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Chapter 13: Problem 59

Cartons having a mass of \(5 \mathrm{~kg}\) are required to move along the assembly line at a constant speed of \(8 \mathrm{~m} / \mathrm{s}\) Determine the smallest radius of curvature, \(\rho,\) for the conveyor so the cartons do not slip. The coefficients of static and kinetic friction between a carton and the conveyor are \(\mu_{s}=0.7\) and \(\mu_{k}=0.5,\) respectively.

Short Answer

Expert verified
The smallest radius of curvature, for the conveyor so the cartons do not slip, is approximately \(9.26 m\).

Step by step solution

01

Identify Forces

Identify the forces acting on the carton. These are the gravitational force (downwards), the normal force (perpendicular to the surface of the conveyor) and the frictional force (tangential to the surface and opposing the motion).
02

Determine the Centrifugal Force

Set up the equation for the centrifugal force acting on the carton in the circular motion, which is \(m*v^{2}/r\), where m is the mass of the carton, v is the velocity and r is the radius of curvature.
03

Calculate the Maximum Static Frictional Force

Calculate the static frictional force that prevents the carton from slipping. This is given by the equation \(F_{s}=\mu_{s}N\), where \(\mu_{s}\) is the coefficient of static friction and N is the normal force (equal to the weight in this case), so \(F_{s}= \mu_{s}*m*g\), where g is the acceleration due to gravity.
04

Set up the Balance of Forces

In order for the carton not to slip, the centrifugal force should be less or equal to the static frictional force. So \(m*v^{2}/r \leq \mu_{s}*m*g\).
05

Solve for the Radius

Solving the equation from Step 4 for the radius: \(r \geq v^{2}/(\mu_{s}*g)\). Substitute the given values: \(r \geq (8ms^{-1})^{2}/(0.7*9.8ms^{-2})\).

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Key Concepts

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

Centrifugal Force
Centrifugal force is an apparent force experienced by an object moving in a circular path, directed outward away from the center of rotation. It's not a real force in classical physics, but rather a result of inertia that tends to pull the object away from the center as it follows a curved path. In the context of the problem, when the cartons move along the curved path of the conveyor, they experience this outward push.

In order to keep the cartons from slipping off the conveyor, the centrifugal force, calculated by the formula \( F_c = m \cdot v^2 / r \), must be countered effectively by static friction. Here, \( m \) is the mass of the carton, \( v \) is the velocity, and \( r \) is the radius of curvature. To ensure safety and efficiency, it's crucial to determine the smallest radius \( \rho \) such that the static friction can still prevent the cartons from slipping due to this 'force'.
Static Friction
Static friction is a resisting force that comes into play when two surfaces are in contact but there is no movement. It's the friction that must be overcome to start moving an object at rest, and it also keeps an object stationary against an applied force. The static frictional force can be calculated using the expression \( F_s = \mu_s \cdot N \), where \( \mu_s \) is the coefficient of static friction, and \( N \) is the normal force.

In our carton-on-conveyor example, \( \mu_s \) is given as 0.7, and since the cartons are moving at a constant speed on a horizontal surface, the normal force is equal to the gravitational force on the cartons (\( m \cdot g \) where \( g \) is the acceleration due to gravity, 9.8 m/s虏). By comparing the static frictional force to the centrifugal force, we can deduce the conditions for which the cartons will not slip, leading to the solution of the radius of curvature problem.
Circular Motion
Circular motion involves any movement of an object along the perimeter of a circle or rotation along a circular path. It's characterized by the object's constant change in direction, which implies a continuous change in velocity even if the object's speed is constant. This is because velocity is a vector quantity, involving both magnitude and direction. The change in velocity creates an acceleration known as centripetal acceleration, which points towards the center of the circle.

The exercise includes cartons moving along a conveyor belt at a constant speed in a circular path. The key to solving for the radius of curvature here lies in ensuring the cartons' circular motion remains constant and unimpeded by slippage, which could occur if the centrifugal 'force' overcomes the static friction between the cartons and the conveyor surface. Hence, understanding the balance of forces involved in circular motion is imperative to solving real-world problems involving curved paths.

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Most popular questions from this chapter

A smooth can \(C\), having a mass of \(3 \mathrm{~kg}\), is lifted from a feed at \(A\) to a ramp at \(B\) by a rotating rod. If the rod maintains a constant angular velocity of \(\dot{\theta}=0.5 \mathrm{rad} / \mathrm{s}\) determine the force which the rod exerts on the can at the instant \(\theta=30^{\circ} .\) Neglect the effects of friction in the calculation and the size of the can so that \(r=(1.2 \cos \theta) \mathrm{m}\). The ramp from \(A\) to \(B\) is circular, having a radius of \(600 \mathrm{~mm}\).

The \(400-\mathrm{kg}\) mine car is hoisted up the incline using the cable and motor \(M\). For a short time, the force in the cable is \(F=\left(3200 t^{2}\right) \mathrm{N},\) where \(t\) is in seconds. If the car has an initial velocity \(v_{1}=2 \mathrm{~m} / \mathrm{s}\) when \(t=0\), determine its velocity when \(t=2 \mathrm{~s}\)

The spring-held follower \(A B\) has a mass of \(0.5 \mathrm{~kg}\) and moves back and forth as its end rolls on the contoured surface of the cam, where \(r=0.15\) mand \(z=(0.02 \cos 2 \theta) \mathrm{m} .\) If the cam is rotating at a constant rate of \(30 \mathrm{rad} / \mathrm{s}\), determine the force component \(F_{z}\) at the end \(A\) of the follower when \(\theta=30^{\circ}\). The spring is uncompressed when \(\theta=90^{\circ} .\) Neglect friction at the bearing \(C .\)

The \(300-\mathrm{kg}\) bar \(B\), originally at rest, is being towed over a series of small rollers. Determine the force in the cable when \(t=5 \mathrm{~s}\), if the motor \(M\) is drawing in the cable for a short time at a rate of \(v=\left(0.4 t^{2}\right) \mathrm{m} / \mathrm{s},\) where \(t\) is in seconds \((0 \leq t \leq 6 \mathrm{~s}) .\) How far does the bar move in \(5 \mathrm{~s} ?\) Neglect the mass of the cable, pulley, and the rollers.

The block \(A\) has a mass \(m_{A}\) and rests on the pan \(B\), which has a mass \(m_{B}\). Both are supported by a spring having a stiffness \(k\) that is attached to the bottom of the pan and to the ground. Determine the distance \(d\) the pan should be pushed down from the equilibrium position and then released from rest so that separation of the block will take place from the surface of the pan at the instant the spring becomes unstretched.
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