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A rope of negligible mass is stretched horizontally between two supports that are 3.44 m apart. When an object of weight 3160 N is hung at the center of the rope, the rope is observed to sag by 35.0 cm. What is the tension in the rope?

Short Answer

Expert verified
The tension in the rope is approximately 3850 N.

Step by step solution

01

Understand the Problem

The setup involves a rope hanging horizontally between two supports 3.44 m apart. When a weight of 3160 N is hung in the middle, the rope sags by 35 cm. The task is to find the tension in the rope.
02

Identify Geometric Relationships

The rope and the sag form a right triangle with the horizontal distance being half of 3.44 m (since the weight is at the center), which is 1.72 m, and the sag forming the vertical side, 0.35 m. Use these to form a right triangle.
03

Calculate Horizontal Tension Components

The horizontal component of tension, denoted as \( T_x \), must satisfy the equation \( 2T_x = T \), where \( T \) is the total tension.
04

Use Trigonometry

Recognize that \( T_x = T \cos(\theta) \), where \( \theta \) is the angle formed by the rope with the horizontal. The vertical component is \( T_y = T \sin(\theta) = 3160/2 \), since the tension is symmetric on both sides.
05

Find the Angle \( \theta \)

Use trigonometry to find \( \theta \). The tangent of the angle is \( \frac{0.35}{1.72} \). Calculate \( \tan^{-1}\left(\frac{0.35}{1.72}\right) \) to find \( \theta \).
06

Apply the Trigonometric Equation

Once \( \theta \) is determined, the formula for the tension is \( T = \frac{3160}{2 \sin(\theta)} \). Calculate this value.
07

Calculate Total Tension

Determine the tension value using the angle and the trigonometric function calculated. Substitute into the equation to find \( T \).
08

Calculation Summary

Using the angle \( \theta = \tan^{-1}\left(\frac{0.35}{1.72}\right) \), and the relationship between the vertical and horizontal components of tension, the overall tension can be calculated as approximately 3850 N.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Statics
Statics is a branch of physics that deals with analyzing forces on objects at rest. When objects are in static equilibrium, the sum of forces and moments acting on them is zero. This means that nothing is moving. In the exercise, we deal with a rope keeping an object of weight 3160 N steady, illustrating static equilibrium.

In statics, we often work with diagrams to help visualize these forces and their effects. Free body diagrams are particularly useful. These diagrams show all the forces acting on an object, breaking them into their components.

For an object to remain at rest, the forces need to be balanced. In our problem, the tension in the rope balances the weight of the object. By calculating the tension, we ensure the rope can hold the object without collapsing.
Tension in Cables
Tension is the force conducted along a rope, string, or cable when it is pulled tight by forces acting from opposite ends. In our exercise, the tension in the rope is the force that counteracts the sag caused by the 3160 N weight.

To determine this tension, we consider both horizontal and vertical components of the force. The horizontal component, denoted as \( T_x \), equals half of the total tension in this symmetric setup, while the vertical component \( T_y \) deals with the weight's effect downwards.

Calculating tension involves understanding these components and their relationship with the forces acting on the rope. Trigonometry is often used to resolve these forces, as the tension force forms a diagonal in our right triangular setup.
Trigonometry in Physics
Trigonometry helps us understand relationships within right-angled triangles, crucial in resolving forces in physics problems. In the exercise, trigonometry links the rope's tension with the geometry formed by the sag and horizontal distance.

We use trigonometric functions like sine, cosine, and tangent for this purpose. For instance, to find the angle \( \theta \) that the rope makes with the horizontal, the tangent function is used: \( \tan(\theta) = \frac{0.35}{1.72} \).\(\)
Once \( \theta \) is known, we can calculate the components of tension using \( T_x = T \cos(\theta) \) and \( T_y = T \sin(\theta) = 3160/2 \).

This understanding allows us to solve for the total tension considering both components, ensuring all forces are in equilibrium. Trigonometry not only bridges mathematical concepts to physics but also simplifies how these forces are analyzed.

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Most popular questions from this chapter

A solid copper cube has an edge length of \(85.5 \mathrm{~cm}\). How much stress must be applied to the cube to reduce the edge length to \(85.0 \mathrm{~cm}\) ? The bulk modulus of copper is \(1.4 \times 10^{11} \mathrm{~N} / \mathrm{m}^{2}\).

A particle is acted on by forces given, in newtons, by \(\vec{F}_{1}=\) \(8.40 \hat{\mathrm{i}}-5.70 \hat{\mathrm{j}}\) and \(\vec{F}_{2}=16.0 \hat{\mathrm{i}}+4.10 \hat{\mathrm{j}} .\) (a) What are the \(x\) component and (b) \(y\) component of the force \(\vec{F}_{3}\) that balances the sum of these forces? (c) What angle does \(\vec{F}_{3}\) have relative to the \(+x\) axis?

A \(73 \mathrm{~kg}\) man stands on a level bridge of length \(L .\) He is at distance \(L / 4\) from one end. The bridge is uniform and weighs \(2.7 \mathrm{kN}\). What are the magnitudes of the vertical forces on the bridge from its supports at (a) the end farther from him and (b) the nearer end?

A makeshift swing is constructed by making a loop in one end of a rope and tying the other end to a tree limb. A child is sitting in the loop with the rope hanging vertically when the child's father pulls on the child with a horizontal force and displaces the child to one side. Just before the child is released from rest, the rope makes an angle of \(15^{\circ}\) with the vertical and the tension in the rope is \(280 \mathrm{~N}\). (a) How much does the child weigh? (b) What is the magnitude of the (horizontal) force of the father on the child just before the child is released? (c) If the maximum horizontal force the father can exert on the child is \(93 \mathrm{~N}\), what is the maximum angle with the vertical the rope can make while the father is pulling horizontally?

A \(75 \mathrm{~kg}\) window cleaner uses a \(10 \mathrm{~kg}\) ladder that is \(5.0 \mathrm{~m}\) long. He places one end on the ground \(2.5 \mathrm{~m}\) from a wall, rests the upper end against a cracked window, and climbs the ladder. He is \(3.0 \mathrm{~m}\) up along the ladder when the window breaks. Neglect friction between the ladder and window and assume that the base of the ladder does not slip. When the window is on the verge of breaking, what are (a) the magnitude of the force on the window from the ladder, (b) the magnitude of the force on the ladder from the ground, and (c) the angle (relative to the horizontal) of that force on the ladder?

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