Solving a Moment Equation: Understanding the Relationship with AD

[goldfish9776]

Homework Statement

Inq this question, I am given that the u dot ( rb x tb + re x w) =0.
I knew that the rb x tb give a resultant moment pointed inside the book , which is parallel to AB... Then it cross with r AD, we will get the resultant moment.
But when re x w , I will gt the moment pointed in dx direction... How can it be related to AD?

Homework Equations

The Attempt at a Solution

 

[haruspex]

I don't understand your question.
$\vec{r_b}\times \vec{T_B}$ is the moment of the tension about D, while $\vec {r_E}\times\vec W$ is the moment of W about D.
To these we could add the moment of the forces at A about D and end up with 0, for equilibrium. However, that moment is orthogonal to $\vec u$, so if we take the dot product with $\vec u$ that moment disappears, leaving the given equation.

 

[goldfish9776]

I don't understand your question.
$\vec{r_b}\times \vec{T_B}$ is the moment of the tension about D, while $\vec {r_E}\times\vec W$ is the moment of W about D.
To these we could add the moment of the forces at A about D and end up with 0, for equilibrium. However, that moment is orthogonal to $\vec u$, so if we take the dot product with $\vec u$ that moment disappears, leaving the given equation.

what do u mean by we could add the moment of the forces at A about D here ? I'm confused...forces at A are not taken into calculation here ...

 

[goldfish9776]

what do u mean by we could add the moment of the forces at A about D here ? I'm confused...forces at A are not taken into calculation here ...

you mean by doing u dot ( rb x tb + re x w) , it means the moment of the forces at A about D

 

[haruspex]

what do u mean by we could add the moment of the forces at A about D here ? I'm confused...forces at A are not taken into calculation here ...

I mean that the full "sum of moments about D = 0" equation looks like :
moment of the tension about D + moment of W about D + moment of the forces at A about D = 0

If we then take the dot product of that with u then the last of those three terms vanishes, leaving:
u.(moment of the tension about D + moment of W about D) = 0

 

[goldfish9776]

I mean that the full "sum of moments about D = 0" equation looks like :
moment of the tension about D + moment of W about D + moment of the forces at A about D = 0

If we then take the dot product of that with u then the last of those three terms vanishes, leaving:
u.(moment of the tension about D + moment of W about D) = 0

why vector rDA is used here? why not rE ? this is weird becoz , moment of the tension about D + moment of W about D only pass thru rE ...

 

[haruspex]

why vector rDA is used here? why not rE ? this is weird becoz , moment of the tension about D + moment of W about D only pass thru rE ...

I don't understand either of those points.
Where is rDA being used that you think is wrong? It would be used for the forces at A, yes, but rDA is in the same direction as u, so when we take the dot product of u with (rDA x forces at A) we get 0.
What do you mean about moments "passing through" rE? Forces pass through points, moments turn about points. All the moments being calculated here are about D.

 

[goldfish9776]

I mean that the full "sum of moments about D = 0" equation looks like :
moment of the tension about D + moment of W about D + moment of the forces at A about D = 0

If we then take the dot product of that with u then the last of those three terms vanishes, leaving:
u.(moment of the tension about D + moment of W about D) = 0

for them , why u is necessary ? aren't them have the vector rB and rE already ?

 

[haruspex]

for them , why u is necessary ? aren't them have the vector rB and rE already ?

Taking the dot product with u is done to eliminate the forces at A from the equation.

 

[goldfish9776]

Taking the dot product with u is done to eliminate the forces at A from the equation.

That's only for force a right? Why tb and w also need to multiply u ?

 

[haruspex]

That's only for force a right? Why tb and w also need to multiply u ?

We have an equation. We cannot multiply some terms of the equation by something and not other terms.

 

[goldfish9776]

We have an equation. We cannot multiply some terms of the equation by something and not other terms.

i still don't understand . why shouldn't thtotal moment about D = rB x tB + rE x W + (resultant force at A) x rDA ?

 

[haruspex]

i still don't understand . why shouldn't thtotal moment about D = rB x tB + rE x W + (resultant force at A) x rDA ?

It does (except that you have the arguments of the last cross product in the wrong order; the distance vector should be on the left). But since the system is in equilibrium that total moment is zero. Write that as an equation. Now take the dot product of each side of that equation with u. What do you get?

 

[goldfish9776]

It does (except that you have the arguments of the last cross product in the wrong order; the distance vector should be on the left). But since the system is in equilibrium that total moment is zero. Write that as an equation. Now take the dot product of each side of that equation with u. What do you get?

sorry , i mean D = rB x tB + rE x W + rDA x (resultant force at A) ... why shouldn't the equation look like this ?

 

[haruspex]

sorry , i mean D = rB x tB + rE x W + rDA x (resultant force at A) ... why shouldn't the equation look like this ?

I just answered that question. That equation is quite correct, but we need to move beyond it.
Step 1: since the system is in equilibrium, total moment about D is zero:
rB x tB + rE x W + rDA x (resultant force at A) = 0.
Step 2:
We want to eliminate the unknown (resultant force at A). We can do that by taking the dot product of both sides of the equation with u:
u.(rB x tB + rE x W + rDA x (resultant force at A)) = 0
u.rB x tB + u.rE x W + u.rDA x (resultant force at A) = u.0 = 0
Now, u.rDA x (resultant force at A) = 0
(can you see why?)

 

[goldfish9776]

I just answered that question. That equation is quite correct, but we need to move beyond it.
Step 1: since the system is in equilibrium, total moment about D is zero:
rB x tB + rE x W + rDA x (resultant force at A) = 0.
Step 2:
We want to eliminate the unknown (resultant force at A). We can do that by taking the dot product of both sides of the equation with u:
u.(rB x tB + rE x W + rDA x (resultant force at A)) = 0
u.rB x tB + u.rE x W + u.rDA x (resultant force at A) = u.0 = 0
Now, u.rDA x (resultant force at A) = 0
(can you see why?)

we already know that the resultant force at A should be parallel to rDA , right , this will bring the rDA x (resultant force at A) = 0 ? So, why would we have to do one more step(step 2 ) ?

 

[haruspex]

we already know that the resultant force at A should be parallel to rDA , right

No. Why do you think that?

 

[goldfish9776]

No. Why do you think that?

the resultant force of Az and Ay should be in perpendicular to Az and Ay(so, it's Ax) , right ? so it's Az . Az x rDA = o , since they are parallel to each other .

 

[haruspex]

the resultant force of Az and Ay should be in perpendicular to Az and Ay(so, it's Ax) , right ? so it's Az . Az x rDA = o , since they are parallel to each other .

Eh?!
The resultant of adding a force in the z direction and a force in the y direction will be somewhere in the YZ plane. It will certainly not be perpendicular to either. Even if it were perpendicular to both, as you claim, that would put it in the X direction, not the Z direction. And even if it were in either the X or Z direction that would not make it parallel to DA.

 

[goldfish9776]

I just answered that question. That equation is quite correct, but we need to move beyond it.
Step 1: since the system is in equilibrium, total moment about D is zero:
rB x tB + rE x W + rDA x (resultant force at A) = 0.
Step 2:
We want to eliminate the unknown (resultant force at A). We can do that by taking the dot product of both sides of the equation with u:
u.(rB x tB + rE x W + rDA x (resultant force at A)) = 0
u.rB x tB + u.rE x W + u.rDA x (resultant force at A) = u.0 = 0
Now, u.rDA x (resultant force at A) = 0
(can you see why?)

yes , becoz u=rDA , they are parallel to each other

 

[goldfish9776]

Eh?!
The resultant of adding a force in the z direction and a force in the y direction will be somewhere in the YZ plane. It will certainly not be perpendicular to either. Even if it were perpendicular to both, as you claim, that would put it in the X direction, not the Z direction. And even if it were in either the X or Z direction that would not make it parallel to DA.

I have mixed up the direction of torque and resultant force.

 

[goldfish9776]

I mean that the full "sum of moments about D = 0" equation looks like :
moment of the tension about D + moment of W about D + moment of the forces at A about D = 0

If we then take the dot product of that with u then the last of those three terms vanishes, leaving:
u.(moment of the tension about D + moment of W about D) = 0

Can you explain why uDA. Moment of forces about D will =0 ?

 

[haruspex]

yes , becoz u=rDA , they are parallel to each other

Yes, because u and rDA are parallel.

 

[goldfish9776]

I just answered that question. That equation is quite correct, but we need to move beyond it.
Step 1: since the system is in equilibrium, total moment about D is zero:
rB x tB + rE x W + rDA x (resultant force at A) = 0.
Step 2:
We want to eliminate the unknown (resultant force at A). We can do that by taking the dot product of both sides of the equation with u:
u.(rB x tB + rE x W + rDA x (resultant force at A)) = 0
u.rB x tB + u.rE x W + u.rDA x (resultant force at A) = u.0 = 0
Now, u.rDA x (resultant force at A) = 0
(can you see why?)

do u mean rDA x (resultant force at A) = 0 ? why it will become 0 ? rDA and resultant force at A are not parallel to each other ?

 

[haruspex]

do u mean rDA x (resultant force at A) = 0 ? why it will become 0 ? rDA and resultant force at A are not parallel to each other ?

No, that cross product is not zero. The dot product of u with it is zero.
If you have a triple product a.(b x c) and a is parallel to either b or c then the triple product is zero. The reason is clear: b x c is perpendicular to b and c; if a is parallel to b then a is perpendicular to b x c.

 

[goldfish9776]

No, that cross product is not zero. The dot product of u with it is zero.
If you have a triple product a.(b x c) and a is parallel to either b or c then the triple product is zero. The reason is clear: b x c is perpendicular to b and c; if a is parallel to b then a is perpendicular to b x c.

since the rE x W and rb xTB are not perpendicular to u , so it,s not = 0 ?
how do u knw that a is parallel to b then a is perpendicular to b x c ??

 

[haruspex]

how do u knw that a is parallel to b then a is perpendicular to b x c ??

The cross product of two vectors (if nonzero) is always perpendicular to the two vectors.
If a is parallel to b then it must be perpendicular to b x c.

 

[goldfish9776]

The cross product of two vectors (if nonzero) is always perpendicular to the two vectors.
If a is parallel to b then it must be perpendicular to b x c.

how do u knw that the rE x W and rb xTB are not perpendicular to u ?

 

[haruspex]

how do u knw that the rE x W and rb xTB are not perpendicular to u ?

Why would that matter? Nothing in the algebra assumes those are not zero.

 

[goldfish9776]

Why would that matter? Nothing in the algebra assumes those are not zero.

because when rE x W and rb xTB are not perpendicular to u , then i can't say that u . (rE x W) and u .(rb xTB ) = 0

 

[haruspex]

because when rE x W and rb xTB are not perpendicular to u , then i can't say that u . (rE x W) and u .(rb xTB ) = 0

Nobody is saying those two dot products are individually zero.
The balance of torques gives us that rE x W + rb x TB + rDA x force_at_A = 0.
We take the dot product of that with u to obtain the equation u.(rE x W) + u.(rb x TB) + u.(rDA x force_at_A) = 0.
We then observe that since u is parallel to rDA the triple product u.(rDA x force_at_A) must be zero.
From that we conclude u.(rE x W) + u.(rb x TB) = 0.
This does not tell us anything about whether u is parallel to any of the four other vectors in that equation.