# Calculating ##F_{DB}##: Equilibrium in a Machine

• Guillem_dlc
In summary, Guillem_dlc argues that the two force components should be applied at D, whereas the OP argues they should be applied at B. While the two force components may be equivalent algebraically, the OP's argument for applying them at B is more convincing.
Guillem_dlc
Homework Statement
The ABC telescopic extension arm is used to lift a worker to the height of overhead, telephone and electrical cables. For the indicated extension, the centre of gravity of the ##1400\, \textrm{lb}##-arm is at the ##G##-point. The worker, the basket and the fixed equipment in the basket weigh ##450\, \textrm{lb}## and their combined centre of gravity is at point ##C##. For the given position, determine the force exerted at ##B## by the single hydraulic jack using when ##\theta =35\, \textrm{º}##.

Solution: ##\vec{F}_{DB}=6490\, \textrm{lb}\,\,\, 62,1\, \textrm{º}##
Relevant Equations
##\sum M=0##
Figure:

My attempt at a solution:
For the calculation of ##F_{DB}## we consider the equilibrium in the whole machine.
$$\left. \begin{array}{r} h\rightarrow h=6\cdot \sin \theta +3=6,44\, \textrm{ft} \\ d\rightarrow d=6\cos \theta -1,5=3,41\, \textrm{ft} \end{array}\right\} \,\, \alpha =\arctan \left( \dfrac{h}{d}\right) =62,1\, \textrm{º}$$
$$\sum M_A=0\rightarrow -9\cos \theta G-20\cos \theta C+F_{DB}\cos \alpha \cdot 3+F_{DB}\sin \alpha \cdot 1,6$$
$$\rightarrow \boxed{F_{DB}=6482,5\, \textrm{lb}}$$
This one is well done, isn't it? It doesn't give me the exact toto but I guess it will be fine.

Guillem_dlc said:
Homework Statement:: The ABC telescopic extension arm is used to lift a worker to the height of overhead, telephone and electrical cables. For the indicated extension, the centre of gravity of the ##1400\, \textrm{lb}##-arm is at the ##G##-point. The worker, the basket and the fixed equipment in the basket weigh ##450\, \textrm{lb}## and their combined centre of gravity is at point ##C##. For the given position, determine the force exerted at ##B## by the single hydraulic jack using when ##\theta =35\, \textrm{º}##.

Solution: ##\vec{F}_{DB}=6490\, \textrm{lb}\,\,\, 62,1\, \textrm{º}##
Relevant Equations:: ##\sum M=0##

Figure:
View attachment 316600
My attempt at a solution:
For the calculation of ##F_{DB}## we consider the equilibrium in the whole machine.
$$\left. \begin{array}{r} h\rightarrow h=6\cdot \sin \theta +3=6,44\, \textrm{ft} \\ d\rightarrow d=6\cos \theta -1,5=3,41\, \textrm{ft} \end{array}\right\} \,\, \alpha =\arctan \left( \dfrac{h}{d}\right) =62,1\, \textrm{º}$$
$$\sum M_A=0\rightarrow -9\cos \theta G-20\cos \theta C+F_{DB}\cos \alpha \cdot 3+F_{DB}\sin \alpha \cdot 1,6$$
$$\rightarrow \boxed{F_{DB}=6482,5\, \textrm{lb}}$$
This one is well done, isn't it? It doesn't give me the exact toto but I guess it will be fine.
I hope I'm not stepping out of line, but I don't believe PF stives to be a homework checking service. Did you have a specific concern about the problem?

erobz said:
I hope I'm not stepping out of line, but I don't believe PF stives to be a homework checking service. Did you have a specific concern about the problem?
I just wanted to make sure I got it right, because I only have the solution. Sorry for the inconvenience.

We'll if you got that close to the official answer, it's probably either some rounding error carried through computation or sig fig consideration. You are within 0.1% of the book result.

Guillem_dlc
Guillem_dlc said:
$$\sum M_A=0\rightarrow -9\cos \theta G-20\cos \theta C+F_{DB}\cos \alpha \cdot 3+F_{DB}\sin \alpha \cdot 1,6$$
I don't understand the lever arm length 1,6 used on the right. Shouldn't it be 1,5? Maybe a typo when posting.

For the answer, I get 6480, almost exactly.

Last edited:
Well, go figure. I get something different from everyone (and the book)

$$\circlearrowleft^+ \sum M_A = 0 = -9 G \cos \theta -20 C \cos \theta + 6 F \sin ( \alpha - \theta)$$

##F \approx 6473 \rm{lb}##

Last edited:
erobz said:
Well, go figure. I get something different from everyone (and the book)

$$\circlearrowleft^+ \sum M_A = 0 = -9 G \cos \theta -20 C \cos \theta + 6 F \sin ( \alpha - \theta)$$

##F \approx 6473 N##
Again, the discrepancy looks like rounding error. Yours is only 0.1% below mine (if you change N to lb).

haruspex said:
Again, the discrepancy looks like rounding error. Yours is only 0.1% below mine (if you change N to lb).
I don't know, something seems fishy.

Taking ##6473 \rm{lb}## as ##6490 \rm{lb}## seem like a bit of a stretch for rounding error. I used stored values for ##\alpha##, and rounded value for ##\alpha ## and I get ## \approx 6474 \rm{lb}##.

Everything else is a single computation with exact values.

Looking carefully at what the OP did, neither of the last two term makes sense to me.

This is the free body diagram of the arm.

Edit: I think I see the intention of the OP now, but if everything is equivalent why the noticeably different results?

Last edited:
erobz said:
Looking carefully at what the OP did, neither of the last two term makes sense to me.

This is the free body diagram of the arm.

View attachment 316628
@Guillem_dlc is taking the two force components as being applied at D, though there may be a small error in the arithmetic; see post #5. I took them as applied at B, which gives a different combination of lever arms but should be equivalent. My form is readily converted to yours, so I think all three algebras are fine.

erobz
haruspex said:
@Guillem_dlc is taking the two force components as being applied at D, though there may be a small error in the arithmetic; see post #5. I took them as applied at B, which gives a different combination of lever arms but should be equivalent. My form is readily converted to yours, so I think all three algebras are fine.
I’m still perplexed by the significance of computational error between ours if you say they are equivalent algebraically? Having a 0.1% error between it an the book is one thing. Having a 0.1% error between two equivalent methods is something else…It’s late, maybe I’m tired and goofing up somehow.

erobz said:
$$\circlearrowleft^+ \sum M_A = 0 = -9 G \cos \theta -20 C \cos \theta + 6 F \sin ( \alpha - \theta)$$

##F \approx 6473 \rm{lb}##
Guillem_dlc said:
$$\sum M_A=0\rightarrow -9\cos \theta G-20\cos \theta C+F_{DB}\cos \alpha \cdot 3+F_{DB}\sin \alpha \cdot 1,5$$
$$\rightarrow \boxed{F_{DB}=6482,5\, \textrm{lb}}$$

So as far as I can see, there are at least ( @haruspex has a third equivalent) three ways of approaching it that should be equivalent. All three should yield the same result. Yet all three give very noticeably different results from each other (none giving the book result)! There has to be something interesting here that explains this. If it was a difference in decimal places, I'd be un-surprised...but a difference of ##10 \, \rm{lb}## between them?

Lnewqban
erobz said:
So as far as I can see, there are at least ( @haruspex has a third equivalent) three ways of approaching it that should be equivalent. All three should yield the same result. Yet all three give very noticeably different results from each other (none giving the book result)! There has to be something interesting here that explains this. If it was a difference in decimal places, I'd be un-surprised...but a difference of ##10 \, \rm{lb}## between them?
The most likely source of rounding error would be in finding alpha then plugging that by hand into trig functions. If you manage to do everything in a single equation typed into a calculator it should keep plenty of digits.
Note that it is not necessary to find alpha as such since the sin and cos can be derived algebraically from tan.

I have now put all my working into a spreadsheet, which keeps 10 or so significant figs, it seems. I get 6480.106…
If we can agree on 6480 then my guess would be that the book's 6490 is a typo.

Are you using a spreadsheet or equivalent? If doing that you still get 6473 please post your values for alpha and its trig functions. I get 1.083323924 radians, cos: 0.4683942052, tan: 1.886273531.

erobz
haruspex said:
The most likely source of rounding error would be in finding alpha then plugging that by hand into trig functions. If you manage to do everything in a single equation typed into a calculator it should keep plenty of digits.
Note that it is not necessary to find alpha as such since the sin and cos can be derived algebraically from tan.

I have now put all my working into a spreadsheet, which keeps 10 or so significant figs, it seems. I get 6480.106…
If we can agree on 6480 then my guess would be that the book's 6490 is a typo.

Are you using a spreadsheet or equivalent? If doing that you still get 6473 please post your values for alpha and its trig functions. I get 1.083323924 radians, cos: 0.4683942052, tan: 1.886273531.
I get the same thing as you when I use Mathcad. I was using a TI-89, I think the rounding on the ##h## and ##d## that I just pulled from the OP was the culprit. I got 6480 on both now. Mamma mia!

haruspex

## 1. How do you calculate ##F_{DB}## in a machine?

To calculate ##F_{DB}## in a machine, you need to first determine the forces acting on the machine. This includes any external forces, such as the weight of the machine, as well as internal forces, such as tension in cables or springs. Once you have identified all the forces, you can use the equations of equilibrium to determine the value of ##F_{DB}##.

## 2. What is the significance of ##F_{DB}## in a machine?

##F_{DB}##, also known as the driving or balancing force, is an important factor in determining the stability and equilibrium of a machine. It represents the force needed to keep the machine in a state of static equilibrium, where all forces and torques are balanced and the machine is not moving or rotating.

## 3. How does the value of ##F_{DB}## affect the operation of a machine?

The value of ##F_{DB}## directly affects the stability and efficiency of a machine. If the force is too low, the machine may not be able to perform its intended function effectively. On the other hand, if the force is too high, it can put unnecessary strain on the machine and potentially lead to damage or malfunction.

## 4. Can the value of ##F_{DB}## change during the operation of a machine?

Yes, the value of ##F_{DB}## can change during the operation of a machine. This can occur if there are changes in the external or internal forces acting on the machine, such as a change in the weight of the load being lifted or a change in the tension of a cable. It is important to regularly reassess and adjust the value of ##F_{DB}## to ensure the machine remains in equilibrium.

## 5. Are there any limitations to using ##F_{DB}## to calculate equilibrium in a machine?

While ##F_{DB}## is a useful tool for calculating equilibrium in a machine, it does have some limitations. It assumes that all forces are acting in a single plane and that the machine is rigid and does not deform under the applied forces. In reality, machines may have more complex forces and may experience some degree of deformation, which can affect the accuracy of using ##F_{DB}## to calculate equilibrium.

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