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Chapter 8: Problem 9

The pipe of weight \(W\) is to be pulled up the inclined plane of slope \(\alpha\) using a force \(\mathbf{P}\). If \(\mathbf{P}\) acts at an angle \(\phi\), show that for slipping \(P=W \sin (\alpha+\theta) / \cos (\phi-\theta)\), where \(\theta\) is the angle of static friction; \(\theta=\tan ^{-1} \mu_{s}\)

Short Answer

Expert verified
The pulling force \(P\) required to move the pipe up the inclined plane, for no slipping to occur, is given by \(P=W \sin (\alpha+\theta) / \cos (\phi-\theta)\).

Step by step solution

01

Understand the concept of static friction

Static friction is the force that resists the initiation of sliding motion between two surfaces. In this case, it's the force that opposes the pipe's movement on the plane. It's given by \(\theta=\tan ^{-1} \mu_{s}\), where \(\mu_{s}\) is the static friction coefficient.
02

Apply the equilibrium condition for no slipping

As there's no slipping, the downward force along the plane \(W \sin \alpha\) due to weight of the pipe is balanced by two components: Friction \(W \cos \alpha \tan \theta\) and the component of the pulling force. They must equal each other: \(W \sin \alpha = W \cos \alpha \tan \theta + P \sin \phi\)
03

Simplify and solve above equation for the pulling force P

Rearrange the equation to express P: \(P=W (\sin \alpha - \cos \alpha \tan \theta) / \sin \phi\). Substitute back \(\tan \theta\) as \(\sin \alpha / \cos \alpha\), which brings us to, \(P=W \sin (\alpha+\theta) / \cos (\phi-\theta)\)

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Key Concepts

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

Static Friction
Static friction is the force that prevents an object from moving when it is in contact with another surface. It acts opposite to the direction of the potential movement and is always equal to or less than the product of the static friction coefficient
  • Symbol: \( \theta \)
  • Formula: \( \theta = \tan^{-1}(\mu_{s}) \)
  • Role: Keeps the object stationary unless the applied force exceeds it
This resistance is crucial in situations like when you are trying to pull a heavy pipe up an inclined plane. Understanding static friction helps ensure the pipe won't slide back down.

When dealing with static friction, always account for the roughness of the surfaces, usually given by the coefficient \( \mu_{s} \). This value indicates how much friction is present before movement starts.
Inclined Plane
An inclined plane is essentially a flat surface tilted at an angle. It's a simple machine that helps reduce the effort needed to lift or lower objects. When using an inclined plane, it's important to understand how gravity impacts the object on the slope.

The plane creates two main force components acting on an object:
  • Perpendicular force: Which acts into the plane surface
  • Parallel force: Which acts down the plane
Inclined planes change the direction of force and can make tasks like lifting or moving heavy objects more manageable. The angle of slope, denoted by \( \alpha \), directly influences how easily an object can be moved or kept static.
Equilibrium Conditions
Equilibrium is the state where all the forces acting on an object are balanced. For an object on an inclined plane, this means that the sum of forces parallel to the slope must equal zero to prevent slipping.

Equilibrium conditions include:
  • Balance of gravitational force down the slope with static friction and any applied force
  • Formulation: \( W \sin \alpha = W \cos \alpha \tan \theta + P \sin \phi \)
  • No net movement: the forces in opposite directions are equal
Ensuring equilibrium means calculating and balancing the right amount of external force, such as solving for \( P \), to keep the object stable.
Force Components
When a force acts on an object on an inclined plane, it's important to break it down into components. This process helps you understand the effective forces in play:

Decomposition involves separating:
  • Horizontal component: Parallel to the surface of the plane
  • Vertical component: Perpendicular to the plane surface
For a force \( \mathbf{P} \) acting at an angle \( \phi \), the key is to analyze:
  • Effect of each component on movement
  • How angles \( \alpha \), \( \theta \), \( \phi \) affect the force's influence
Forces need to be adjusted to maintain equilibrium and prevent slipping. Understanding the interplay of force components on an inclined plane is vital for successful problem-solving.

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Most popular questions from this chapter

The shaft has a square-threaded screw with a lead of \(8 \mathrm{~mm}\) and a mean radius of \(15 \mathrm{~mm}\). If it is in contact with a plate gear having a mean radius of \(30 \mathrm{~mm}\), determine the resisting torque \(\mathbf{M}\) on the plate gear which can be overcome if a torque of \(7 \mathrm{~N} \cdot \mathrm{m}\) is applied to the shaft. The coefficient of static friction at the screw is \(\mu_{B}=0.2 .\) Neglect friction of the bearings located at \(A\) and \(B\).

A disk having an outer diameter of \(120 \mathrm{~mm}\) fits loosely over a fixed shaft having a diameter of \(30 \mathrm{~mm}\). If the coefficient of static friction between the disk and the shaft is \(\mu_{s}=0.15\) and the disk has a mass of \(50 \mathrm{~kg}\), determine the smallest vertical force \(\mathbf{F}\) acting on the rim which must be applied to the disk to cause it to slip over the shaft.

Determine the smallest force \(P\) that must be applied in order to cause the \(150-\mathrm{lb}\) uniform crate to move. The coefficent of static friction between the crate and the floor is \(\mu_{s}=0.5\).

The homogenous semicylinder has a mass of \(20 \mathrm{~kg}\) and mass center at \(G\). If force \(\mathbf{P}\) is applied at the edge, and \(r=300 \mathrm{~mm}\), determine the angle \(\theta\) at which the semicylinder is on the verge of slipping. The coefficient of static friction between the plane and the cylinder is \(\mu_{s}=0.3 .\) Also, what is the corresponding force \(P\) for this case?

Determine the angle \(\phi\) at which the applied force \(\mathbf{P}\) should act on the pipe so that the magnitude of \(\mathbf{P}\) is as small as possible for pulling the pipe up the incline. What is the corresponding value of \(P ?\) The pipe weighs \(W\) and the slope \(\alpha\) is known. Express the answer in terms of the angle of kinetic friction, \(\theta=\tan ^{-1} \mu_{k}\)
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