Free Body Diagram and directions

[Bauxiet]

Hi everyone,
I am an electromechanical engineering student. The last few days I am trapping into some confusion about some things. I am just going to state my question underneath.

Question:

First question:

As they learned me at the university. When you draw a free body diagram it doesn't mind in which direction you draw a force. In the end, you get the same magnitude but with a different sign, what means that the force will be pointing in the opposite direction. This is correct?

For example the exercise below. What if a drew the N-force in the opposite direction? Would it give me the same numeric solution but with a different sign? For my experience, the angular acceleration that i had to calculate was a different numeric solution when I had drew the N-force in the opposite direction.

For example I had to calculate the angular acceleration. I drew the N-force counter the y-axis. So the equation would be: -N - 20*g = 0 <=> N = -20*g ! This means I have drawn my force in the wrong direction. Do I have to change the value to N= 200 N or do I have to calculate with the N = -200 N until the end when the exercise is done? (and use the - 200N in momentum equations for example).

My second question:

Must a (acceleration) and alfa (angular acceleration) always go in the same direction? For example: My a is pointing to the right (positive) direction. Can my alfa turn counter clockwise or must it turn clockwise (and does it care which direction positive is? or must i choose clockwise positive because the right direction for a is positive?)

I hope my questions are a little bit clear. I am experiencing some problems with this topic. I just have to know if the directions of forces/angular accelerations are important or it will all give the same result if you just follow the convention until the end of the exercise. Or do i have to change the value during the exercise. (for example (do i have to change N = -200 N to N = 200 N during the exercise and fill in the positive number in the other equations?)

Thank you very much in advance.
Greetings,
Bauxiet

When having a dynamic or static problem. Take for example a dynamic problem. Like underneath:

The exercise is written in dutch, sorry for this.

This is my solution:

 

[BvU]

Hello Bauxiet, (welkom dus)

Please do use the template, don't delete it. (see guidelines). It helps you formulate a clearer question and it makes answering a lot simpler.

1)
Forces have to be consistent: if you want to solve $\sum \vec F =0$ for equilibrium you can't have $m\vec g$ positive if pointing down and $\vec N$ positive if pointing up (*). Apart from that, the choice of a positive direction is free (but you can be asking for trouble, of course ).

2)
Check your definition of $\vec \alpha$ and the relationship with $\vec a$. A body can accelerate one way and have an angular acceleration in the other direction. Only when $\vec\alpha$ and $\vec a$ are connected (e.g. through a no-slip condition) things become different.

(*) And you deal with that in the right way by writing a minus sign in front of the 20 g ! Idem in 2T - 10g . I think that is a good way to do it.

 

[Bauxiet]

Hello Bauxiet, (welkom dus)

Please do use the template, don't delete it. (see guidelines). It helps you formulate a clearer question and it makes answering a lot simpler.

1)
Forces have to be consistent: if you want to solve $\sum \vec F =0$ for equilibrium you can't have $m\vec g$ positive if pointing down and $\vec N$ positive if pointing up. Apart from that, the choice of a positive direction is free (but you can be asking for trouble, of course ).

2)
Check your definition of $\vec \alpha$ and the relationship with $\vec a$. A body can accelerate one way and have an angular acceleration in the other direction. Only when $\vec\alpha$ and $\vec a$ are connected (e.g. through a no-slip condition) things become different.

Thanks for your reply. Sorry I didn't use the template, will do it in the future!

1) When: positive direction (up/y) and positive direction (right/x). Imagine I am searching the N-force. I don't know the direction (down or up).
The equation will be: -N - 25*g = 0 (imagine I drew the N force down in negative direction) solving will give: N = -25*g. So far, this means that the force will be pointing up and not down (because of the negative sign). Does it matter in which direction I draw an unknow force (in all cases)? And what if I find out in an exercise that my force is drawn in the wrong direction. Do I have to calculate with the N = -25*g till the end of the exercise or do I have to use N = 25*g in further calculations during the exercise?

2) The disk has a no-slip condition. The acceleration (a) is pointing to the right (positive x direction). My angular acceleration MUST be clockwise? (otherwise there would be slip...). Is this correct? Extra question: what if I don't know if there is slip/no-slip...
So far I the case of the no-slip condition. Imagine I draw the angular acceleration in counter clockwise direction and the acceleration (a) in the positive x direction.
Would my calculations point out I drew the angular acceleration in the wrong direction? (will my result be negative...?)

Thanks!

 

[BvU]

1)

The equation will be: -N - 25*g = 0

Yes, that is $\sum \vec F = 0$. You adopted $y^+$ is up, so $\vec g = (0, -9.81)$ m/s² . For the y components $\sum \vec F = 0 \Rightarrow \vec N_y - 9.81 m = 0$ ( I use $\vec N_y$ for the y component of $\vec N$). Outcome: $\vec N_y = 9.81 m$ Newton. A positive value, so pointing upwards.

You use the letter N for the vertical component in downward direction and compensate by writing a minus sign in front.
And you do the same with mg.
It's OK, but you risk confusion, of course. Either way you get $\vec N = -m\vec g$.

Most of the times the cause of confusion is in the symbol for g. Some people understand intuitively where to use a minus sign and where not. When you use $\vec g$ that is more unambiguous.

2)
The no-slip condition means that the point P on the rim that touches the table does not have a horizontal velocity component $\vec v_{P, x}$.
With $$0 = \vec v_{P, x} = \vec v_{{\rm c.o.m.}, x} + (\vec \omega \times \vec r_P)_x $$ you get $$
0 = \vec v_{{\rm c.o.m.}, x} - \vec \omega_z \;\vec r_{P,y} $$ Now, $\vec r_{P,y}$ is negative so that the z component of $\vec \omega$ comes out negative too: it points into the paper away from you. Corresponding to clockwise in the drawing. And likewise $\vec a$ and $\vec \alpha$.

 

[David Lewis]

The magnitude of a vector is always positive. When you prefix a vector with a plus/minus sign, it just connotes the sense of the vector (right/left, up/down, forward/backward). This converts parallel vectors to scalars so you can manipulate them with algebraic addition.

Alternatively, you can arbitrarily assign a positive sense to all vectors, and then you would follow the rules of vector addition.

With angular acceleration vectors, there is no obvious sense so you use the right-hand rule or some similar convention.