# Thread: I need some help

1. ## 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.

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

2. Question: 2.148.

Resolve all three forces in to horizontal and vertical components:

$B=1500(\cos (50^{\circ}),\sin (50^{\circ}))$
$A=X(-\cos (\alpha ), -\sin (\alpha))$
$W=600(0,-1)$

Then as $A+B+W=\vec 0$, we have a pair of simultaneous equations:

$1500 \cos (50^{\circ}) - X\cos (\alpha )=0$,
$1500 \sin (50^{\circ}) - X\sin (\alpha )-600=0$,

which can be solved fairly easily.

RonL

3. I am not getting the answer of 1110

1500sin(50)-600= Xsin(a)
1149-600=549sin(a)

I can not wait to finish this class.

4. Originally Posted by jaycas21
I am not getting the answer of 1110

1500sin(50)-600= Xsin(a)
1149-600=549sin(a)

I can not wait to finish this class.
$X\cos(\alpha)=1500\cos(50^{\circ})\approx 964.18$
$X\sin(\alpha)=1500\sin(50^{\circ})-600\approx 549.07$

Dividing the second by the first gives:

$\tan (\alpha)=\frac{549.07}{964.18}$,

so:

$\alpha = 29.66^{\circ}$

Substituting $\alpha$ back into either of the equations above gives:

$X=1109.55$.

RonL

5. 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

6. Originally Posted by jaycas21
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
Its exactly the same method, except that you are not told the angle, but
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

7. Thus far I got

Sqrt of100^2+50^2= 112

Cosa=26.7
Sina= 26.5

280sin(26.5)=124.9
but the answer in the book is F=140

8. Originally Posted by jaycas21
Thus far I got

Sqrt of100^2+50^2= 112

Cosa=26.7
Sina= 26.5

280sin(26.5)=124.9
but the answer in the book is F=140
I have $\tan (\alpha)=50/100=0.5$, so $\alpha\approx26.57^{\circ}$, and so $\cos(\alpha ) \approx 0.8944$
and $\sin(\alpha ) \approx 0.4473$.

Then:

$D=280(-1,0)$
$F=X(0,1)$
$E=Y(\cos (\alpha ), -\sin (\alpha ))$,

so $D+E+F=0$ gives us:

$-280+\cos (\alpha )Y=0$
$X-\sin (\alpha)Y=0$.

Rearranging:

$\cos (\alpha )Y=280$
$\sin (\alpha ) Y =X$

Dividing the second by the first gives:

$\tan (\alpha ) =X/280$,

so $X=140$ pounds, and $Y=X/\sin (\alpha) \approx 140/0.8944=156.5$pounds

Opps, that last one should be $Y=X/\sin (\alpha) \approx 140/0.4473=313.0$pounds

RonL

9. 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.