I hate to do this

I hate to do this

. . .but I need some help on physics. I have a quiz tomorrow and no one can figure out the last three example problems. It gets even better though, the teacher is gone(damn football) tomorrow so I can't even go in early to ask for help. I worked in a group today to figure out the first 15 problems then we all came home and worked some more, we are still stumped on these last few.


Anyway, if any of you are willing to do physics this late or even just point us in the right direction it would be great.

http://img144.imageshack.us/my.php?image=img2ne8.jpg

I need help with 21, 23, and 25. Any help is appreciated.
 
I forgot to post the answers from the back of the book.

21. Tleftwire = 1/3w, Trightwire = 2/3w

23. T1 =501N, T2 = 672 N, T3 = 384N

25. 6.15m
 

Rius666

Diabloii.Net Member
For number 21 you first have to find the centroid of the weight of the sign. Since the sign is a semicircle x-bar will be at the half-way point and y-bar will be 4*r/(3*pi) where r = radius. This is where the center of mass of the sign is. Once you have that you will then have a total of 3 forces: the weight (at its centroid) and the two tension forces.

At point A (the first rope on the left) there will be a reaction force T_1 going up. At point B there will also be a reaction force T_2 going up. The sum of the forces in the y-direction will equal to zero due to static equilibrium, thus 0 = T_1 + T_2 - W. Take the moment (right hand ccw direction) about point A to get Moment = T_2*(.75) - W(.5). Because of static equilibrium, the moment will equal to zero. Now you have two equations and 2 unknowns and simply just use algebra to solve.

Problem 23 is solved in a similar manner. First take the moment about point A (the point of T_2 and T_3). Since T_1 is angular you will have to break it down to x and y coordinates. Thus Moment = 0 = sin(40)*T_1*2.00m - 700N*(.5m). This will letyou solve for T_1. Now you can use the sum of forces to solve for the rest. Forces in the x direction must equal to zero, thus: -T_3 + cos(40)*T_1 = 0. Do the same for the y direction: T_2 +sin(40)*T_1 = 0.

Since it's late and I don't feel like actually solving for it, I'll just letyou know whaty ou need to do to solve for 25. First it helps if you draw a free body diagram and label all the forces. That means the reaction force (x direction only) of the ladder against the wall, the weight of the man, the reaction forces of the ladder touching the ground, and the weight of the ladder. Once you have these labeled you have to solve for the distance d up the ladder. Again this is a static equilibrium problem so set the forces in both the x and y directions equal to zero and take the moment about the point on the ground where the ladder touches the ground. Use the angle to calculate the distance of the moment arm since you know the ladder is 8m long and the angle is 50 degrees find the x and y distances and use those as the moment arm.

I'm actually a bit surprised you are doing these kind of problems in your physics class since these are more mechanical statics than kinematics. Hope this helped.
 

Dondrei

Diabloii.Net Member
Don't tell him, sell the answers to him.

*EDIT*

Too late. Damn forums are full of commies, while I'm here trying to make a profit...
 

Savage

Diabloii.Net Member
So easy >.<

No, I'm kidding. Well, no I'm not, but of course it's easy once you've (I've) been there. I always found I understood material the best after the final exam. Not after studying for the final. No, no. After the exam itself.

I'm aware this post helps not at all.
 

Gertlex

Banned
Rius666 said:
Problem 23 is solved in a similar manner. First take the moment about point A (the point of T_2 and T_3). Since T_1 is angular you will have to break it down to x and y coordinates. Thus Moment = 0 = sin(40)*T_1*2.00m - 700N*(.5m). This will letyou solve for T_1. Now you can use the sum of forces to solve for the rest. Forces in the x direction must equal to zero, thus: -T_3 + cos(40)*T_1 = 0. Do the same for the y direction: T_2 +sin(40)*T_1 = 0.
Wrong. :thumbsup:

You need to account for the adjusted center of mass which turns out to be about 0.65 from the left, 1.35 from the right. Multiply those distances by the vertical tensions that match each one, and then you have a set of equal value. (The torques are equal).

I should get ready for class now...
 

Rius666

Diabloii.Net Member
Gertlex said:
Wrong. :thumbsup:

You need to account for the adjusted center of mass which turns out to be about 0.65 from the left, 1.35 from the right. Multiply those distances by the vertical tensions that match each one, and then you have a set of equal value. (The torques are equal).

I should get ready for class now...
Oops, didn't see the part about the board actually having a mass of 30kg. That's what I get for doing 'em at 2 am. But yeah since you can't neglect the weight of the board you have to do the centroid thing again like Gertlex said.
 

MadMachine

Diabloii.Net Member
dondrei said:
Don't tell him, sell the answers to him.

*EDIT*

Too late. Damn forums are full of commies, while I'm here trying to make a profit...
Funny. I thought you were here just to spamzor the forums.

*jab*
 

Merick

Diabloii.Net Member
I hate to do this, but you're a pain in the neck.
I thought you knew this, you're handing me a canceled check.
You're so helpless. Your girlfriends think you're a saint.
I'll give you a quarter. I'll keep my judgments to myself.
And I get caught up in the moonlight.
Reaching out for a rotten egg, I don't want to beg.
It's crystal clear, your time is nearly gone.

Count your blessings and do the things that you should.
All the has-beens have never had it so good.
Stormy weather, the kids are making a racket.
In the wilderness, the wild lives are so mild.
And I get caught up in the moonlight.
Reaching out for a rotten egg, I don't want to beg.
It's crystal clear, your time is nearly gone.

And I get caught up in the moonlight.
Reaching out for a rotten egg, I don't want to beg.
It's crystal clear, your time is nearly gone.

That's what I think of when I see the subject.
 

Rius666

Diabloii.Net Member
Actually for that plank problem it would be easier to not combine the weights of the man and the plank and find the centroid of that, since that way you would have to set up a ratio to account for the different masses and positions. It would probably be easier to just leave them alone as is and take the moments.
 

Gertlex

Banned
Rius666 said:
Actually for that plank problem it would be easier to not combine the weights of the man and the plank and find the centroid of that, since that way you would have to set up a ratio to account for the different masses and positions. It would probably be easier to just leave them alone as is and take the moments.
Neither Kow or I have been taught that yet, however.
 

Silent Shaddow

Diabloii.Net Member
noobs, ha ragnar try this...

by differentiating and intergrating prove that:
the intergrul of XarccoshX dX = .25(2Xsquared -1)arccoshx-.25X(xsquared-1)to the power of a half

seriously if the physics question took u 5 min :rolleyes: ha,
-silent
 

ragnar_ii

Diabloii.Net Member
silent shaddow said:
noobs, ha ragnar try this...

by differentiating and intergrating prove that:
the intergrul of XarccoshX dX = .25(2Xsquared -1)arccoshx-.25X(xsquared-1)to the power of a half

seriously if the physics question took u 5 min :rolleyes: ha,
-silent

It did take me 5-10 mins. All you do is sum the forces and moments. Your problem is a calculus problem, and that is by far my weakest point in anything math/engineering/physics related. I can do calculus, but just enough for my engineering purposes. I am by no means a math guru.
 
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