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Attach a mass to the ring
You tie a short cord to the bottom of the ring and tie a 1.02 kg mass to the cord. (A 1.02 kg mass has a weight of approximately 10 newtons.). Then you allowthe two cords that are tied to the pipe to support the mass, which pulls straight down with a force of 10 newtons.
The structure of the system
Now, let's think about the structure of this system. It consists of three cords emanating out from a small ring with each cord exerting a force on thering.
Compression or tension?
To begin with, a flexible cord cannot support a compressive force. If you push on one end of a cord in an attempt to push the object attached to the otherend, the cord will simply bend. Therefore, the cords in this scenario cannot possibly exert a pushing force on the ring.
However, a cord can support a tension or pulling force. In other words, any or all of the cords can support a pulling force on the ring.
Nothing else is touching the ring
The ring is not in contact with anything other than the three cords at this point in time, and the ring is in equilibrium. By that I mean that the ring haszero velocity and no acceleration.
(We are also assuming that the ring is, for all practical purposes, a point and it is not experiencing a moment or a torque. Once again, we'll get intomoments and torques in a future module.)
Thee pulling forces
So, we know that the ring is being subjected to three pulling forces. A pulling force of 10 newtons is pulling straight down toward the center of theearth. The forces associated with the two upper cords must be along the lengths of the cords. Otherwise, the cords and the ring would be moving.
Tactile graphics
The file named Phy1100c1.svg contains a force vector diagram for this scenario.
Figure 7 shows the mirror image that is contained in that file for the benefit of your assistant who will create the tactile graphicfor this exercise.
| Figure 7 . Mirror image from the file named Phy1100c1.svg. |
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Figure 8 shows a non-mirror-image version of the same image.
| Figure 8 . Non-mirror-image version of the image from the file named Phy1100c1.svg. |
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Figure 9 shows the key-value pairs that go with the image in the file namedPhy1100c1.svg.
| Figure 9 . Key-value pairs for the image in the file named Phy1100c1.svg. |
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m: Force vector diagram.
n: Parallelogram lineso: 7.3 newtons
p: Vector Cq: Resultant
r: 10 newtonss: Vector B
t: 5.3 newtonsu: 60 degrees
v: 45 degreesw: 10 newtons
x: Vector Ay: File: Phy1100c1.svg |
Directions of the forces
The fact that the cords can only support tension tells us that one of the two forces pulling in an upward direction is actually pulling at an angle of 45 degrees north of east(the direction of thecord). The other force is pulling at angle of 60 degrees north of west.
What is the tension in each cord?
So, three interesting questions are:
We can determine the answers to those questions using the parallelogram rule. At least we can find the tension in each of the cords and determine ifthose tensions are sufficient to carry the 10-newton load.
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