Rotational equilibrium equation - Moment of Forces (2D prob)

[masterchiefo]

Homework Statement

The landing gear of an aircraft is composed of a main leg OA (with a weight including the wheel is
1000N to G) and two secondary links BC and 50cm long each CD. The DC link forms an angle
θ of 30 degrees with the vertical.

a) Add the rotational equilibrium equation with respect to fulcrum A of
forces acting (only) on the leg OA and determine the force
exerted by the DC link D.

Pic in attachment.

Homework Equations

The Attempt at a Solution

sin(x)=40/80 = 30 degree

Total forces X : DC *cos(30) - w*sin(30) = 0
Total forces in Y: DA+DC*sin(30)-w*cos(30)=0
Moment of force(total) = 0.80m*DC*cos(30)+0-w*sin(30) = 0

I have a problem with my Moment of force equation, I am missing something or something is wrong because my calculator gives me an error "FALSE".

Took a good 2 hours to try to figure out by my own but I did not find my error.

Thank you for helping.

 

[haruspex]

DC *cos(30) - w*sin(30) = 0

You're defining the X direction as perpendicular to AD?
w is the 1000N, yes?
You're overlooking any X component of the reaction at joint A.

0.80m*DC*cos(30)+0-w*sin(30) = 0

Isn't this the only equation you need?
Please explain exactly what you are typing into your calculator.

 

[masterchiefo]

You're defining the X direction as perpendicular to AD?
w is the 1000N, yes?
You're overlooking any X component of the reaction at joint A.

Isn't this the only equation you need?
Please explain exactly what you are typing into your calculator.

Basically yes only this equation gives me the correct answer, but I used all the 3 equations to make sure that I got all the X and Y correctly and to fully answer the question as it ask for all the questions when its in equilibrium.

 

[masterchiefo]

Basically yes only this equation gives me the correct answer, but I used all the 3 equations to make sure that I got all the X and Y correctly and to fully answer the question as it ask for all the questions when its in equilibrium.

my Y axe is the same as my AD
yes w = 1000N

 

[masterchiefo]

You're defining the X direction as perpendicular to AD?
w is the 1000N, yes?
You're overlooking any X component of the reaction at joint A.

Isn't this the only equation you need?
Please explain exactly what you are typing into your calculator.

I am not sure which X or Y component I am overlooking I checked again and can't seem to know :(

 

[haruspex]

I am not sure which X or Y component I am overlooking I checked again and can't seem to know :(

Your first equation is for the forces on the arm+wheel perpendicular to the arm. That may include a force from the joint at A. But you don't care about that, because you don't need that equation, so just throw that equation away.

 

[BvU]

a) Add the rotational equilibrium equation with respect to fulcrum A of
forces acting (only) on the leg OA and determine the force
exerted by the DC link D.

I don't understand. The exercise wants a torque equilibrium wrt A. How come I don't see a $1000\ {\rm N} \times 100 \ {\rm cm} = 100 \sin({\pi\over 6})\,{\rm Nm}$ appearing that has to be compensated by the force DC exerts on D ?

[edit]
Ah, yes, I understand: you omit the 100 cm in the torque equation . But you can not assume the force DC (I do assume with DC you mean the force DC exerts) has a direction of $\pi/6$ wrt AD. Why should it ? Not $\pi /3$ either: it is not a cable but a rigid link.

And: apparently you look ahead at parts (b) and perhaps more that you don't show us. Concentrate on part (a) and realize it only yields one component of the force DC exerts.

 

[haruspex]

But you can not assume the force DC (I do assume with DC you mean the force DC exerts) has a direction of π/6\pi/6 wrt AD. Why should it ?

I'm not sure whether masterchiefo has thought through this or merely assumed it (given that it was wrongly assumed in respect of the forces at A), but it is not hard to show that for DC the force must be in the direction of DC. Think about torque on that member, and bear in mind it is considered massless.

 

[BvU]

Haru's right! Excuse me; dumb error. C is a hinge of course - the gear is retractable. I missed that in the problem statement, but it's probably obvious from the word 'link'

 

[masterchiefo]

I'm not sure whether masterchiefo has thought through this or merely assumed it (given that it was wrongly assumed in respect of the forces at A), but it is not hard to show that for DC the force must be in the direction of DC. Think about torque on that member, and bear in mind it is considered massless.

Here is my drawing of the forces (in attachment). Anything am I missing? I still don't understand what other forces there could be.

 

[masterchiefo]

Here is my drawing of the forces (in attachment). Anything am I missing? I still don't understand what other forces there could be. or why do I have to trow away that equation? In the question it ask me to write down the equations in equilibrium.

 

[haruspex]

Here is my drawing of the forces (in attachment). Anything am I missing? I still don't understand what other forces there could be, or why do I have to trow away that equation? In the question it ask me to write down the equations in equilibrium..

Clearly there is a force on DA from the joint A, and you have that, but you cannot assume its direction is along DA. Indeed, it cannot be.
Since G is to the right of D, the weight of the arm+wheel has a clockwise moment about D. This can only be balanced by an anticlockwise moment from the force at A.
You can keep the equation provided you put in an unknown for the X component of the force at A (or, equivalently, an unknown angle for the force at A).
Note that you made the same assumption about the forces on CD, but in that case it's ok because the link CD is consider massless. Any force component on CD at D, say, that's orthogonal to CD would have a moment about C with nothing to balance it.

 

[masterchiefo]

Clearly there is a force on DA from the joint A, and you have that, but you cannot assume its direction is along DA. Indeed, it cannot be.
Since G is to the right of D, the weight of the arm+wheel has a clockwise moment about D. This can only be balanced by an anticlockwise moment from the force at A.
You can keep the equation provided you put in an unknown for the X component of the force at A (or, equivalently, an unknown angle for the force at A).
Note that you made the same assumption about the forces on CD, but in that case it's ok because the link CD is consider massless. Any force component on CD at D, say, that's orthogonal to CD would have a moment about C with nothing to balance it.

You are right, the question that its asked is to write down the rotational equilibrium equation only, that doesn't include the totalFx = 0 and totalFy = 0. I would use those in translation equilibrium equation. Sorry yesterday was kinda very tired and I just now realized that I only need the last equation of Moment.

You are the best mate.
So the answer would be 722N

 

[masterchiefo]

Clearly there is a force on DA from the joint A, and you have that, but you cannot assume its direction is along DA. Indeed, it cannot be.
Since G is to the right of D, the weight of the arm+wheel has a clockwise moment about D. This can only be balanced by an anticlockwise moment from the force at A.
You can keep the equation provided you put in an unknown for the X component of the force at A (or, equivalently, an unknown angle for the force at A).
Note that you made the same assumption about the forces on CD, but in that case it's ok because the link CD is consider massless. Any force component on CD at D, say, that's orthogonal to CD would have a moment about C with nothing to balance it.

Question B)
Write the rotational equilibrium equation of the link relative to BC
point B and determine the magnitude of the moment of force M(B) required
to raise the landing gear when θ = 30 degree (note that the
tire does not touch the ground )

50cm*sin(30) = 25cm
25+50=75cm = 0.75m

0 = M(B) - 1000*sin(30)*1 + 722*cos(30)*0.75m
= 31.2N

is that right ?

 

[BvU]

Where and how is the 1000*sin(30)*1 torque acting on B ?
And I don't see a point at 75 cm from B where the 722 N is acting either ?

 

[masterchiefo]

Where and how is the 1000*sin(30)*1 torque acting on B ?
And I don't see a point at 75 cm from B where the 722 N is acting either ?

well the gravity act everywhere ? this is how I see it.

The 722 would be acting on 50cm?

 

[BvU]

well the gravity act everywhere ? this is how I see it

It does act everywhere. In part a you calculated the force of gravity is compensated for by the force in AD and in DC. So that's over and done with. DC translates the force to link C. You can't let the gravity force tag along with that: that's overcounting!

The 722 would be acting on 50cm ?

Doesn't sound very self-assured ! Where else can it act ? Any further and you're in the air !

Buy the way, are we convinced the various arms are massless as Haru proposes ? AO clearly isn't, or else G would be at O.
(I can live with BC and DC being massless or else you can't do a sensible thing in this exercise, but I do find it somewhat inconsistent! -- from the exercise composer, that is).
(And AO has to be a lot stronger than the other two, so perhaps I'm nitpicking).