TEC 111

TEC 111 QUESTIONS

1. The elements A, B, X, Y and Z have atoms with outer electron shell configurations containing 1, 2, 4, 7 and 8 electrons respectively. State and describe the type of bonding which is likely to occur in the following cases:

a) Between a large number of atoms of A.

Metallic bond; the outer electron is readily

IRD 107 CAT1 TAKE AWAY

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MOI UNIVERSITY

SCHOOL OF ENGINEERING

NAME: CAVIN OLUOCH OJIJO

ADM NO: EC/35/12

COURSE TITLE: COMMUNICATIONS SKILLS

COURSE CODE: IRD 107

LECTURER: MR. KHWALIA

DATE GIVEN: 07/11/11

DATE DUE: 05/12/11

TASK: EVALUATE THE IMPORTANCE OF STUDY

SKILLS IN THE LIFE OF A STUDENT IN THE

UNIVERSITY.

1.1.0 INTRODUCTION

Study skills are strategies or approaches applied to learning. They are

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NUMBER SYSTEMS

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A number can

ANGLE OF REPOSE

The angle of repose or, more precisely, the critical angle of repose,[1] of a granular material is the steepest angle of descent or dip of the slope relative to the horizontal plane when material on the slope face is on the verge of sliding. This angle is in the range 0°–90°.

When bulk granular materials are poured onto a horizontal surface, a conical pile will form. The internal angle between the surface of the pile and the horizontal surface is known as the angle of repose and is related to the density, surface area and shapes of the particles, and the coefficient of friction of the material. Material with a low angle of repose forms flatter piles than material with a high angle of repose.

The term has a related usage in mechanics, where it refers to the maximum angle at which an object can rest on an inclined plane without sliding down. This angle is equal to the arctangent of the coefficient of static friction μs between the surfaces.

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INCLINED PLANE

Inclined Planes

An object placed on a tilted surface will often slide down the surface. The rate at which the object slides down the surface is dependent upon how tilted the surface is; the greater the tilt of the surface, the faster the rate at which the object will slide down it. In physics, a tilted surface is called an inclined plane. Objects are known to accelerate down inclined planes because of an unbalanced force. To understand this type of motion, it is important to analyze the forces acting upon an object on an inclined plane. The diagram at the right depicts the two forces acting upon a crate that is positioned on an inclined plane (assumed to be friction-free). As shown in the diagram, there are always at least two forces acting upon any object that is positioned on an inclined plane - the force of gravity and the normal force. The force of gravity (also known as weight) acts in a downward direction; yet the normal force acts in a direction perpendicular to the surface (in fact, normal means "perpendicular").
The first peculiarity of inclined plane problems is that the normal force is not directed in the direction that we are accustomed to. Up to this point in the course, we have always seen normal forces acting in an upward direction, opposite the direction of the force of gravity. But this is only because the objects were always on horizontal surfaces and never upon inclined planes. The truth about normal forces is not that they are always upwards, but rather that they are always directed perpendicular to the surface that the object is on.

The task of determining the net force acting upon an object on an inclined plane is a difficult manner since the two (or more) forces are not directed in opposite directions. Thus, one (or more) of the forces will have to be resolved into perpendicular components so that they can be easily added to the other forces acting upon the object. Usually, any force directed at an angle to the horizontal is resolved into horizontal and vertical components. However, this is not the process that we will pursue with inclined planes. Instead, the process of analyzing the forces acting upon objects on inclined planes will involve resolving the weight vector (F_grav) into two perpendicular components. This is the second peculiarity of inclined plane problems. The force of gravity will be resolved into two components of force - one directed parallel to the inclined surface and the other directed perpendicular to the inclined surface. The diagram below shows how the force of gravity has been replaced by two components - a parallel and a perpendicular component of force.

  The perpendicular component of the force of gravity is directed opposite the normal force and as such balances the normal force. The parallel component of the force of gravity is not balanced by any other force. This object will subsequently accelerate down the inclined plane due to the presence of an unbalanced force. It is the parallel component of the force of gravity that causes this acceleration. The parallel component of the force of gravity is the net force.

The task of determining the magnitude of the two components of the force of gravity is a mere manner of using the equations. The equations for the parallel and perpendicular components are:

In the absence of friction and other forces (tension, applied, etc.), the acceleration of an object on an incline is the value of the parallel component (m*g*sine of angle) divided by the mass (m). This yields the equation

(in the absence of friction and other forces)
In the presence of friction or other forces (applied force, tensional forces, etc.), the situation is slightly more complicated. Consider the diagram shown at the right. The perpendicular component of force still balances the normal force since objects do not accelerate perpendicular to the incline. Yet the frictional force must also be considered when determining the net force. As in all net force problems, the net force is the vector sum of all the forces. That is, all the individual forces are added together as vectors. The perpendicular component and the normal force add to 0 N. The parallel component and the friction force add together to yield 5 N. The net force is 5 N, directed along the incline towards the floor.
The above problem (and all inclined plane problems) can be simplified through a useful trick known as "tilting the head." An inclined plane problem is in every way like any other net force problem with the sole exception that the surface has been tilted. Thus, to transform the problem back into the form with which you are more comfortable, merely tilt your head in the same direction that the incline was tilted. Or better yet, merely tilt the page of paper (a sure remedy for TNS - "tilted neck syndrome" or "taco neck syndrome") so that the surface no longer appears level. This is illustrated below.

Once the force of gravity has been resolved into its two components and the inclined plane has been tilted, the problem should look very familiar. Merely ignore the force of gravity (since it has been replaced by its two components) and solve for the net force and acceleration.
As an example consider the situation depicted in the diagram at the right. The free-body diagram shows the forces acting upon a 100-kg crate that is sliding down an inclined plane. The plane is inclined at an angle of 30 degrees. The coefficient of friction between the crate and the incline is 0.3. Determine the net force and acceleration of the crate.
Begin the above problem by finding the force of gravity acting upon the crate and the components of this force parallel and perpendicular to the incline. The force of gravity is 980 N and the components of this force are F_parallel = 490 N (980 N • sin 30 degrees) and F_{perpendicular} = 849 N (980 N • cos30 degrees). Now the normal force can be determined to be 849 N (it must balance the perpendicular component of the weight vector). The force of friction can be determined from the value of the normal force and the coefficient of friction; F_frict is 255 N (F_frict = "mu"*F_norm= 0.3 • 849 N). The net force is the vector sum of all the forces. The forces directed perpendicular to the incline balance; the forces directed parallel to the incline do not balance. The net force is 235 N (490 N - 255 N). The acceleration is 2.35 m/s/s (F_net/m = 235 N/100 kg).

 

Practice

The two diagrams below depict the free-body diagram for a 1000-kg roller coaster on the first drop of two different roller coaster rides. Use the above principles of vector resolution to determine the net force and acceleration of the roller coaster cars. Assume a negligible affect of friction and air resistance. When done, click the button to view the answers.





The affects of the incline angle on the acceleration of a roller coaster (or any object on an incline) can be observed in the two practice problems above. As the angle is increased, the acceleration of the object is increased. The explanation of this relates to the components that we have been drawing. As the angle increases, the component of force parallel to the incline increases and the component of force perpendicular to the incline decreases. It is the parallel component of the weight vector that causes the acceleration. Thus, accelerations are greater at greater angles of incline. The diagram below depicts this relationship for three different angles of increasing magnitude.


Roller coasters produce two thrills associated with the initial drop down a steep incline. The thrill of acceleration is produced by using large angles of incline on the first drop; such large angles increase the value of the parallel component of the weight vector (the component that causes acceleration). The thrill of weightlessness is produced by reducing the magnitude of the normal force to values less than their usual values. It is important to recognize that the thrill of weightlessness is a feeling associated with a lower than usual normal force. Typically, a person weighing 700 N will experience a 700 N normal force when sitting in a chair. However, if the chair is accelerating down a 60-degrees incline, then the person will experience a 350 Newton normal force. This value is less than normal and contributes to the feeling of weighing less than one's normal weight - i.e., weightlessness.
Angles of Friction

---

Equations

tan φs = μs

tan φk = μk

Nomenclature

symbol	description
φs	angle of static friction
φk	angle of kinetic friction
μs	coefficient of static friction
μk	coefficient of kinetic friction
N	normal component of the reaction surface

Explanation

It is sometimes convenient to replace the normal force N and the friction force F by a resultant R. Consider a block of weight W resting on a horizontal plane surface:

No friction	R = N		If no horizontal force is applied to the block, the resultant R reduces to the normal force N.
No motion	F = Px φ < φs		However, if the applied force P has a horizontal component Px which tends to move the block, the force R will have a horizonatal component F and, thus, will form an angle φ with the normal to the surface as shown.
Motion impending	Fm = Px φ = φs		If Px is increased until motion becomes impending, the angle between R and the vertical grows and reaches a maximum value.

This maximum value is called the angle of static friction and is denoted by φ_s. From the geometry of the previous figure, the equation is:

tan φs =

Fm N

=

μsN N

tan φs = μs

Motion

F = Px φ < φs

If motion actually takes place, the magnitudes of the friction force drops to Fk; similarly, the angle φ between R and N drops to a lower value φk, called the angle of kinetic friction.

From the geometry of the previous figure, the equation is:

tan φk =

Fk N

=

μkN N

tan φk = μk

Another example may show how the angle of friction can be used to advantage in the analysis of certain types of problems. Consider a block resting on a base and subjected to no other force than its weight W and the reaction R of the board. The board can be given any desired inclination.

No friction	R = N		If the base is horizontal, the force R exerted by the board on the block is perpendicular to the base and balances the weight W.
No motion	F = Px φ < φs		If the base is given a small angle of inclination, θ, the force R will deviate from the perpendicular to the base by the angle θ and will keep balancing W; it will then have a normal componenet N of magnitude N = W cos θ and a tangential component F of magnitude F = W sin θ.
Motion impending	Fm = Px φ = φs		If the angle of inclination continues to increase, motion will soon become impending. At that time, the angle between R and the normal will have reached its maximum value φs. The value of the angle of inclination corresponding to impending motion is called the angle of repose. Clearly, the angle of repose is equal to the angle of static friction φ.
Motion impending	Fm = Px φ = φs		If the angle of inclination θ is further increased, motion starts and the angle between R and the normal drops to the lower value φk. The reaction R is not vertical any more, and the forces acting on the block are unbalanced.

LAWS OF FRICTION

Laws of Dry Friction and Coefficients of Friction

---

Equations

(Eq1)     Fm = μsN	Magnitude of the maximum static friction force
(Eq2)     Fk = μkN	Magnitude of the kinetic friction force

Nomenclature

symbol	description
Fm	maximum force
Fk	kinetic friction force
μs	coefficient of static friction
μk	coefficient of kinetic friction
N	normal component of the reaction surface

Explanation

Friction is both good and bad: it allows objects to move, such as in walking or driving, yet can result in a lot of wasted energy through the generation of heat, such as in engines, and in other areas.

Friction can be understood simply by observing the following picture of a cinder block sitting on asphalt:



Zooming in, it can be seen that there are irregularities in the surfaces of both materials, and when they are moved against each other, these small bits are pushing against each other. It is the summation of all of these horizontal forces causing resistance that make it hard to push an object with a lot of friction. When an object is pushed the bits may break off, or they may bump over each other. The weight of the object, or any vertical forces that are pushing down on the object, will influence the friction because as vertical downward force is increased, it becomes harder to overcome those horizontal forces between all the little small bits interconnected with each other between the contacting surfaces.

The laws of dry friction may be explained with a wooden block of weight W that is placed on a horizontal concrete plane surface as shown below.



The forces acting on the block are its weight W and the reaction of the surface. Since the weight has no horizontal component, the reaction of the surface also has no horizontal component. The reaction is therefore normal to the surface and is represented by N as shown above.

Now a horizontal force P is applied to the block as shown in the following figure:



If P is small enough, the block will not move. This only means that some other horizontal force must therefore exist, which balances P. This other force is the static-friction force F, which is actually the resultant of numerous forces acting over the entire surface of contact between the block and the plane, as indicated in the first picture in this lesson. It is generally assumed that these forces are due to the irregularities of the surfaces in contact and, to a certain extent, to molecular attraction.

If the force P is increased, the friction force F also increases, continuing to oppose P until its magnitude reaches a certain maximum value F_m, as shown in the following plot:



If P is further increased, the friction force cannot balance it any more and the block starts sliding. A more accurate representation of the previous plot may be the following:



F and P are relatively proportional until F_m is reached, and motion begins. F_k may be shown as a wavy and bumpy line because the frictional force varies at the micro-level interactions, as seen in the first picture in this lesson, which may not be consistent.

Notice in the following figure that, as the magnitude F of the friction force increases from 0 to F_m, the point of application a of the resultant N of the normal forces of contact moves to the right, so that the couples formed, respectively, by P and F and by W and N remain balanced.



If N reaches point b, which is fixed at the bottom corner of the block, before F reaches its maximum value F_m, the couples will no longer be balanced and the block will tip about point b before the block can start sliding.

As soon as the block has been set in motion, the magnitude of F drops from F_m to a lower value F_k as shown below and in the above plots:



This is because there is less interpenetration between the irregularities of the surfaces in contact when these surfaces move with respect to each other. From then on, the block keeps sliding with increasing velocity while the friction force is now denoted by F_k, which is called the kinetic friction force. This can be observed in the plot above. The kinetic friction force remains approximately constant throughout the motion of the block.

Experimental evidence shows that the maximum value F_m of the static friction force is proportional to the normal component N of the reaction of the surface. The equation is:

(Eq1)

Fm = μsN

Where μ_s is a constant called the coefficient of static friction. Similarly, the magnitude F_k of the kinetic friction force may be put in the form:

(Eq2)

Fk = μkN

Where μ_k is a constant called the coefficient of kinetic friction. The coefficients of friction μ_s and μ_k do not depend upon the area of the surfaces in contact. Both coefficients, however, depend strongly on the nature of the surfaces in contact. Since they also depend upon the exact condition of the surfaces, their value is seldom known with an accuracy greater than 5 percent. The corresponding values of the coefficient of kinetic friction would be about 25 percent smaller.

It is presumed that four different situations can occur when a rigid body is in contact with a horizontal surface:

1.	No friction (Px = 0)	F = 0 N = P + W		The forces applied to the body do not tend to move it along the surface of contact; there is not friction force.
2.	No motion (Px < Fm)	F = Px F < μsN N = Py + W		The applied forces tend to move the body along the surface of contact but are not large enough to set it in motion. The friction force F which has developed can be found by solving the equations of equilibrium for the body. Since there is no evidence that F has reached its maximum value, the equation Fm = μsN cannot be used to determine the friction force.
3.	Motion impending (Px = Fm)	Fm = Px Fm = μsN N = Py + W		The applied forces are such that the body is just about to slide, that is, motion is impending. The friction force F has reached its maximum value Fm and, together with the normal force N, balances the applied forces. Both the equations of equilibrium and the equation Fm = μsN can be used. Also note that the friction force has a sense opposite to the sense of impending motion.
4.	Motion (Px > Fm)	Fk < Px Fk = μkN N = Py + W		The body is sliding under the action of the applied forces, and the equations of equilibrium do not apply any more. However, F is now equal to Fk and the equation Fk = μkN may be used. The sense of Fk is opposite to the sense of motion.

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