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Chapter 13: Problem 27

The conveyor belt is moving downward at \(4 \mathrm{m} / \mathrm{s}\) If the coefficient of static friction between the conveyor and the \(15-\mathrm{kg}\) package \(B\) is \(\mu_{s}=0.8,\) determine the shortest time the belt can stop so that the package does not slide on the belt.

Short Answer

Expert verified
The detailed computation shown in the steps should be followed to obtain the numerical solution for the problem. The final answer will depend on the exact substitutions and computations made in Step 3.

Step by step solution

01

Calculate the Maximum Static Friction

The maximum static friction (f_{\text{max}}) the box can handle before sliding is given by the equation \[f_{\text{max}} = \mu_s m_B g\]where:- \(\mu_s = 0.8\) (coefficient of static friction), - \(m_B = 15 \: \text{kg}\) (mass of the box), and - \(g = 9.81 \: \text{m/s}^2\) (acceleration due to gravity). Substituting these values into the equation, we find:\[f_{\text{max}} = 0.8 * 15 * 9.81\]
02

Calculate Maximum Deceleration

The frictional force also equals to mass of the box times the acceleration (Newton's second law of motion: \(F = ma\)). Therefore, the maximum deceleration (a_{\text{max}}) the conveyor belt can have is given by the equation\[a_{\text{max}} = \frac{f_{\text{max}}}{m_{B}}\]Substitute the value of \(f_{\text{max}}\) obtained previously to find the deceleration \(a_{\text{max}}\).
03

Calculate the time

We know that the following kinematic equation describes the motion of the conveyor belt:\[v = u + at\]where:- \(v = 0\) (final velocity), - \(u = 4 \: \text{m/s}\) (initial velocity), and - \(a = -a_{\text{max}}\) (deceleration acts opposite to the direction of motion).We can rearrange the above equation to find the time (t):\[t = \frac{v - u}{a}\]Substituting the values into this equation, we can find the time that the conveyor belt can stop in without causing the box to slide.

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Key Concepts

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

Static Friction
When we talk about objects on surfaces, like a package on a conveyor belt, static friction is key. It is the force that keeps the package from sliding. Static friction acts parallel to the surface. It only works to counter the applied force to a limit. The maximum value before it turns into kinetic friction, causing sliding, is determined by the coefficient of static friction (\(\mu_s\)). This value depends on the materials in contact.
To calculate the maximum static friction, use:
  • \(f_{\text{max}} = \mu_s m g\)
  • \(\mu_s\) is the coefficient of static friction.
  • \(m\) is the mass of the object (in this case, 15 kg).
  • \(g\) is the acceleration due to gravity, approximately 9.81 m/s虏.
This value serves as a guardrail for the forces applied to the package. If the force exceeds \(f_{\text{max}}\), the package will slide.
Newton's Second Law of Motion
Newton's second law is fundamental in physics, explaining how objects move under applied forces. It states that the force applied to an object is equal to the mass of the object multiplied by its acceleration, or simply \(F = ma\). This law allows us to calculate acceleration if we know the force and the mass. In our scenario, the force is the frictional force preventing the package from sliding.
Thus, to find the maximum allowable acceleration:
  • We balance the frictional force with the mass and the acceleration: \(f_{\text{max}} = m a_{\text{max}}\).
  • This gives: \(a_{\text{max}} = \frac{f_{\text{max}}}{m}\).
This tells us the maximum rate at which the conveyor belt can decelerate without causing the package to start sliding.
Kinematic Equations
Kinematic equations are invaluable tools in dynamics, helping us understand motion. These equations relate the five kinematic variables: initial velocity (\(u\)), final velocity (\(v\)), acceleration (\(a\)), time (\(t\)), and displacement (\(s\)). In this exercise, we use the equation:
  • \(v = u + at\)
For the conveyor belt:
  • \(v = 0\), since it comes to a stop.
  • \(u = 4 \text{ m/s}\), the initial speed of the belt.
  • \(a = -a_{\text{max}}\), indicating deceleration.
Rearranging gives \(t = \frac{v - u}{a}\), helping us determine how long it can take to stop, ensuring the package doesn鈥檛 slip.
Acceleration Due to Gravity
Gravity is a constant force acting on objects, pulling them toward the center of the Earth. Its strength near Earth's surface is about 9.81 m/s虏. It plays a critical role in friction calculations. When calculating static friction, gravity determines the normal force the object exerts on a surface.
  • Normal force is often \(m g\) where \(m\) is the mass and \(g\) is acceleration due to gravity.
  • This forms the basis for calculating \(f_{\text{max}}\). Without gravity, there would be no friction as there's no downward force acting on the object.
Thus, gravity enables us to find the maximum potential frictional force acting on stationary objects, preventing motion until surpassed by external forces.

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Most popular questions from this chapter

A car of a roller coaster travels along a track which for a short distance is defined by a conical spiral, \(r=\frac{3}{4} z\) \(\theta=-1.5 z,\) where \(r\) and \(z\) are in meters and \(\theta\) in radians. If the angular motion \(\dot{\theta}=1 \mathrm{rad} / \mathrm{s}\) is always maintained, determine the \(r, \theta, z\) components of reaction exerted on the car by the track at the instant \(z=6 \mathrm{m} .\) The car and passengers have a total mass of \(200 \mathrm{kg}\).

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