Classical mechanic problem/pulley-mass

arbolis · #1 08-17-2008, 12:57 PM

I'm really clueless about how to start to solve this problem.
Two cords A and B support a body of mass m=100 kg. The B cord pass by a pulley (we don't consider friction) and the 2 extremes of B are attached to the body. An extreme of A is attached to a wall and the other part is attached to the body. (well, take a look at the picture).
Determine the tension in the cords.

My attempt : I plant a system of reference in the center of the body. So its weight is [LaTeX Error: Can't write to directory] is worth [LaTeX Error: Can't write to directory] j. As it is at equilibrium, this means that the cords must suffer a force of [LaTeX Error: Can't write to directory] j.
After this, I'm confused about many things. First, the B cord really confuses me. Can I consider it as a cord forming an angle of 45° with the horizontal? If yes, how can I be rigorous to explain it. I cannot say "my intuition says so". I must be rigorous and I'm not able to say why I think the B cord could be simplified as one cord without a pulley and forming such an angle.
And then, even if I consider the B cord as such, I don't know how to do the exercise. I see like 2 right triangles in which the sum of the 2 sides on the y-axis must equal [LaTeX Error: Can't write to directory]. When it does, I just have to calculate the value of the hypotenuses and I'm done, but I don't know how to do this mathematically.

topsquark · #2 08-17-2008, 04:57 PM

Quote:

Originally Posted by arbolis

I'm really clueless about how to start to solve this problem.
Two cords A and B support a body of mass m=100 kg. The B cord pass by a pulley (we don't consider friction) and the 2 extremes of B are attached to the body. An extreme of A is attached to a wall and the other part is attached to the body. (well, take a look at the picture).
Determine the tension in the cords.

My attempt : I plant a system of reference in the center of the body. So its weight is [LaTeX Error: Can't write to directory] is worth [LaTeX Error: Can't write to directory] j. As it is at equilibrium, this means that the cords must suffer a force of [LaTeX Error: Can't write to directory] j.
After this, I'm confused about many things. First, the B cord really confuses me. Can I consider it as a cord forming an angle of 45° with the horizontal? If yes, how can I be rigorous to explain it. I cannot say "my intuition says so". I must be rigorous and I'm not able to say why I think the B cord could be simplified as one cord without a pulley and forming such an angle.
And then, even if I consider the B cord as such, I don't know how to do the exercise. I see like 2 right triangles in which the sum of the 2 sides on the y-axis must equal [LaTeX Error: Can't write to directory]. When it does, I just have to calculate the value of the hypotenuses and I'm done, but I don't know how to do this mathematically.

Okay, tensions in a Free-Body Diagram are always along the strings. So you have "three" tensions: one in the direction of A and two along the string directions on string B. However the neat point here is that the string around the pulley is an uncut ideal string (ideal unless otherwise stated) so the tension in the "two pieces" are the same.

So you've got a tension T1 at 30 degrees above the horizontal to the left, a tension T2 at 60 degrees above the horizontal to the right and a tension of equal magnitude (T2) above the horizontal to the right.

You are correct about the sum of the tensions in the vertical direction. Remember also that the sum of the tensions in the horizontal direction must be 0 N. That gives you enough information to finish the problem.

-Dan

arbolis · #3 08-17-2008, 05:45 PM

Glad to see you back!

Quote:

You are correct about the sum of the tensions in the vertical direction. Remember also that the sum of the tensions in the horizontal direction must be 0 N. That gives you enough information to finish the problem.

Thanks! I didn't think about that. If I understand well, I cannot simplify the B cord as a cord without a pulley, forming an angle of 45° with the horizontal.
Now I have 3 triangles but I'm still stuck... It's hard to say for me what I know and where exactly I'm stuck. I've done quite a lot of draws to find out the values of the sides of the triangles but I can't solve them.
I've reached something like [LaTeX Error: Can't write to directory]. Where a is the hypotenuse of the 2 right triangles formed by the B cord (since from what you've told me, the tension in the 2 parts of the cord is the same) and c is the side along the y-axis of the right triangle formed by the A cord. I used a draw of the forces, I mean the triangle I used are composed by the component of the forces. (which differs from the picture in magnitude but not in direction).
Ahm.............. I just could solve this, but I reach a non-sens answer... I got that [LaTeX Error: Can't write to directory] is approximately worth [LaTeX Error: Can't write to directory] and [LaTeX Error: Can't write to directory] approximately [LaTeX Error: Can't write to directory]. Having said that, it means that the tension [LaTeX Error: Can't write to directory] is worth [LaTeX Error: Can't write to directory] which is greater than the weight of the body. I'm totally lost.

topsquark · #4 08-17-2008, 06:03 PM

You've got the Free-Body Diagram on the hanging mass and you know that it is in equilibrium. Now set two positive directions, say, +x to the right and +y upward. Now use Newton's 2nd in each coordinate direction:
[LaTeX Error: Can't write to directory]

[LaTeX Error: Can't write to directory]

Two equations, two unknowns, no funky triangles.

-Dan

arbolis · #5 08-17-2008, 06:33 PM

I don't know how to thank you, that helps me a lot. I don't know why I had such difficulties to understand the problem. Next time I'll try to think harder and more accurately.

topsquark · #6 08-18-2008, 07:33 AM

Quote:

Originally Posted by arbolis

I don't know how to thank you, that helps me a lot. I don't know why I had such difficulties to understand the problem. Next time I'll try to think harder and more accurately.

It's merely a matter of practice. I've had some twenty years to perfect doing these so I make it look easy. Give yourself some time.

For the record I recommend always sketching a Free-Body Diagram, then setting two positive directions (or one if the problem is simple enough), then resolving all of the forces into components in those directions. Do that before you do anything else with the problem.

-Dan

arbolis · #7 08-18-2008, 08:23 AM

Quote:

It's merely a matter of practice. I've had some twenty years to perfect doing these so I make it look easy. Give yourself some time.

For the record I recommend always sketching a Free-Body Diagram, then setting two positive directions (or one if the problem is simple enough), then resolving all of the forces into components in those directions. Do that before you do anything else with the problem.

You're right. I need more training so I'll do many more exercises. Generally drawing a free-body diagram helps, but in my case my intuition told me to simplify the b-cord into one cord forming an angle of 45° upper the horizontal. I should have drawn a diagram even before to use my intuition because when it's not accurate then your diagram is false at the very beginning.
It's hard for me to "believe" or better say "understand" that in this problem the b-cord is equivalent to 2 separated cords but with the restriction that it must have the same tension. If I see a problem of the same type, I won't forget it.