Physics help

[Piccokuh]

Oy, I'm back again with need for Physics help. I have a test tomorrow and I was working through problems to get ready, but I can't figure out a few. (Why, oh why, did my Physics test have to come after my Advanced Calculus one?)

1. A solid disk of mass 10.0 kg and radius 40.0 cm is rotating about an axis through its center. If a force of 50.0 N is applied tangentially, calculate the angular acceleration. (I have no idea.)

2. Two painters stand on a uniform plank of weight 200 N and length 4m suspended by ropes at each end. Calculate the tensions T1 and T2 in the two ropes. (I know this will have to do with equilibrium of torques, but we haven't done one problem similar to this at all.)

3. A uniform meter stick pivoted at its center has a 100-g mass suspended at the 25.0 cm position. A) At what position should a 75.0 g mass be suspsended to put the system in equilibrium? B) What mass would have to be suspended at teh 80.0 cm position for the system to be in equilibrium? (I have a vague idea with it, but we haven't done anything with a centered gravity and opposing masses like this.)

Not that I know what ya'll could say, but any advice to help me with my test on torque, inertia, gravity (planetary and satellite related), or such (the chapter titles are Rotational Motion and the Law of Gravity & Rotational Equilibrium and Rotational Dynamics) would help.

Thanks to anyone who takes the time to read and help. :D

[theundeclinable]

hmm.. my maths days are well behind me but here is my best shot on questions 2&3. cant help q1. too long ago and quite complicated.

q2 we need to know the weight of the painters otherwise we are lacking some figures. then i can do that one.. its got nothing to do with equilibrium of torques, or i have never ever heard that before, its just about balancing forces...

q3.

you need to take moments about the centre point...

1) 100*0.25=0.75*n

n= 25/75

n=33 1/3 cm. i think.

2) 100*0.25= x*.80

25/80= x

x= 31 1/4 g

i think thats right, its been over a year since i did any mechanics but thats roughly how it works..

like i said before get me the weights of the painters and i will do the other one...

[Piccokuh]

Whoops, and I thought I doublechecked to make sure I had taht right.

Okay, #2 has a 600N painter .75 m from the left and a 800N painter 1m from the right end.

[theundeclinable]

hmm.. i am struggling with that one since i can take moments but the 200n weight of the board i dont get where the force is acting, cos surely it is evenly distributed... hmm.. sorry.

[Piccokuh]

Well, thanks anyway. You at least got me thinking in the right direction. For some reason, I had just hit a mental block. I suppose my Advanced Calculus test had drained me too much. But you at least clued me in to what I should have been thinking along.

[theundeclinable]

your welcome. best part of the syllabus :)

[Diamond187]

For the second question, the board exerts a force at it's centre of mass. And for the first question, you need to calculate the moment of inertia for the disk, I'm sure you have that as a formula, cuz otherwise you have to do the integral around the disk, which would be a waste, since you can just look the damn thing up.
And the biggest help I can offer anyone in physics: DRAW THE FORCE DIAGRAM. It sounds really simple, but it works like you wouldn't believe.

[cheesecake]

The thing that helps me with physics, apart from drawing the force diagram as tha Diamond said, is to apply it to a situation. This should only be done on really hard questions (when you actually have a reason to spend time checking it).

What I mean is to try to think of an every day situation where this question might be seen, and to think what the result would be. So, when you have come upon your answer, chack back to the question to make sure that you havn't done something silly like accidently written up the forces the wrong way around, as most of the time I stuff up in maths test, it is due to little mistakes like that.

[Fiz]

Sorry guys, I haven't been here for a while. :) Too much to do with finals coming up. :(

If you still need help understanding these things (I know its a little late now) then let me know.

I can tell you that any time you see a question about equilibrium, remember that when an object is in equilibrium, the sum of all the forces acting on the object equal zero. The hard part is recognizing what forces are at work and setting all the equations to equal zero. For example, if you have a system where a block is sitting on a table, you know that the only forces that are at work are the normal force and the gravitational force. So:

Normal Force + Gravitational Force = 0

You just have to then plug in the info you have and solve for the variable you're looking for by isolating it.

[The First]

Correct me if I'm wrong here... but isn't the normal force 0 on a horizontal surface? Wouldn't there have to be a force working into the opposite direction of the gravitational force (as in the table pushing upwards, or rather preventing being pushed downwards...)?

Or of course is could be confusing forces. I know the normal force as the force that an object exerts on a surface that's perpendicular to it. So the normal force would be the fraction of the gravitational force that a child sliding down a slide exerts on the slide...

The word force was used 9 times in this post.

May the force be with you.

10 times.

[Fiz]

You've got the right idea, however, the normal force is not zero on a horizontal surface. The normal force, as you said, is always perpendicular to the surface, but it is the force applied by the surface upon the object resting on it to keep it from accelerating. Remember though, that this is only for objects that are not moving. An object sitting on a flat table experiences the force of gravity pulling it down, and therefore the table must exert an equal and opposite force on the object to keep it on the table.

The normal force on a slide is a little different. Remember that the gravitational force is always down to the center of the Earth. In a convential diagram, its always an arrow pointing straight down. Now, the normal force on a slide (or inclined plane) forms an angle with the gravitational force, and we must use angles to determine the rest of the free body diagram. There are other forces at work as well, like static or kinetic frictional forces and the acceleration of the actual object itself down the slide.

Hope that clears it up a bit!