How to solve this kind of problem? (Mechanics Theory)

[nightsailor833]

Homework Statement

I am currently learning Theory Mechanics, and I am always troubled on this kind of problem for don't know how to add 3D torques. Can anyone tell me how to solve this kind of problem / i.e. a general step-by-step solution to such problems?

Relevant Equations

M=r×F

 

[PhDeezNutz]

I believe you calculate 3 different torques (about x,y, and z axis) and arrange them in a vector.

You do this for each picture.

Compare and contrast.

 

[PhDeezNutz]

Actually it would be much easier to use the cross product and since the torque is about point $A$ the individual torques are

$\vec{\tau} = \left( \vec{r} - \vec{r}_A \right) \times \vec{F}$

I hope you don't mind if I call $\left( \vec{r} - \vec{r}_A \right)$ by a different letter to condense it. Let $\vec{R} = \left( \vec{r} - \vec{r}_A \right)$ so $\vec{\tau} = \left(\vec{R} \right) \times \vec{F}$.

I'll get you started on the first torque to calculate in diagram $\left. a \right)$. I'm going to calculate the torque due to the upward $3 \cdot 10^3 \text{ kN}$ force at the back top left corner of the rectangular prism.

$\vec{R} = \vec{r} - \vec{r}_A = \left(0,0,3 \right) - \left(4,5,3\right) = \left( -4,-5, 0\right)$

For that particular $\vec{F}$ we have $\vec{F} = \left(0,0,3 \cdot 10^3 \right)$

Find the cross product of the two above quantities to get your first torque about point $A$.

$\vec{\tau} = \vec{R} \times \vec{F} = \left( -4,-5, 0\right) \times \left(0,0,3 \cdot 10^3 \right) = \left(-1.5 \cdot 10^4 , 1.2 \cdot 10^4 , 0 \right)$ (hopefully I did that right)

Let me know if you need help remembering how to do cross products.

 

[PhDeezNutz]

@nightsailor833

I'm getting an answer that corroborates the answer shown as correct on your homework.


for situation $\left. a \right)$ I'm getting

$\sum \vec{F} = \left( 0,0,0 \right)$
$\sum \vec{\tau} = \left(-1.5 \cdot 10^4 , 8 \cdot 10^3 , 0 \right)$

for situation $\left. b \right)$ I'm getting the same as I got for situation $\left. a \right)$
$\sum \vec{F} = \left( 0,0,0 \right)$
$\sum \vec{\tau} = \left(-1.5 \cdot 10^4 , 8 \cdot 10^3 , 0 \right)$


for situation $\left. c \right)$ I'm getting

$\sum \vec{F} = \left( 0,0,0 \right)$
$\sum \vec{\tau} = \left(0, 0 , 0 \right)$

Thus corroborating the given correct answer that situation $\left. a \right)$ and $\left. b \right)$ are equivalent whereas $\left. c \right)$ is not.

Instead of doing each cross product out by hand I used this site https://www.wolframalpha.com/input/?i=cross+product+calculator

 

[Lnewqban]

Welcome, @nightsailor833 !

Did anybody told you the meaning of those consecutive two arrow heads that are shown in systems a and c?

 

[PhDeezNutz]

Welcome, @nightsailor833 !

Did anybody told you the meaning of those consecutive two arrow heads that are shown in systems a and c?

I believe they are torques as noted by their stated units of $\text{kN}\cdot \text{m}$.

I don't know if that question was for OP specifically or a general question.

 

[haruspex]

Homework Statement: I am currently learning Theory Mechanics, and I am always troubled on this kind of problem for don't know how to add 3D torques. Can anyone tell me how to solve this kind of problem / i.e. a general step-by-step solution to such problems?
Relevant Equations: M=r×F

View attachment 353085

Given a force $\vec F$, its torque $\vec \tau$ about a point P, and the displacement $\vec s$ from P to another point P' we can calculate the torque about P'. Consequently, if we pick the same origin O for each force system then knowing the net force and the net torque about O for each is enough to determine whether they are the same.
It may be convenient to choose O such that several forces pass through O. This makes their torques zero.
I assume you know how to write each torque, force and displacement as a vector and compute the vector products to find the torques of the forces.

 

[Lnewqban]

I believe they are torques as noted by their stated units of $\text{kN}\cdot \text{m}$.

I don't know if that question was for OP specifically or a general question.

Correct, but considering the direction of the represented torque is important.

That question was for OP specifically.

Please, see:
http://hyperphysics.phy-astr.gsu.edu/hbase/vvec.html#vvc1

 

[PhDeezNutz]

@Lnewqban I assume this is a statics/dynamics class question. I’ve never taken those classes myself. In the course of answering this question I watched some YouTube videos and pieced together what I thought was reasonable and it turned out to corroborate “situations a and b are equivalent whereas c is not” (maybe it’s a coincidence)

If you worked it out did you get the same answer as me in #4 ?

May I PM you my work to see if my approach is valid?

The long and short of it is that I assumed.

Forces produce torques that you must take into account in the summation of torques.

BUT

Although some torques are given you don’t have to figure out the force that produced them and account for them in the summation of forces.


This seems paradoxical but it gave a qualitative correct answer.

 

[haruspex]

Forces produce torques that you must take into account in the summation of torques.

BUT

Although some torques are given you don’t have to figure out the force that produced them and account for them in the summation of forces.

Right. You do not have to suppose that torques arise from forces. Mathematically, you can just take it that torques can exist by themselves. But if you are more comfortable with the "everything is forces" view, you can represent a pure torque by a pair of equal and opposite forces that are antiparallel.
Note that a pure torque exerts the same torque about every point in space. If $\vec F$ and $-\vec F$ are $\vec r$ apart then the torque about point $\vec s$ relative to the former is $\vec s\times\vec F+(\vec s-\vec r)\times(-\vec F)= \vec r\times\vec F$

 

[PhDeezNutz]

Mind blown. Should have seen that. Thanks for your response. It's like when a (physical) dipole in a uniform electric field experiences zero net force but a net torque.

$\vec{\tau}_{net} = \vec{r}_1 \times \vec{F}_1 + \vec{r}_2 \times \vec{F}_2 = \vec{d} \times q \vec{E} + \vec{-d} \times -q \vec{E} \neq 0$

but $\vec{F}_1 + \vec{F}_2 = q \vec{E} - q \vec{E} = \vec{0}$



I wish OP would show up.

 

[Lnewqban]

If you worked it out did you get the same answer as me in #4 ?

I did not attempt solving the problem myself, sorry.

May I PM you my work to see if my approach is valid?

Of course!
I am always willing to learn.

 

[PhDeezNutz]

I did not attempt solving the problem myself, sorry.

Of course!
I am always willing to learn.

Lol I think you'd be the one teaching me if anything. I'll PM you semi-shortly.

@haruspex I hope you don't mind if I include you in the discussion per your comments in #10.

namely.....you don't have to ascertain where the torques came from (force wise) to include it in the summation of forces.

That said all forces exert a torque (whether zero or nonzero) and they have to be included in the summation of torques.

 

[haruspex]

you don't have to ascertain where the torques came from (force wise) to include it in the summation of forces.

You mean, in the summation of torques, right?

 

[PhDeezNutz]

You mean, in the summation of torques, right?

Maybe I should have been more clear

"because each torque can be represented as through 2 equal and opposite forces at the same distance in different directions the forces cancel out and don't have to be included in the summation of forces"

torques still have to be included in the summation of torques.

 

[PhDeezNutz]

Basically there are no extra forces to include from provided torques because they can be conveniently represented in a way such that they (the forces provided the torque) cancel out.

Also this might take awhile to LaTeX up but I will provide it to both of you by tomorrow.

 

[haruspex]

Maybe I should have been more clear

"because each torque can be represented as through 2 equal and opposite forces at the same distance in different directions the forces cancel out and don't have to be included in the summation of forces"

torques still have to be included in the summation of torques.

Right.