Physics concept (please explain)





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Discus: SAT/ACT Tests and Test Preparation: July 2004 Archive: Physics concept (please explain)
By Best_Wr (Best_Wr) on Saturday, July 24, 2004 - 01:11 pm: Edit

For physics B:
i covered tension, and they keep apprearing over and over again. But I don't quite understand tension. How can I know which way the Ftension is exerted? It's not the math part that's hard-- I just can't draw free-body diagram, especially Ftension. For example, in giancoli's book, there's a problem with elevators, and tension on two sides (one side with elevator and the other side with some kinda engine) both point upward. How could that be? I hope you get what i'm saying. Sometimes force of tension points upward or downward (or left or right), and i just don't know why. Could you explain clearly? It matters where Ftension points, right?

By Joshjmgs (Joshjmgs) on Saturday, July 24, 2004 - 01:44 pm: Edit

I love the Giancoli book! Cherish it! It was one of my 5 books for Honors Physics, and it was by far the best one.

Ftension is the force that a rope or string that is applied to an object. The vector is in the same direction that string is connected to the object. A string connected to an object that is parallel to the y-axis of the Cartesian plane would be pointing straight up. This would be simple to solve for tension, because:
EFy = Ftension - Fg = ma = m(0) = 0
And in this case, Ftension = Fg.

An example of a more complex problem, you have two strings attached to your wind chime at two different angles. The Ftension of each string is the direction of the string. You can find both Ftensions by knowing the gravitational force that is being acted on the chime and the angles that the strings form. You would break up Ftension1 and Ftension2 into Ftenion1x, Ftension1y, Ftension2x, and Ftension2y. It is at equilibrium, so:
EFy = Ftension1y + Ftension2y - Fg = 0
EFx = Ftension2x - Ftension1x = 0
Solve.

By Davidn08 (Davidn08) on Saturday, July 24, 2004 - 02:36 pm: Edit

The tension points in the direction opposite to the pull/push (that's kinda implied by Newton's third law).

Think about it, if you're holding a rope with a weight hanging, it stays still, right? The system must be balanced. There are two forces applied to the system, 1. the weight of the weight (down), 2. the tension in the rope (up).

if that rope were to snap, the upper part of the rope would momentarily flop upwards (just after the snap)... because of the tension being in the upwards direction.

By Muawan (Muawan) on Saturday, July 24, 2004 - 03:57 pm: Edit

tension is actually nondirectional, and constant through the entire rope, it must simply balance all forces

By Best_Wr (Best_Wr) on Saturday, July 24, 2004 - 10:23 pm: Edit

actually, tension is directional b/c it's force, and force is a vector.

By Mathninja (Mathninja) on Saturday, July 24, 2004 - 11:04 pm: Edit

In addition, tension is not constant throughout the entire rope. If we hang a weight from a vertical rope hanging from a ceiling, tension will be greater at the top than of the rope the bottom of it, assuming the rope has a mass, which all ropes do last time I checked.

By Conker (Conker) on Sunday, July 25, 2004 - 04:10 am: Edit

If tension is not constant throughout a rope, then how do you do those pulley problems with blocks on either end of the rope? One of the first things you do in a pulley problem is set up a system of equations and cancel out F_t.

By Skapoor1 (Skapoor1) on Sunday, July 25, 2004 - 11:20 am: Edit

In most physics problems I remember we assumed that the rope was weightless. However, ropes do have mass, so as you go up the rope the mass being held becomes greater, therefore force of gravity becomes greater and tension must be greater.

By Joshjmgs (Joshjmgs) on Sunday, July 25, 2004 - 12:08 pm: Edit

"In most physics problems I remember we assumed that the rope was weightless. However, ropes do have mass, so as you go up the rope the mass being held becomes greater, therefore force of gravity becomes greater and tension must be greater."

Actually you take into account the gravitational force on the rope by using the rope's linear density and length to find the torque of the rope's mass.
So its *.5(u)(L)(g) = .5*mass*gravity = Fg. Thus, gravity is calculated, and tension is constant throughout the rope is constant, or is at least considered constant because the center of gravity is the average of all of the g-forces acting upon each segment of the rope.

*Note: Since the center of gravity is in the middle of the rope, half way down the rope, Fg only affects that point, so Fg = .5(u)(L)(g).

By Mathninja (Mathninja) on Friday, July 30, 2004 - 02:45 pm: Edit

i'm siding with skapoor1


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