2.3 Exercises from the Feyman Lectures on Physics

[new90]

Homework Statement

a body is acted upon by n forces and is in static equilibrium.Use the principle of virtual work t prove that
if n = 1 the magnitude of the force must be zero

Relevant Equations

no equation

i don't know what's the answer but I think that the force needs to be gravity because is going down to the floor

 

[DrClaude]

i don't know what's the answer but I think that the force needs to be gravity because is going down to the floor

What if the body is the middle of outer space?

 

[new90]

then there willl be work because there there will be a force pusing in some direction

 

[sysprog]

Please try to understand what Professor Feynman meant by the words 'static equilibrium'.

 

[new90]

i think it means that the body its not moving

 

[sysprog]

i think it means that the body its not moving

It means that any force acting on the body in any direction is counter-balanced by a force acting in an oppositional direction.

 

[kuruman]

Somewhere in your analysis you must use the principle of virtual work as the problem directs you to do. Write an expression in the case of a body at equilibrium when acted upon by $n$ forces and then see what happens when you set $n=1$.

 

[etotheipi]

Also remember that the principle of virtual work says the virtual work must be zero for any infinitesimal displacement. Because you could easily choose $d\vec{r}$ to be orthogonal to the resultant force!

 

[sysprog]

Also remember that the principle of virtual work says the virtual work must be zero for any infinitesimal displacement. Because you could easily choose $d\vec{r}$ to be orthogonal to the resultant force!

Well, equally true is that it must be non-zero for any non-zero force -- if it is infinitesimal it is still infinitesimally non-zero. Let's please not abuse zero while we are petting our pet inconsistencies . . .

 

[Adesh]

Are you able to do it just by condition implied by static equilibrium? I mean do you know that $\Sigma$ notation for statics?

 

[etotheipi]

Well, equally true is that it must be non-zero for any non-zero force -- if it is infinitesimal it is still infinitesimally non-zero. Let's please not abuse zero while we are petting our pet inconsistencies . . .

My apologies, I was being careless. I meant to say "for any virtual displacement".

 

[sysprog]

Are you able to do it just by condition implied by static equilibrium? I mean do you know that $\Sigma$ notation for statics?

In my view, the sigma notation, in this instance, refers to sum, I think that I don't quite fully understand your question . . .

 

[new90]

It means that any force acting on the body in any direction is counter-balanced by a force acting in an oppositional direction.
but there's its says only one force

 

[Adesh]

In my view, the sigma notation, in this instance, refers to sum, I think that I don't quite fully understand your question . . .

Does OP know this $\sum_{i=1}^{n} \vec{F_i} =0$ ?

 

[new90]

whos OP

 

[Adesh]

whos OP

OP : Original Poster.

 

[new90]

OK

 

[new90]

I DINT KNOW THANKS

 

[Adesh]

I DINT KNOW THANKS

Do you know for Static Equilibrium we should have $$ \sum_{i =1}^{n} \vec{F_i} =0 $$ ?

 

[new90]

i don't know what it means the n

 

[Adesh]

i don't know what it means the n

$$\sum_{i=1}^{n} \vec{F_i} = \vec{F_1}+ \vec{F_2} + \vec{F_3} +... + \vec{F_n} =0$$ Are things a little clearer now?

 

[new90]

yes thanks

 

[new90]

but in this case its just
Fi= 0

 

[Adesh]

but in this case its just
Fi= 0

If $n=1$ we have $$ \sum_{i=1}^{1} \vec{F_i} = F_1 = 0$$ How about now?

 

[new90]

yes its clear
thanks

 

[Adesh]

yes its clear
thanks

But you don’t have to do it like this
You have to do it by the Principle of virtual work.

 

[new90]

Principle of virtual work = the force x the virtual
displacement?

 

[new90]

thanks I answered with principle of virtual work

 

[sysprog]

$$\sum_{i=1}^{n} \vec{F_i} = \vec{F_1}+ \vec{F_2} + \vec{F_3} +\dots+ \vec{F_n} =0$$

I 'corrected' your $\LaTeX$ $-$ you had 4 dots instead of 3 $-$ your math was just fine $\dots$

 

[Adesh]

I 'corrected' your $\LaTeX$ $-$ you had 4 dots instead of 3 $-$ your math was just fine $\dots$

It seems that you have adjusted it’s height too!

 

[sysprog]

It seems that you have adjusted it’s height too!

I used \dots instead of typing period marks in -- oh, and, 'its', in the possessive sense, doesn't take the apostrophe -- that's for a contraction of 'it is'; not for a possession of something that belongs to something that we call 'it'.

 

[Adesh]

I used \dots instead of typing period marks in -- oh, and, 'its', in the possessive sense, doesn't take the apostrophe -- that's for a contraction of 'it is'; not for a possession of something that belongs to something that we call 'it'.

Thank you for teaching me that. Something like that occurs in Shakespearen Sonnets too, “ ‘Tis “ , Shakespeare did it for his iambic feet (In my Opinion).

 

[wrobel]

Assume that a rigid body consists of $N$ particles and i-th particle has a mss $m_i$ and a radius-vector $\boldsymbol r_i$.
Principle of virtual work for equilibrium is written as follows $$\sum_{i=1}^N(\boldsymbol F_i,\delta\boldsymbol r_i)=0,\qquad (1)$$ here $\delta \boldsymbol r_i$ is a virtual displacement of the i-th particle.

Fix any point $A$ in the rigid body. Then $$\delta\boldsymbol r_i=\delta\boldsymbol r_A+\delta\boldsymbol\psi\times\boldsymbol\rho_i,\quad \boldsymbol\rho_i=\boldsymbol r_i-\boldsymbol r_A.\qquad (2)$$
Here $\delta\boldsymbol\psi$ is a virtual rotation of the rigid body (perhaps someone prefers another term).

Now plug formula (2) in (1). After some transformations we get:
$$(\boldsymbol F,\delta \boldsymbol r_A)+(\boldsymbol\tau_A,\delta\boldsymbol\psi)=0,$$
here $\boldsymbol F=\sum_i\boldsymbol F_i$ is the net force and $\boldsymbol\tau_A=\sum_i\boldsymbol\rho_i\times \boldsymbol F_i$ is the total torque about the point $A$.

If $\delta \boldsymbol r_A$ and $\delta\boldsymbol\psi$
are independent then we obtain the standard condition of equilibrium: $\boldsymbol F=0,\quad \boldsymbol \tau_A=0$.

 

[sysprog]

Thank you for teaching me that. Something like that occurs in Shakespearen Sonnets too, “ ‘Tis “ , Shakespeare did it for his iambic feet (In my Opinion).

Using 'tis as an abbreviation for 'it is' is different from (incorrectly) using 'it's' for something possessed by an 'it'. Yes, a lot of of Shakespeare is written in iambic pentameter.

Friends, Romans, countrymen, lend me your ears.
I come to bury Caesar, not to praise him.
The evil that men do lives after them;
The good is oft interrèd with their bones.
So let it be with Caesar. The noble Brutus
Hath told you Caesar was ambitious.
If it were so, it was a grievous fault,
And grievously hath Caesar answered it.
Here, under leave of Brutus and the rest—
For Brutus is an honorable man;
So are they all, all honorable men—
Come I to speak in Caesar’s funeral.
He was my friend, faithful and just to me.