Question: 2.148.

Resolve all three forces in to horizontal and vertical components:

Then as , we have a pair of simultaneous equations:

,

,

which can be solved fairly easily.

RonL

Results 1 to 9 of 9

- Feb 1st 2006, 05:49 AM #1

- Joined
- Jan 2006
- Posts
- 17

## I need some help

I am currently working on some of these review problems in the section I have been able to solve some. I am having a problem setting up these three problems. On the attachment below you will find problem 2.148, 2.150 & 2.154 these are the ones which I am kind of stuck on.

Answers are:

2.148= 1110 & 29.7 deg

2.150 = 313 & 140

2.154 = 8.78 -6.59j-8.78k

Can some help me set up the problems?

Thanks

Jay

Problem 2.148 and 2.150

Problem 2.154

- Feb 1st 2006, 10:07 AM #2

- Joined
- Nov 2005
- From
- someplace
- Posts
- 14,972
- Thanks
- 5

- Feb 1st 2006, 05:43 PM #3

- Joined
- Jan 2006
- Posts
- 17

- Feb 1st 2006, 09:34 PM #4

- Joined
- Nov 2005
- From
- someplace
- Posts
- 14,972
- Thanks
- 5

- Feb 2nd 2006, 09:46 AM #5

- Joined
- Jan 2006
- Posts
- 17

With question 2.150 would I basically set it up the same way.

I was going to place coordinates based on the horizontal and vertical axis's then figure out the components and find the magitude for one and basically subtract from D+ E+F=0

what do you think.

That a lot for the help its been great

- Feb 2nd 2006, 11:15 AM #6

- Joined
- Nov 2005
- From
- someplace
- Posts
- 14,972
- Thanks
- 5

Originally Posted by**jaycas21**

you can find it form the dimensions of the truss (in fact you don't need the

angle itself but the cos and sin of the angle which can be found directly

with the aide of Pythagoras's theorem).

So resolve the forces into horizontal and vertical components, then the

sum of the horizontal components is zero, as also is the sum of vertical

components.

This will give you two equations in the magnitudes of forces E and F.

These equations should be easier to solve than those we did before.

RonL

- Feb 2nd 2006, 03:27 PM #7

- Joined
- Jan 2006
- Posts
- 17

- Feb 2nd 2006, 09:30 PM #8

- Joined
- Nov 2005
- From
- someplace
- Posts
- 14,972
- Thanks
- 5

Originally Posted by**jaycas21**

and .

Then:

,

so gives us:

.

Rearranging:

Dividing the second by the first gives:

,

so pounds, and pounds

Opps, that last one should be pounds

RonL

- Feb 3rd 2006, 05:28 AM #9

- Joined
- Jan 2006
- Posts
- 17

Right! I was able to get X=140 as I was working on it last night, but it still doesn’t make sense to me because the if the D=280 and F= 140 then would E also equal 140 to make the equation zero. I was trying it out this morning and I was trying to figure why the give the answer of E as 313

Again thank so much for the help I am really trying hard on this class but it is kicking my butt. I have dynamics next.