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Chapter 2: Problem 27

A lightweight rope of area \(A\) and modulus of elasticity \(E\) is hung over a stationary shaft. A weight \(W\) is attached to the longer end, and, at the same time, the rope is forced against the shaft with a horizontal force \(P\) just sufficient to prevent the weight from dropping. Find the value of \(P\) if the static coefficient of friction between the rope and the shaft is \(f\).

Short Answer

Expert verified
The horizontal force \(P\) required to prevent the weight from dropping is calculated by the equation \(P = fW\), where \(f\) is the static friction coefficient between the rope and the shaft, and \(W\) is the weight of the attached object.

Step by step solution

01

Calculate Normal Force

Firstly an understanding of what's happening must be achieved. A rope is hung over a shaft, with one end having a weight \(W\) hanging from it. The rope is being pushed against the shaft with a force \(P\), just enough to stop the weight from dropping. This force \(P\) is acting horizontally. Now, the weight \(W\) is acting vertically downwards, and this is countered by the normal force acting upwards from the shaft. The weight and the normal force balance each other out, hence, the normal force is equal to \(W\).
02

Apply Static Friction Equation

Now, it is known that the force \(P\) is just enough to ensure that the weight doesn't drop. This means that \(P\) is countering the force due to friction, which prevents motion. This static frictional force can be calculated by the formula \(fF_N\), where \(f\) is the coefficient of static friction, and \(F_N\) is the normal force. As the normal force \(F_N\) is equal to \(W\), the static frictional force is equal to \(fW\).
03

Evaluate the Result

Given that \(P\) is balancing the frictional force, the value of \(P\) is equal to the frictional force. As a result, \(P = fW\). Therefore, the value of \(P\) is directly dependent on the weight of the object and the static friction between the rope and the shaft.

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Key Concepts

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

Coefficient of Friction
The coefficient of friction (\(f\)) is a measure of how much frictional force exists between two surfaces that are in contact. Friction is what keeps things from sliding on surfaces.
  • Static Friction vs. Kinetic Friction: Static friction occurs when surfaces are not moving relative to each other, while kinetic friction happens when they are moving. In this case, we are dealing with static friction because the rope is neither sliding nor dropping.
  • Formula: The static frictional force is given by the product of the coefficient of static friction (\(f\)) and the normal force (\(F_N\)). It can be expressed as \( fF_N \).
Here, the coefficient of static friction is the key factor in determining how large force \( P \) needs to be to keep the weight from falling. By understanding this concept, you can predict and control how objects behave when they are on the verge of slipping.
Normal Force
Normal force (\(F_N\)) is the force exerted by a surface in a perpendicular direction to the object resting on it. In the context of statics problems, it is crucial to understand as it directly affects frictional interactions.
  • Direction of Normal Force: In our problem, the normal force acts upwards against the weight \(W\) hanging downwards. The shaft exerts this force to balance the rope's weight.
  • Equilibrium Condition: In static equilibrium, the normal force should be equal to the applied vertical force; hence \(F_N = W\).
For this exercise, normal force is pivotal in calculating the frictional force that resists slipping, as it is part of the friction formula itself.
Statics Problems
Statics problems involve analyzing forces on objects that are in equilibrium 鈥� meaning they are not moving. Understanding how different forces interact and balance one another is at the heart of these problems.
  • Equilibrium: When forces are balanced, the object does not move. In this exercise, the objective is to find the horizontal force \(P\) that counteracts weight \(W\), using friction.
  • The Role of Forces: Forces such as weight, tension, and friction must be carefully considered to ensure they balance properly. This involves using vector components and often setting up equations to solve for unknowns.
Successfully solving statics problems like this one requires breaking down the forces into understandable parts and using principles such as equilibrium to find the solutions. By focusing on how forces like \(P\), friction, and \(W\) relate, you'll get the full picture of what keeps the rope in place without moving.

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Most popular questions from this chapter

Illustrated is a schematic diagram of a cable-control system for the rudder of a subsonic jet trainer aircraft. The rudder lever arm is connected to the pilot's foot control by \(4.5\)-mm-diameter extra-flexible stainless-steel cable, a 25 -mm length of which has a spring constant of \(60 \mathrm{MN} / \mathrm{m}\) (including the effect of untwisting). The cables have an initial tension of \(1.4 \mathrm{kN}\). Cable length from rudder lever arm to the pilot's foot control is about \(6 \mathrm{~m}\). The pilot can push on his foot control with a force of about \(700 \mathrm{~N}\). In a static test of the rudder control system a force was exerted on the rudder and gradually increased until the pilot could no longer hold his foot control stationary. Through what angle had the rudder rotated when the force reached the level which just caused the pilot's foot control to move? Would this angle have been different if there had been no initial tension in the cables?

An operator of a punch press operates part of the press by pushing a foot lever. The lever has a spring to return it to position after each push. The operator has complained that he gets tired pushing the lever. Can you suggest a change in the spring which may make the operator's job easier?

A steel cable hangs under its own weight. The diameter of the cable is not constant but varies in a manner that makes the tensile stress at all points along the cable the same. Derive the differential equation that describes the variation of cable diameter with position along the cable. Solve the equation and find the expression for \(\mathrm{d}_{1}\) in terms of \(\mathrm{d}_{2}, \mathrm{~L}, \sigma_{0}\) and \(\gamma(\) the weight per unit volume of the cable).

A \(2 \mathrm{~cm}\)-outside-diameter brass tube is to be compressed \(0.1 \mathrm{~cm}\) by means of a steel screw clamp, each screw of which has 10 threads per \(\mathrm{cm}\) and an effective cross-sectional area of \(0.5 \mathrm{~cm}^{2}\). It is known that it will take \(5 \mathrm{kN}\) to compress the brass tube \(0.1 \mathrm{~cm}\). The tube is put into the clamp with the jaws parallel and just touching the tube. How many turns must be given the screw \(C\) to compress the tube \(0.1 \mathrm{~cm}\) ?

Some miners are trapped \(2000 \mathrm{~m}\) below the surface. They make their way to the bottom of an abandoned shaft. At the surface is a hoist with \(1990 \mathrm{~m}\) of \(2.5 \mathrm{~cm}\) diameter standard plow-steel hoisting rope. A 30\(\mathrm{cm}\) length of this rope weighs \(10 \mathrm{~N}\) and has a spring constant (including the effect of untwisting) of about \(2 \times 10^{6} \mathrm{~N} / \mathrm{cm}\). If you think the miners can be hoisted to the surface, explain quantitatively how this can be done.
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