# Static Equilibrium problems

1. Nov 19, 2005

### jrd007

Okay I cannot figure out the answers to any of these question.

1) A uniform steel beam has a mass of 940 kg. On it is resting half of an identical beam. What is the vertical support force at each end? Answers:5.8 x 10^3 N & 8.1 x 10^3N

------- (1/2)beam
--------------- 940 kg beam
|...................|
|...................|
|...................| <~~~ supports

This problem does not give me a length at all, so I have no idea about how to go about solving this. All I am given is that the mass of the steel beam is 940 kg and the 1/2identical one is 470 kg. I have no idea how I can use my T1 + T2 + T3 = 0 or my F1 + F2 +F3 = 0 Equations. Can some one please help.

~~~~~~~~~~~~~

2) A 75 kg aduly sits at one end of a 9.0 m long board. His 25 kg child sits on the other end. (a)Where should the pivot be placed so the board is balanced? (b) Find the pivot point if the board is uniform and has a mass of 15 kg. Answers: (a)2.3 m (b)2.5 m

2. Nov 19, 2005

### Staff: Mentor

The sum of the forces of the supports must oppose the weights of the beams, so FL + FR = (940 kg + 470 kg) 9.8 m/sec2.

Then pick either end and apply $\Sigma$M = 0.

Assume the moment arm of each beam is the distance between either support and the CM of the beam.

In problem 2, the same, the sum of the moments. If the board has length L, then one moment arm is x and the other is L-x.

3. Nov 19, 2005

### jrd007

FL + FR = (940 kg + 470 kg) 9.8 m/sec2 = 13818 N

I picked one of the ends, so don't I just divided by 9.8 m/s^2 since I am looking for the weight? = 1,410 which is not correct. How do I set up the $\Sigma$M = 0.

4. Nov 19, 2005

### Staff: Mentor

OK, the big beam B1 has length L, and B2 (smaller beam) has length, L/2.

Pick the left side as the pivot. If the moment arms are located at the center of masses of each beam, what are the moment arms, x1 and x2, of B1 and B2, respectively.

The Moments are just the products of the forces and moment arms.

So -W1*x1 - W2*x2 + ?? = 0, where W1 is weight of B1 and W2 is weight of B2.

I use the convention that a positive moment would cause a counterclockwise rotation.

5. Nov 19, 2005

### jrd007

W1*x1 - W2*x2 + ?? = 0 that gives us 4606...

6. Nov 19, 2005

### Staff: Mentor

What did you use for x1 and x2?

7. Nov 20, 2005

### jrd007

I do not know. What do I use for x1 and x2?

8. Nov 20, 2005

### jrd007

No length is given so I do not know how we are expected to get a position, x.

9. Nov 20, 2005

### Staff: Mentor

One does not need a numerical value for this problem.

Take the length of the long beam (940 kg) as L, and so the half beam has length L/2.

Now, taking the left side as 0, what the is the location of the CM's of the long beam and the half beam? This will give the moment arms about the left support for the long beam and half beam. The moment arm of the right support is L.

Last edited: Nov 20, 2005
10. Nov 20, 2005

### marlon

For the first question, keep it simple. Just assume that the length of the longest bar is l (you do not need to know this value).

You have 4 forces : F+F' - 4700 - 9400 = 0
I took 10 m/s² for g and F is the upward (along Y axis) support force on the left side, F' is the one on the right hand side.

Now, we calculate the moments (+ for counterclockwise rotation). The clue is to take the pivot point in the origin (there were the left hand side support force F originates). F originates in the origin, 4700 N at distance l/4, 9400N at l/2 and F' at l.

Now, what is the equation for the sum of the moments in this case ?

Once you have that, you will be able to get rid of the distance l and together with the equation for the forces, you have in total 2 equations with 2 unkowns F and F'

marlon

11. Nov 20, 2005

### jrd007

That is the problem I am having. Forming the sum of the moments equations..... would it be 4700 N + 9400 N + F = 0

12. Nov 20, 2005

### marlon

Not at all. Do you know what a moment is ?

It is defined as the vector product of vector r and force-vector F. Vector r denotes the position where the force originates. Now, in your case, you are lucky because the position vector r and the force F are perpendicular to each other. So, you can just multiply the two. beware of the signs, though, make sure you follow the convention (+ for counterclockwise rotation). For each force, i already gave you the r vector...

marlon

13. Nov 20, 2005

### jrd007

Okay, can you atleast set up the equation for me? Maybe that will help me better understand? I am lost.

14. Nov 20, 2005

### Staff: Mentor

To analyze this problem (as others have explained) start by identifying the forces acting on the beams. I see 4 forces:
(1) the weight of the 940 kg beam (where does it act?)
(2) the weight of the 470 kg beam (where does it act?)
(3) the upward force on the left support; call it F1
(4) the upward force on the right support; call it F2

Now apply the conditions for equilibrium. First the force equation: The net force must add to zero. Set up that equation using the four forces above.

Then the torque equation: The net torque about any point must add to zero. Torque is defined as Force X perpendicular distance to the axis. Choose the left support as your axis. What's the torque due to force (1) above about that axis? (Don't get hung up on the length of the beam: Call it L. It will cancel out.) Call torques that tend to rotate the beams in one direction to be positive and the other direction to be negative. (It doesn't matter which is which.) Find the torques due to all 4 forces (about the left axis) and set up the torque equation. Try it.

15. Nov 21, 2005

### marlon

i gave you nearly everything man. I gave you the formula to calculate each moment separately. Just apply this formula for each force and add up the 4 momenta you will get (one of them will be 0).

Besides, read Doc Al's last post in this thread and think things over. It is really very simple.

16. Nov 21, 2005

### jrd007

I get it now, Doc! Thanks.

It would be (1/2)mg(1/2)L + mg(1/2L) = F2(L)
L cancel out. So (5/8)mg = F2
F2 = 5757

Then you use F1 = F(m) + F(1/2m) - F2
And you get F1 = 8096! Thanks for all the help everyone.

Now solving for where to place my pivot point, would I do something like
F1(adult) + F2(child) = 0 (but where do I put my x = pivot point)?

17. Nov 21, 2005

### Staff: Mentor

You need to solve for the point where the torques (from the three forces acting) would balance. (Call the pivot point a distance X from one end and solve for it by writing the torque equation about that pivot point.)

18. Nov 21, 2005

### jrd007

T1 + T2 + T3?

but in the first one we do not know the weight of the board. So would it be

(75 kg x 9.8 x 1m) + (25 kg x 9.8 x 9m) + F(x) ?

19. Nov 21, 2005

### Staff: Mentor

Even though it is not clear from the way the problem is worded, I believe that for part (a) they want you to ignore the mass of the board. Thus for part (a) only two forces act: the weights of the adult and child. To find the balance point, set up the torque equation so that clockwise torques balance counterclockwise torques. Call the distance from the adult to the pivot point x. (What's the distance from the child to the pivot point?) Answer these questions:

(1) What's the torque due to the adult's weight?
(2) What's the torque due to the child's weight?

Set them equal and solve for x.

20. Nov 22, 2005

### jrd007

The adults T would be (75 kg x 9.8 m/s^2)(x) = (25 kg x 9.8 m/s^2)(childs distance)
Would the child be 9 m away or 4.5?