

### civil_dude

No, m = m period. Don't read too much into it. Only the beams are massless, not the weight hanging from the truss. Another thing, I would recalc your reaction at C. Why would it be mg? Aren't there two reactions? One at C and one at E and aren't they equidistant?

5. Jan 23, 2006

### KoGs

Well C is symetric with E. So calculating one automatically gives you the other by symetry.

I talked to my prof today. He said the part of my equation with T2 should be using the angle of 120. That is, the T2cos(60) part should be T2cos(120) and the T2sin(60) part should be T2sin(120). Even after explaining to me why, for the life of me I cannot figure out what he is saying. Someone else want to give it a shot and explain to me why? All I did was drew a perpendicular line straight down to intersect line segment CD. And then simply using cos(60) = T2_x / T2. So T2_X = T2Cos(60). Where 60 is the angle of C. I also tried making a right angle triangle the other side on T2 (ie. outside the ABC triangle), and again I got T2Cos(60).

6. Jan 23, 2006

### civil_dude

First find your reactions at C & E.

Next sum the forces in the y direction which lets the Vertical component of T2 = that reaction, e.g. T2sin60=mg/2, so T2 = mg/2sin60

then the horizontal component of T2 will equal T6, e.g. T6 = T2 cos60 and so on.