Cross Sectional Area to Newtons

In summary, the issue discussed is with a hot water pipe in an upstairs bathroom that makes 18 "ticks" as it heats up and expands. The distance the pipe moves with each tick is calculated to be 0.000146 meters. The second part of the problem involves finding the force required to unstick the pipe, given the distance it moves and the cross-sectional area of the copper in the pipe. To solve this, one needs to find the force required to compress the pipe by the length of one tick, which can be calculated by multiplying the elasticity of copper by the cross-sectional area and the distance the pipe moves, and then dividing by the length of the pipe. This information is sufficient to solve the problem, although it is
  • #1
PhysicsMiester
5
0
1. When the hot water in a certain upstairs bathroom is turned on, a series of 18 "ticks" is heard as the copper hot-water pipe slowly heats up and increases in length. The pipe runs vertically from the hot-water heater in the basement, through a hole in the floor 5 m above the water heater. The "ticks" are caused by the pipe sticking in the hole in the floor until the tension in the expanding pipe is great enough to unstick the pipe, enabling it to jump a short distance through the hole. If the hot-water temperature is 50°C and room temperature is 20°C

This is a two part problem:
1st is a) the distance the pipe moves with each "tick"
Coefficient of Expansion is 17 x 10 ^-6 [(°C)^-1]
So to solve this all I did is following:
17 x 10 ^-6 [(°C)^-1] by 5 by 50°C = a
17 x 10 ^-6 [(°C)^-1] by 5 by 20°C = b
and then (a - b)/18 = 0.000146 meters

So here comes my question:
b) the force required to unstick the pipe if the cross-sectional area of the copper in the pipe is 3 x 10-5 m^2

I don't have an idea how to even approach this problem. If anyone can give me any clue it would be great



Thank you
 
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  • #2
For part b you need to know something about the force required to compress the pipe by the length of one tick movement. Note that the area given is the cross-sectional area of the copper in the pipe, not the cross-sectional area of the pipe.
 
Last edited:
  • #3
Well I have found that the typical value for the elastic modulus of copper is 1.1 x 10^11 Pa. Now I have length of the pipe which is 5 meters, Cross-sectional area of the copper in the pipe in m^2, how much it moves with each tick (18 ticks as well) and typical value of elasticity of copper in N/m^2. I am still a little confused on how to solve this problem.
 
  • #4
PhysicsMiester said:
Well I have found that the typical value for the elastic modulus of copper is 1.1 x 10^11 Pa. Now I have length of the pipe which is 5 meters, Cross-sectional area of the copper in the pipe in m^2, how much it moves with each tick (18 ticks as well) and typical value of elasticity of copper in N/m^2. I am still a little confused on how to solve this problem.

Assume that after each tick, the pipe achieves a relaxed state (no compression) As it heats up, its natual length increases by thermal expansion, but it is physically constrained to stay at one length. This requires a force, and the amount of force increases with temperature because the amount of compression from natural length is steadily increasing. When the force is no longer strong enough to keep the pipe compressed, it suddenly moves one tick to a new relaxed position.

The assumption of achieving a relaxed postion at each movement may not be completely valid, but you have not been given enough information to know exactly how much force is already acting after each move.
 
  • #5
what else do you think I would need to solve this?
 
  • #6
PhysicsMiester said:
what else do you think I would need to solve this?

Nothing. You have all that is needed. Find out how much force it would take to compress the pipe by the length of one tick.
 
  • #7
Should I multiply
elasticity of copper (N/m^2) by
cross-sectional area of the copper in the pipe (3 x 10-5 m^2)
multiple by the distance each tick the pipe moves
and divide by the distance of the pipe (5m)?
 
  • #8
I would like to thank you for your help.

Thank you
 

1. What is the relationship between cross sectional area and newtons?

The cross sectional area of an object refers to the area of a slice of that object taken perpendicular to its longest axis. The force of newtons is a unit of measurement for force. The two are related because the cross sectional area can affect how much force is needed to move or change the direction of the object.

2. How does changing the cross sectional area affect newtons?

If the cross sectional area of an object increases, the force of newtons required to move or change its direction will also increase. This is because a larger cross sectional area means there is more surface area for the force to act on, making it more difficult to move the object.

3. Can the cross sectional area to newtons ratio be used to calculate force?

Yes, the ratio of cross sectional area to newtons can be used to calculate force. This is because force is directly proportional to the cross sectional area, meaning that as one increases, the other also increases. This relationship can be expressed mathematically as F=ma, where F is force, m is mass, and a is acceleration.

4. How does the shape of an object affect the cross sectional area to newtons ratio?

The shape of an object can greatly affect the cross sectional area to newtons ratio. For example, a long, thin object will have a smaller cross sectional area compared to a shorter, wider object with the same mass. This means that more force will be required to move or change the direction of the long, thin object compared to the shorter, wider object.

5. Why is the cross sectional area to newtons ratio important in engineering?

The cross sectional area to newtons ratio is important in engineering because it helps determine the strength and stability of structures. Engineers must consider the cross sectional area of materials when designing structures to ensure they can withstand the forces acting upon them. For example, a bridge must have a large enough cross sectional area to support the weight of cars and other vehicles passing over it without collapsing.

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