/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 81 An engineer is designing a conve... [FREE SOLUTION] | 91Ó°ÊÓ

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An engineer is designing a conveyor system for loading hay bales into a wagon (Fig. P11.81). Each bale is 0.25 \(\mathrm{m}\) wide, 0.50 \(\mathrm{m}\) high, and 0.80 \(\mathrm{m}\) long (the dimension perpendicular to the plane of the figure), with mass 30.0 kg. The center of gravity of each bale is at its geometrical center. The coefficient of static friction between a bale and the conveyor belt is 0.60 , and the belt moves with constant speed. (a) The angle \(\beta\) of the conveyor is slowly increased. At some critical angle a bale will tip (if it doesn't slip first), and at some different critical angle it will slip (if it doesn't tip first). Find the two critical angles and determine which happens at the smaller angle. (b) Would the outcome of part (a) be different if the coefficient of friction were 0.40\(?\)

Short Answer

Expert verified
(a) The bale tips at a smaller angle, \(14.04^\circ\). (b) Tipping still occurs first with \(0.40\) friction.

Step by step solution

01

Understand the Problem

We have a hay bale resting on an inclined conveyor. We must find two critical angles: the angle at which the bale begins to tip, and the angle where it begins to slip down the conveyor belt.
02

Find Critical Angle for Tipping

For tipping, consider the point at which the gravitational torque causes the hay bale to tip about its bottom edge. The bale will tip when the line of action of gravitational force passes outside this edge. This occurs when \( \tan(\beta_{tip}) = \frac{d}{h} \), where \(d = 0.25/2 = 0.125\ \text{m}\) (half the width) and \(h = 0.50\ \text{m}\). Thus, \(\beta_{tip} = \tan^{-1}(0.125/0.50)\).
03

Calculate Critical Angle for Tipping

Calculate \(\beta_{tip}\) using the relation from Step 2: \[\beta_{tip} = \tan^{-1}(0.125/0.50) \approx \tan^{-1}(0.25) \approx 14.04^\circ\]
04

Find Critical Angle for Slipping

For the bale to start slipping, the component of gravitational force parallel to the inclined plane must equal the maximum static friction force: \(mg \sin(\beta_{slip}) = \mu_s mg \cos(\beta_{slip})\), where \(\mu_s = 0.60\). Thus, \[\tan(\beta_{slip}) = \mu_s = 0.60\] and \[\beta_{slip} = \tan^{-1}(0.60)\].
05

Calculate Critical Angle for Sliping

Calculate \(\beta_{slip}\) using the relation from Step 4: \[\beta_{slip} = \tan^{-1}(0.60) \approx 30.96^\circ\]
06

Determine Which Happens First

Compare the calculated critical angles: \(\beta_{tip} \approx 14.04^\circ\) and \(\beta_{slip} \approx 30.96^\circ\). The bale tips before it slips because \(14.04^\circ < 30.96^\circ\).
07

Consider Different Coefficient of Friction

If the coefficient was \(0.40\), recalculate just \(\beta_{slip}: \tan(\beta_{slip}') = 0.40\), thus \(\beta_{slip}' = \tan^{-1}(0.40) \approx 21.80^\circ\). Since \(\beta_{tip} = 14.04^\circ\) remains the same, the bale will still tip first as \(14.04^\circ < 21.80^\circ\).

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Key Concepts

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

Critical Angles
When dealing with an inclined plane and an object like a hay bale, critical angles are essential for understanding when the object will tip over or slip down. Two main forces act on the object: gravity and friction.

The critical angle for tipping, known as \( \beta_{tip} \), occurs when the line of gravitational force shifts beyond the bale's pivot edge. This hinge effect causes the bale to tip. To calculate this, we use:
  • \( \tan(\beta_{tip}) = \frac{d}{h} \)
  • \( d \) is half of the bale's width, and \( h \) is its height
  • For the given bale dimensions, \( \beta_{tip} \approx 14.04^{\circ} \)
The second critical angle, \( \beta_{slip} \), defines when the parallel component of gravitational force overcomes static friction, causing the object to slip. This angle is determined by:
  • \( \tan(\beta_{slip}) = \mu_s \)
  • \( \mu_s \) is the static friction coefficient between the surfaces
  • For \( \mu_s = 0.60 \), \( \beta_{slip} \approx 30.96^{\circ} \)
These critical values help predict object behaviour as the incline steepens.
Tipping vs Slipping
Understanding tipping versus slipping involves analyzing how an object behaves under the influence of gravity and friction on an inclined plane.

**Tipping** arises from a rotational effect. When the incline angle reaches the critical tipping angle, the force of gravity creates a torque that can rotate the object around its edge, causing it to tip over. This happens when:
  • The center of gravity moves beyond the base support.
  • The object’s shape and dimensions play crucial roles here. Height and base width determine how easily an object tips.
On the other hand, **slipping** occurs due to a translational motion when static friction is no longer sufficient to hold the object stationary:
  • Friction resists but when incline angle increases past \( \beta_{slip} \), movement begins.
  • The coefficient of static friction is a key factor; greater friction restricts slipping.
In scenarios like the conveyor system for hay bales, predicting whether the bale tips or slips first helps in designing safer and more efficient systems. For the hay bale, it tips at a lower angle \( (14.04^{\circ}) \) than it slips \( (30.96^{\circ}) \). Reducing the friction coefficient would make slipping angle lower, yet in this case, the bale still tips first.
Inclined Plane Physics
Inclined plane physics describes the mechanics of objects on tilted surfaces and is a classic topic in engineering and physics education.

When we analyze objects on an inclined plane, such as hay bales on a conveyor belt, several essential components come into play:
  • The component of gravitational force parallel to the plane: \( mg \sin \beta \)
  • The component of gravitational force perpendicular to the plane: \( mg \cos \beta \)
  • Static friction: opposes the parallel component to prevent slipping
As the angle of incline \( \beta \) increases, the parallel force component increases, while the frictional force remains constant:
  • When parallel force exceeds static friction, slipping occurs.
  • When tipping is considered, the bale rotates about its edge when the gravitational torque becomes sufficient.
These considerations help optimize conveyor systems and other inclined devices to ensure safety and efficiency by understanding the balance between friction, gravitational force, and torque. Efficient design anticipates the balance point where an object transits from stable to unstable equilibrium, thus preventing accidents.

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Most popular questions from this chapter

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