More Physics Help (I really should get it stickied :P)

[Deathslayer7]

Ok. So below is the picture with the following.

r1 = 1.30m
r2= 2.15 m
F1= 4.20 N
F2 = 4.90 N
theta1 = 75 degrees (on the left side of the pic)
theta2 = 58 degrees (on the right side)

So whats the torque?

I know T(net)= T1 + T2

T1 = F1*d1(perpendicular)
T2= f2*d2(perpendicular)

I'm not sure though what d1 and d2 are. Can someone explain how to find d, for any problem.

Thanks.

PIC:

http://i198.photobucket.com/albums/aa125/death_slayer7/Physics2.jpg

[averagejoe]

This confuses me a little; r1 and r2 are the lengths of the long lines? Also, they're rotating around zero, yes?

[Deathslayer7]

r1 ends at the point where F1 starts.
r2 ends at the point where F2 starts.

And the axis of revolution is through O and perpendicular to the plane.

and yes they rotate around O (the letter o) (which is at the origin)

[averagejoe]

r1 ends at the point where F1 starts.
r2 ends at the point where F2 starts.

And the axis of revolution is through O and perpendicular to the plane.

and yes they rotate around O (the letter o) (which is at the origin)

So then d1 and d2 are just r1 and r2.

HOWEVER, you need to be careful with your forces. This is like last time, when you need to draw triangles and make components. Only the force component in the theta (polar coordinates-wise) direction counts for any torque.

A more detailed explanation to follow.

[Deathslayer7]

while a detailed explnation is following here is a better pic.

http://i198.photobucket.com/albums/aa125/death_slayer7/Physics2-1.jpg

[averagejoe]

Okay. Draw a line straight out from r1. Draw another line perpendicular to that one, pointing in the clockwise direction. These are your two components for r and theta. Do the same thing with r2, but draw theta pointing the other way, because it should point with the force. Now it's just a matter of trigonometry to figure out how much force points along the "r" line, and how much points along the "theta" line. You should be somewhat familiar with this.

Edit: Actually reverse the direction of theta; I thought the forces were pointing the other way.

[Deathslayer7]

Yea. See. That's the problem. I'm not sure what to draw....

[Deathslayer7]

Something like this? :smallconfused:


http://i198.photobucket.com/albums/aa125/death_slayer7/Physics2a.jpg

[averagejoe]

Sort of. I find it easier to color code.

http://i171.photobucket.com/albums/u282/teddifunken/physics2.jpg

[Deathslayer7]

can you explain what you did there? :smallconfused:

[averagejoe]

I drew a close up of r1 as well, and that's what I want to focus on now. Note that the gray line should be in a straight line with the black one, and the orange line should be perpendicular to them both. The red and gray lines make an angle of 70 degrees.

Now, you should have done similar stuff doing momentum and forces, except now instead of splitting the force into the "x" and "y" direction, you want to split them into the "r" and "theta" direction, which are the gray and orange lines respectively. The only tricky thing about r and theta, as opposed to x and y, is they change direction when you shift your point, whereas x and y have a constant direction. I drew in r and theta for r1, but for r2 r and theta will be different.

[Deathslayer7]

ok. I understand that. Then what did you get for the torque?

[averagejoe]

For T1 I got 5.27.

[Deathslayer7]

Ok, I understand why you want to spilt it into two components, but what exactly does the theta component do for you and what does the r component do for you?

[averagejoe]

The r component does nothing; the theta component is the force that matters. Remember, you can always think of theta as the instantaneous slope of a circle centered at the origin (or a tangent line), in other words, the direction your object is going to rotate.

Edit: If, however, you were interested in pulling the object outward (away from the origin), for example, then the r component would matter. It just doesn't matter for rotations because the direction called "theta" perfectly describes something that is just rotating but has no changing r.

[Deathslayer7]

ok. so following what you said.

we get this:

Torque= F*d(perpendicular)

F= 4.2*cos(20)=3.947

so then your d is the perpendicular distance from the rotational axis (origin in this case)

so d = I have no idea. :smalltongue:

[averagejoe]

ok. so following what you said.

we get this:

Torque= F*d(perpendicular)

F= 4.2*cos(20)=3.947

so then your d is the perpendicular distance from the rotational axis (origin in this case)

so d = I have no idea. :smalltongue:

Well, for one, it would be cos(15), because the angle given is 75 and 75+15=90. For two, d=r1. It's just the distance away that the force is applied. If you think of it in terms of, say, pulling a lever, and d would be the distance from the base of the lever (aka where the lever is rotating) to your hand.

[Deathslayer7]

ok. that made sense. And now i got 5.27 :smallbiggrin:

now to try T2 on my own.

[Deathslayer7]

T=F*d

F=4.9*cos(32)= 4.16

d=2.15

so T=4.16*2.15= 8.93

Did i do that right? :smallconfused:

[averagejoe]

Yep. That's what I got. Now, what's the total torque?

[Deathslayer7]

Well, the torques would be acting in opposite directions, so the torque with the biggest amount wins.

So CCW = 5.27
CW = 8.93

So torque = 3.66 N*m in the CW direction.

[averagejoe]

Yep, exactly. *gives thumbs up*

[Deathslayer7]

Now to start problem 3. :smalltongue::smallsigh:

[averagejoe]

Now to start problem 3. :smalltongue::smallsigh:

Heh, I'm in the same situation, basically. Good luck.

[Deathslayer7]

thank you. :smallsmile:

[Deathslayer7]

i should clarify though that this is not assigned hw, but hw that I am doing on my own, so I understand the concepts.

The hw that the teacher assigned were only like 7 problems, out of about 40ish. But i need the extra practice. It helps with me forming ideas, as well as knowing what to do.

Like number 3 asks when F1= -F1 and so on. I know how to do it. Yay!. :smallbiggrin: