Question

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.

Answer 1

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).

                                    $tan\alpha =\frac{30}{30}\\ \alpha =45^0$

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

                                 $y-direction\\T_2sin\alpha =mg\\\\T_2=\frac{mg}{sin\alpha } =\frac{17.1*10^{-3}*9.8}{sin45} \\\\T_2=0.236N$

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

                                 $x-direction\\T_1-T_2cos\alpha =0\\T_1=T_2cos\alpha \\T_1=0.236*cos45=0.167N$

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

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Answer 2

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.

                              $\alpha =tan^{-1}(\frac{30}{30})=45^0$

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

                                $T_2sin\alpha =mg\\T_2=\frac{mg}{sin\alpha } =0.236N$

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

                                     $T_1=T_2cos\alpha \\T_1=0.167N$

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

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

brainly.com/question/28106866

#SPJ1