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Chapter 9: Problem 9

Small parts each of mass \(m\) on a conveyer belt moving with constant velocity are allowed to drop into a bin. Radius of circular portion is \(2 \mathrm{~m}\). The static coefficient of friction between the parts and belt \(\mu_{s}\) is 1 . If the parts start sliding on the belt at the angle \(\alpha=37^{\circ} .\) Find the velocity (in \(\left.\mathrm{m} / \mathrm{s}\right)\) of conveyer belt.

Short Answer

Expert verified
The velocity of the conveyor belt is approximately 3.83 m/s.

Step by step solution

01

Identify the forces

When the part starts sliding, the forces acting on it are the gravitational force and the force of static friction. Since it starts sliding at an angle \( \alpha = 37^{\circ} \), the frictional force reaches its maximum, which is \( \mu_s \times N \), where \( N \) is the normal force.
02

Resolve gravitational force components

The gravitational force \(mg\) can be resolved into two components: one parallel to the belt and the other perpendicular to the belt. The parallel component is \( mg \sin \alpha \), and the perpendicular (normal) component is \( mg \cos \alpha \).
03

Use friction condition for sliding

Sliding occurs when the static frictional force equals the component of gravitational force along the belt. At the angle \( \alpha \), \( \mu_s N = mg \sin \alpha \). Since \( N = mg \cos \alpha \), we have \( \mu_s mg \cos \alpha = mg \sin \alpha \).
04

Solve for velocity

As per the condition for sliding, \( \mu_s = \tan \alpha \). Given \( \mu_s = 1 \) and \( \alpha = 37^{\circ} \), we check if \( \tan 37^{\circ} \approx 0.75 \), which suggests \( \mu_s \) should approximately equal 0.75, indicating a precise point for sliding. Hence, we must consider the velocity component:Use \( v^2 = Rg\tan \alpha \), where \( R \) is the radius. Therefore, the velocity \( v \) is:\[ v = \sqrt{g R \tan \alpha} \]Substitute \( g = 9.8 \ m/s^2 \), \( R = 2 \ m \), and \( \tan 37^{\circ} = 0.75 \):\[ v = \sqrt{9.8 \times 2 \times 0.75} \approx 3.83 \ m/s \].

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Key Concepts

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

Friction
Friction is a force that resists the relative motion of surfaces in contact. In this exercise, we're specifically interested in static friction, which prevents surfaces from sliding over each other. Static friction acts in response to the forces applied to an object and can adjust its magnitude up to a maximum limit determined by the static coefficient of friction, denoted as \( \mu_s \). The maximum static frictional force can be expressed as \( \mu_s \cdot N \), where \( N \) is the normal force.

Consider when an object starts to slide on an inclined surface, the friction reaches this maximum. We can relate this directly to a scenario with a conveyor belt and bins. The exercise examines this concept by determining not when mere motion happens, but when the sliding is initiated due to reaching the peak static friction value.

Some key points to remember about static friction:
  • It acts to prevent motion until its threshold.
  • The threshold is determined by both the coefficient of friction and the normal force.
  • When exceeded, sliding begins, and kinetic friction typically takes over, which is generally weaker than static friction.
Circular Motion
Circular motion occurs when an object travels along a circular path. This exercise discusses parts on a conveyor belt that involves a radius, here given as 2 meters. In circular motion, speed, timing, and forces interplay in fascinating ways. To maintain circular motion, a centripetal force must consistently act toward the center of the circle.

In scenarios involving dynamics where slipping and movement on circular paths occur, such as on a moving belt, understanding how this connects to centripetal forces is paramount. The centripetal force ensuring the object stays on its path in this context derives directly from the frictional forces operating against gravitational forces as the belt turns, hinting at the belt's designed velocity to oppose sliding by employing the calculated centripetal force: \( v^2 = Rg\tan(\alpha) \).

Some vital insight into circular motion in this setup includes:
  • A continuous inward force must be present to enable circular motion.
  • Velocity needs careful adjustment 鈥� moving too fast or too slow may disrupt the desired path.
  • Equilibrium in forces ensures stable motion.
Normal Force
The normal force is the supporting force exerted by a surface perpendicular to the object in contact. In our scenario with the conveyor belt, the normal force plays a pivotal role in counteracting other forces, particularly the component of gravitational force pulling the parts downward.

Understanding the concept of normal force here means looking at how a balance of forces prevents an object from simply falling through the surface. It matches the perpendicular part of gravitational force \( mg\cos(\alpha) \), key in calculating the maximum static friction the surface can offer.

In practical terms, here are a few critical points about normal force relevant to the exercise:
  • It's always perpendicular to the surface with which an object is in contact.
  • It directly affects the frictional capacity of a surface, being crucial in calculating static and kinetic friction.
  • Clarity about its role ensures accurate analysis of forces on inclined or curved surfaces.
By grasping normal force's influence, we can understand more complex motions and restraining forces, such as those on a conveyor belt system.

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Most popular questions from this chapter

A particle is moving in \(x y\) plane. Its velocity and acceleration at a certain instant is given by \(\vec{v}=3 \hat{i}+4 \hat{j} \mathrm{~m} / \mathrm{s}\) and \(\vec{a}=5 \hat{i}+\frac{15}{4} \hat{j}\). Find the rate of change of speed \(\left(\mathrm{in} \mathrm{m} / \mathrm{s}^{2}\right)\) of particle at that instant.

The kinetic energy \(K\) of a particle moving along a circle of radius \(R\) depends upon the distance \(s\) as \(K=a s^{2} .\) The force acting on the particle is (1) \(2 a \frac{s^{2}}{R}\) (2) \(2 a s\left[1+\frac{s^{2}}{R^{2}}\right]^{1 / 2}\) (3) \(2 a s\) (4) \(2 a\)

A force of constant magnitude \(F=10 \mathrm{~N}\) acts on a particle moving in a plane such that it is perpendicular to the velocity \(\vec{v}(|\vec{v}|=v=5 \mathrm{~m} / \mathrm{s})\) of the body, and the force is always directed towards a fixed point. Then the angle (in radian) turned by the velocity vector of the particle as it covers a distance \(S=10 \mathrm{~m}\) is (take: mass of the particle as \(m=2 \mathrm{~kg}\) )

Two particles describe the same circle of radius \(R\) in the same direction with the same speed \(v\), then at the given instant relative angular velocity of 2 with respect to 1 will be (1) \(\frac{2 v \sin \frac{\theta}{2}}{R}\) (2) \(\frac{v}{2 R \sin \frac{\theta}{2}}\) (3) \(\frac{v}{R}\) (4) \(\frac{v \cos \frac{\theta}{2}}{R}\)

A particle tied at one end of an inextensible string of length \(R\), whose another end is fixed. Particle is performing circular motion in vertical plane. Ratio of maximum to minimum speed is \(\lambda\). Choose the correct option(s). (1) Speed at the top most point is \(\sqrt{\left(\frac{4 \lambda^{2}}{\lambda^{2}-1}\right) g R}\) (2) Speed at the top most point is \(\sqrt{\left(\frac{4}{\lambda^{2}-1}\right) g R}\) (3) Ratio of maximum to minimum tension is \(\left(\frac{5 \lambda^{2}-1}{5-\lambda^{2}}\right)\) (4) Ratio of tension in string to weight when string is horizontal is \(2\left(\frac{\lambda^{2}-1}{\lambda^{2}+1}\right)\)
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