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A thin flexible gold chain of uniform linear density has a mass of 17.1 g. It hangs between two 30.0 cm long vertical sticks (vertical axes) which are a distance of 30.0 cm apart horizontally (x-axis), as shown in the figure below which is drawn to scale.
Evaluate the magnitude of the force on the left hand pole.

A thin flexible gold chain of uniform linear density has a mass of 171 g It hangs between two 300 cm long vertical sticks vertical axes which are a distance of class=

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Nandhitha1098

Answer: A thin flexible gold chain of uniform linear density has a mass of 17.1 g. It hangs between two 30.0 cm long vertical sticks (vertical axes) which are a distance of 30.0 cm apart horizontally (x-axis), as shown in the figure below which is drawn to scale. Then, the magnitude of the force on the left-hand pole will be, 0.167N.

Explanation: To find the correct answer, we have to know more about the Basic forces that acts upon a body.

What is force and which are the basic forces that acts upon a body?

  • A push or a pull which changes or tends to change the state or rest, or motion of a body is called Force.
  • Force is a polar vector as it has a point of application.
  • Positive force represents repulsion and the negative force represented attraction.
  • There are 3 main forces acting on a body, such as, weight mg, normal reaction N, and the Tension or pulling force.

How to solve the problem?

  • We have given that, the gold chain hangs between the vertical sticks of 30cm and the horizontal distance between then is 30cm.
  • From the given data, we can find the angle [tex]\alpha[/tex] (in the free body diagram, it is given as θ).

                                   [tex]tan\alpha =\frac{30}{30}[/tex]

                                   [tex]\alpha =tan^{-1}(1)=45[/tex] degree.

  • From the free body diagram given, we can write the balanced equations of total force along y direction as,

                                  [tex]y- direction,\\T_2sin\alpha =mg\\T_2=\frac{mg}{sin \alpha } =\frac{17.1*10^{-3}kg*9.8m/s^2}{sin 45}=0.236 N[/tex]

  • From the free body diagram given, we can write the balanced equations of total force along x direction as,

                                  [tex]x- direction\\T_1-T_2cos\alpha =0\\T_1=0.236*cos45=0.167N[/tex]

Thus, we can conclude that, the magnitude of force on the left-hand pole will be 0.167N.

Learn more about the Basic forces that acts upon a body here:

https://brainly.com/question/28105677

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sqdancefan

Answer:

  0.1426 N

Explanation:

A chain of uniform mass density is suspended between two poles 30 cm apart. The geometry of the problem is such that the left support only supplies a horizontal force on the chain. The right support must both balance that horizontal force and supply a vertical force that balances the weight of the chain.

Magnitude of forces

For some tension T in the chain at the right support, the vertical force will be ...

  vertical force = T·sin(α) = W . . . . . matches the weight (W) of the chain

for some angle α between the horizontal and the chain at the right pole.

The corresponding horizontal force is ...

  horizontal force = T·cos(α)

This force balances the horizontal force at the left support pole. In terms of W, this force is ...

  horizontal force = W/sin(α)·cos(α) = W/tan(α)

Angle

The curve assumed by a chain of uniform mass density can be demonstrated to be a catenary. For supports 30 cm apart, its equation can be described by ...

  y = 30·cosh(x/30)

The diagram shows that y=4 for x=0, so we need to subtract 26 cm from this:

  y = 30·cosh(x/30) -26

The slope of the curve at any point is the derivative of this function:

  y' = 30(1/30)(sinh(x/30)) = sinh(x/30)

At the right support, the slope of the curve is ...

  y' = sinh(30/30) = sinh(1) ≈ 1.1752012

This is the tangent of the angle that the curve makes with the horizontal at the right support.

  tan(α) = 1.1752012

Note, you can see from the grid squares on the graph that the slope at the right support is slightly more than 1.

Weight

The weight of the chain is the product of its mass and the acceleration due to gravity:

  W = ma = (0.0171 kg)(9.8 m/s²) = 0.16758 N

Force on the Pole

Then the force on the left-side pole is ...

  horizontal force = W/tan(α) = (0.16758 N)/1.1752012

  horizontal force ≈ 0.1426 N

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Additional comment

The attached graph is a plot of the catenary curve we have assumed for the gold chain. We have attempted to match the vertical height on the left side, but we note that there seems to be a small discrepancy at the right side. The graph in the problem statement seems to show the right attach point at about y=21, not 20.3.

Ver imagen sqdancefan

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