# Argh Statics is a pain.

1. Apr 10, 2004

### faust9

Argh!!! Statics is a pain.

Argh, this problem is killing me. I've tried a couple of approaches to this proble. First approach was to break it into 4 'Y' shaped vectors with with points A,B,C,D as the centers of the 'Y'. No avail. The next approach was to couple the forces to the origin but the forces cancel out when I do that. The third approach was to divide the quadralateral into 4 triangles with the y-axis equal to 1/2 y = 30lbf, and x-axis = (2-s)=1/2(s)(40). Again to no avail.

Any help would be appreciated. Thanks a lot.

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2. Apr 10, 2004

### arildno

The symmetry of the problem makes it sufficient to require balance of forces on the line segment AC.
Split the work in two steps:
1. Let x be the distance from A to the centre. Then, using Pythagoras, you have:
x^(2)+(y/2)^2=4.
By making use of this expression, you may find the the additional stetch of the spring as a function of y, and hence, the magnitude of the spring force.
2. Decompose the spring force and F along AC, and require equilibrium of forces.

3. Apr 10, 2004

### arildno

Just to rectify a flawed answer:
If a is the angle between the horizontal axis and AC, the component of F along AC is Fsin(a), and this entails a forcecomponent at A in the horizontal direction Fsin(a)cos(a).
Requiring balancing of forces at A in the horizontal direction yields then:
2Fsin(a)cos(a)-K=0, where K is the spring force.
Note that sin(a)=(y/2)/2.

4. Apr 10, 2004

### arildno

It's getting late; the rectified answer was also wrong:
The unit vector AC is cos(a)i+sin(a)j. The unit vector DC is -cos(a)i +sin(a)j.
Hence, Fj =1/(2sin(a))*(AC+DC), or the component of F along AC is F/(2sin(a))AC.
The horizontal comp of F is therefore Fcotan(a)/2.

5. Apr 10, 2004

### faust9

I had a similar problem. Ever time I thought I had the answer, I found a problem. Thanks for you insight and help.

6. Apr 10, 2004

### arildno

To make it perfectly clear:
Step 1.
You should find that the total stretch of the spring is 1.5+(2-x), where x is the length from A to the centre (expressed by Pyth. in y).

Step 2.
A careful balance of forces :
a)
Equilibrium of A requires the balance between the stresses T in rods AC and AB together with the spring force K.
The vertical force balance at A requires that the magnitudes of the stresses in AC and AB is equal (T).
The horizontal balance yields T in terms of the angle and K, T=K/(2cos(a)).

b)
By requiring equilibrium at C, you will find 2Tsin(a)=F, or finally

F=Ktan(a)

(This agrees with the last post, since the equilibrium at A required there must include the contribution of the force -F transmitted as stress from B through rod AB).

7. Apr 10, 2004

### HallsofIvy

If you think "statics" is a pain, wait until you get to "dynamics"!

At least in statics everything stands still!

8. Apr 11, 2004

### faust9

I guess I'm not the brightest bulb on the christmas tree, but I still don't get it. Arildno, ar you using the correct line segment notation (i.e. when you say AB do you mean AB or AD) because what your saying doesn't match up with the diagram, so I don't know what you are using as your point of reference. Also, when you say F=Ktan(a) where K is the spring force do you mean K is the spring constant or K is k(1.5+$\Delta$s) where $\Delta$s is the change in length along the x-axis? Also, you stated the spring force was F, so how are you accounting for the verticle forces F and -F from the diagram?

Thanks.

9. Apr 11, 2004

### arildno

Oh dear..
1. I have used "A" for "C" and vice versa (terribly sorry)
(Hence, my AC is really from C to A in the diagram (in ordinary terms, CA))
2. Spring force K is the force, not spring constant.
3. I cannot see I have termed the spring force K as F anywhere;
I'll look over it.

10. Apr 11, 2004

### arildno

I hope the following is clearer:
a)
Force equilibrium at C (on the diagram):
3 forces are involved:
Spring force Kvec, in vector form: -Ki, where K is magnitude of Kvec.
Forces due to stresses in CA and CB:
In the j'th direction, we find that the magnitudes of the stresses in CA and CB must be equal, T, so in the horizontal direction, we have:
2Tcos(a)-K=0 (a is the angle CA makes with the horizontal)
This entails that the stress in each rod is given by T=K/(2cos(a))

b) Force equilibrium at A (at the diagram)
3 forces are involved:
External force Fvec, in vector form: Fi, where F is magnitude of Fvec.
Forces due to stresses in CA and CD:
In the i'th direction, we find that the magnitudes of the stresses in CA and CD must be equal.
In addition the stress in CA is equal along the whole rod (T), so the stress in CD must also equal T.
In the vertical, j'th direction, we have:
-2Tsin(a)+F=0 (a is the angle CA makes with the horizontal)
This entails that the stress in each rod is given by T=F/(2sin(a)).
c)
However, in CA, we now have two expressions for T (from a) and b)).
These must be equal:
F/(2sin(a))=K/(2cos(a)), or F=Ktan(a)

11. Apr 11, 2004

### faust9

If k is the spring force i still have two unknowns (angle and delta s) and one equation.

12. Apr 11, 2004

### arildno

You are to determine the displacement y given F.
By Pythagoras, you have x^(2)+(y/2)^(2)=4. (Right?)
cos(a)=x/2, sin(a)=y/2, and delta s=2-x.
Hence, you really have only one unknown, (y), and one equation F=Ktan(a)
(K=k(1.5+(2-x))).

13. Apr 11, 2004

### arildno

Oops :sin(a)=(y/2)/2

14. Apr 11, 2004

### faust9

Clear as glass now thanks a lot.