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.

The magnitude of the force on the left-hand pole of the thin flexible gold chain of uniform linear density with a mass of 17.1 and, hangs between two 30.0 cm long vertical sticks, which are a distance of 30.0 cm apart will be, 0.167N.

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 (in the free body diagram, it is given as θ).

[tex]tan\alpha =\frac{30}{30}\\ \alpha =45^0[/tex]

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}*9.8}{sin45} \\\\T_2=0.236N[/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=T_2cos\alpha \\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:

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aadildhillon023 aadildhillon023 · Answer 2 · 2022-07-30T12:51:18+02:00

The force acting on the left-hand pole of a thin, flexible gold chain of uniform linear density weighing 17.1 grams that is suspended between two vertical sticks that are 30.0 cm apart will be 0.167 N in magnitude.

We need to learn more about the fundamental forces that affect a body in order to arrive at the right answer.

What exactly is force, and what are the fundamental forces that affect a body?

Force is a push or a pull that modifies or tends to modify the condition, rest, or motion of a body.

Given that it has a point of application, force is a polar vector.

Repulsion is represented by positive force, and attraction by negative force.

A body is subject to three main forces: weight mg, normal response N, and tension or pulling force.

How can the issue be resolved?

We have given that, the gold chain hangs between the vertical sticks of 30cm and the horizontal space between then is 30cm.

We can determine the angle (shown as in the free body figure) using the information provided.

[tex]\alpha =tan^{-1}(\frac{30}{30})=45^0[/tex]

We may get the balanced equations for the total force in the y direction from the provided free body diagram as follows:

[tex]T_2sin\alpha =mg\\T_2=\frac{mg}{sin\alpha } =0.236N[/tex]

We can create the balanced equations for the total force in the x direction using the provided free body diagram:

[tex]T_1=T_2cos\alpha \\T_1=0.167N[/tex]

Thus, we may infer that the force acting on the left-hand pole will have a value of 0.167N.

Learn more about the tension that acts upon a body here:

https://brainly.com/question/13052160

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