# Homework Help: Easy statics question

1. Jan 18, 2004

### formulajoe

There are three cables holding a beam in place. one cable, AC, is horizontal and coming off the left top of the beam. it has a value of 4KN. Another cable,AD, makes a 30 deg angle with AC and has a value of 5.2 KN. I need to find the magnitude of the third cable,AE, and the total force on the beam,AB. The beam makes a 65 deg angle, and AE makes a 50 deg angle. The beam and AE make a triangle, and AC and AD make a triangle.

I can find the total value of the addition of AC and AD, which is 8.9 KN. but i cant figure out any way to find the value of AE. If i can get AE, AB will be easy to find.

I made a rough pic with paint, but its too big to attach.
this is in the component force section of the book.

2. Jan 18, 2004

### jamesrc

The picture would help. I'm a little confused about how it looks. Is point A the left end of the beam? If so, it seems like all three cables are attached to that end (nothing is attached to B). When you give the angles, could you specify where they are measured from? Like when you say 65 degrees, is that 65 degrees above the horizontal? If you can elaborate a little I will try to help you out.

3. Jan 18, 2004

### formulajoe

point a is the top of the beam. point b is the bottom of the beam.
point c is on a horizontal line with point a. point d is at a 30 deg downward angle with the line ac. both of these are to the left. point e is to the right and down from point a. it makes a 50 deg angle with the ground. the angle is point up and to the left. AB is pointing up and to the right, and it makes a 65 deg angle with the ground.

4. Jan 18, 2004

### jamesrc

I've got to assume that the third cable is BE rather than AE, since I don't see how the beam can be held up if nothing is attached to point B. To find the tension in BE, you can just check the sum of the forces in the horizontal direction (the units of force in all cases beloe is kN):

$$\Sigma F_x = -4 - 5.2\cos(30^\circ) + T_{\rm BE}\cos(50^\circ) = 0$$

You can solve for TBE directly. You can also solve for the weight of the beam by checking the sum of forces in the y-direction.

The resulting tension in the beam can be found by resolving the force components along the beam. (TAB = 5.2cos(65-30) + 4cos(65) = TBEcos(65-50))

Last edited: Jan 18, 2004
5. Jan 18, 2004

### formulajoe

point B is where the beam rests on the ground. points C,D are attached to a wall that is perpendicular to the ground. point E is on the ground.

6. Jan 18, 2004

### jamesrc

Oh, let me look at it one more time and I'll post in a minute or three...

7. Jan 18, 2004

### jamesrc

So do you know what the beam weighs?

If you write out 2 force balance and 1 moment balance equation, you should get something like:

$$\Sigma F_x = 0 = -4 -5.2\cos(30^\circ) + T_{\rm AE}\cos(50^\circ) + R_{\rm Bx}$$
$$\Sigma F_y = 0 = -W - 5.2\sin(30^\circ) - T_{\rm AE}\sin(50^\circ) + R_{\rm By}$$
$$\Sigma M_A = 0 = -\frac{WL}{2}\cos(65^\circ) + R_{\rm By}L\cos(65^\circ) + R_{\rm Bx}L\sin(65^\circ)$$

(I hope the subscripts are clear here; the reaction forces at B are assumed to be positive. If the ground is frictionless and the beam is not attached to the ground (merely laying there), then you can set RBx = 0.)

Then resolve the components of force along AB to find the tension there:

$$T_{\rm AB} = 4\cos(65^\circ) - 5.2\cos(85^\circ) - T_{\rm AE}\cos(15^\circ) = R_{\rm Bx}\cos(65^\circ) + R_{\rm By}\cos(25^\circ)$$

(you can calculate the beam tension either way)

Last edited: Jan 18, 2004
8. Jan 18, 2004

### formulajoe

no idea what the beam weighs. just given the tension in AC and AD. theres two unknowns, so thats why i cant figure it out.

9. Jan 18, 2004

### formulajoe

those answers dont come out right. the tension in ae should be 7.29 and the tension in ab 9.03.

10. Jan 18, 2004

### himanshu121

Why don't u put it on one of the webpages, the problem is not clear to me

11. Jan 18, 2004

### jamesrc

Hmmm... Those are definitely the right answers then? Unless I've still got the picture wrong, the equations I gave should be true, though without the weight handy, you should probably find the sum of moments about the cg to solve the problem. If you assume the ground is frictionless so that RBx = 0, then I get the tension in AE as 13.2 kN. I'm sorry I couldn't be more helpful.

12. Jan 18, 2004

### himanshu121

For an Equilibrium to hold true whatever the fig is The following hold true at any point

$$\Sigma{\vec F}=\vec0$$ AND
$$\Sigma{\vec\tau}=\vec0$$

Last edited: Jan 18, 2004