Problems Exam Materials

Newton’s First Law of Motion- a body at rest remains at rest and a body in uniform motion will continue to move with such motion

unless it is acted upon by an unbalanced force will cause it to move or accelerate.- a body at rest or moving with uniform motion has zero acceleration

Σx = 0Σy = 0

Force – push or pull

Types:1. Weight (W) – force acting at the center of gravity of a body or system which is directed vertically

downward.

2. Normal force (N) – force that is acting almost always upward perpendicular to surface contact.

3. Tension (T) – force acting along the line but away from the point of consideration.

4. Compression (C) – force acting along the line towards the point of consideration.

5. Friction (f) – force that tends to resist the impending motion or the motion of a body.

Classification of force1. Concurrent force – acting at a common point within a body.2. Nonconcurrent force – force acting at different points within a body.

First condition of equilibriumFor a body to remain at rest the resultant or unbalanced force must be equal to zero. Therefore,

Σx = 0Σy = 0R = 0

Procedure;1. Sketch the problem showing all necessary data.2. Draw the equivalent “free body diagram”

- isolated sketch of the problem showing only the forces acting on the system.

Sample problems1. Solve for the tension on the cord and the compression on the wooden bar. (neglect the weight of the

wooden bar)

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600N

60O

45O

2. A cylinder weighing 400 N is held against a smooth inclined plane by means of weightless rod AB as shown in the figure. Determine the force P and N exerted on the cylinder b the rod an inclined plane respectively.

3. Determine the tension on the cords.

Friction and inclined plane

Friction – force that exists always parallel to the surfaces of bodies in contact– force that resists the motion or impending motion of a body– force that is always opposite the motion of a body.

Two types of friction1. Static friction (fs) – the frictional force needed to keep an object from sliding.2. Kinetic friction (fk) – the frictional force that exists when the body is in contact with it.

Coefficient of friction (µ)- refers to the condition of the surface (roughness or smoothness)

0 < µ < 1

µ = F/N2

55O25O

W

B

A

200N

400N

40O

80O

50O

60O

Laws of friction1. Friction is directly proportional to the normal force.2. Friction is roughly independent of the velocity or speed of the body.3. Friction is roughly independent of the surface area in contact.4. Static friction is always greater than kinetic friction.

Angle of repose- also called limiting angle- the angle between a horizontal floor and a plane surface wherein an object will slide or move with a

uniform velocity (v = k)

Problems1. In the figure, suppose that a block weighs 20N, that the tension P can be increased to 8N before the

block starts to slide, and that the force of 4N will keep the block moving at constant speed once it has been set in motion. Find the coefficient of static and kinetic friction. What is the friction force if the block is at rest on the surface and a horizontal force 5N is exerted on it?

2. What force T at an angle 30o above the horizontal is required to drag a block weighing 20N to the right as the constant speed, if the coefficient of kinetic friction between the block and the surface is 0.20?

3. In the given figure, what must be the value of F so that the block will move up the plane with uniform velocity µk=0.4.

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Newton’s Second Law- if an unbalanced force acts on a body, it is accelerated by an amount proportional to the unbalanced

force and in the same direction but inversely proportional to its mass.

Where:F - force (N, dynes)m - mass (kg, g)a – acceleration (m/sec, cm/sec)

Dynamics of Translation- a study of motion and the forces causing the motion.

Sample Problems1. A constant horizontal force at 40N acts on a body on a smooth horizontal plane. The body starts from

rest and is observed to move 100m in 5 sec. What is the mass of the body?

2. A body of mass 15 kg rests on a frictionless surface and is acted on by a horizontal force of 30 N.a. What is the acceleration produced?b. How far will the body travel in 10 sec?c. What will be its velocity at the end of 10 sec?

3. In the given figure, determine the acceleration if a constant force of 100 N acts on the body, the coefficient of friction between the block and the floor is 0.2.

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10 s

Vo = 0distance

Force

5 s

Vo = 0distance

Force

Friction

37o

Direction of Acceleration

Force

Sir Isaac Newton: The Universal Law of Gravitation

There is a popular story that Newton was sitting under an apple tree, an apple fell on his head, and he suddenly thought of the Universal Law of Gravitation. As in all such legends, this is almost certainly not true in its details, but the story contains elements of what actually happened.

What Really Happened with the Apple?

Probably the more correct version of the story is that Newton, upon observing an apple fall from a tree, began to think along the following lines: The apple is accelerated, since its velocity changes from zero as it is hanging on the tree and moves toward the ground. Thus, by Newton's 2nd Law there must be a force that acts on the apple to cause this acceleration. Let's call this force "gravity", and the associated acceleration the "accleration due to gravity". Then imagine the apple tree is twice as high. Again, we expect the apple to be accelerated toward the ground, so this suggests that this force that we call gravity reaches to the top of the tallest apple tree.

Sir Isaac's Most Excellent Idea

Now came Newton's truly brilliant insight: if the force of gravity reaches to the top of the highest tree, might it not reach even further; in particular, might it not reach all the way to the orbit of the Moon! Then, the orbit of the Moon about the Earth could be a consequence of the gravitational force, because the acceleration due to gravity could change the velocity of the Moon in just such a way that it followed an orbit around the earth.

This can be illustrated with the thought experiment shown in the following figure. Suppose we fire a cannon horizontally from a high mountain; the projectile will eventually fall to earth, as indicated by the shortest trajectory in the figure, because of the gravitational force directed toward the center of the Earth and the associated acceleration. (Remember that an acceleration is a change in velocity and that velocity is a vector, so it has both a magnitude and a direction. Thus, an acceleration occurs if either or both the magnitude and the direction of the velocity change.)

But as we increase the muzzle velocity for our imaginary cannon, the projectile will travel further and further before returning to earth. Finally, Newton reasoned that if the cannon projected the cannon ball with exactly the right velocity, the projectile would travel completely around the Earth, always falling in the gravitational field but never reaching the Earth, which is curving away at the same rate that the projectile falls. That is, the cannon ball would have been put into orbit around the Earth. Newton concluded that the orbit of the Moon was of exactly the same nature: the Moon continuously "fell" in its path around the Earth because of the acceleration due to gravity, thus producing its orbit.

By such reasoning, Newton came to the conclusion that any two objects in the Universe exert gravitational attraction on each other, with the force having a universal form:

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The constant of proportionality G is known as the universal gravitational constant. It is termed a "universal constant" because it is thought to be the same at all places and all times, and thus universally characterizes the intrinsic strength of the gravitational force.Universal law of Gravitation

Mathematically

;

Where:- mass of the first body (kg, g) - mass of the second body (kg, g)

r – distance between the two bodies (m, cm)G – Universal Gravitation Constant

Sample Problem1. Two sphere balls whose masses are 3.4 kg and 2.5 kg are placed with their centers 60 cm apart. With

what force do they attract each other?

2. Determine the force of attraction between the earth and the moon if the distance between the earth and the moon is 3.84 x 108 m. Let the mass of the earth be 5.98 x 1024 kg and the mass of the moon to be 7.35 x 1028 kg.

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m2m1

60 m

Earth

moon

60 m

3. If the center between two 30 kg lead balls is 70 cm, with what force do they attract?

Uniform Circular Motion (UCM)- motion along a circular path at constant speed.

Characteristic of UCM1. The speed is constant.2. The direction of motion is continually or uniformly changing.3. A constant in magnitude acceleration is present. This acceleration is always directed at the center of

the circular path and is known as radial acceleration or centripetal central acceleration.

Mathematically;Acp = v2/r

WhereAcp = centripetal accelerationv = tangential velocityr = radius of the circle

Principle of circle

Relationship between s, r,

s s s = r

Linear quantitiess = linear displacement (cm, m)v = linear velocity (cm/s, m/s)a = linear acceleration (cm/s2, m/s2)

Angular quantities = angular displacement (rad)ω = angular velocity (rad/s)α = angular acceleration (rad/s2)

Relationship between v and ω(s = r) /tv = ωr

Derivation of centripetal accelerationConsider a moving body along a circular path with angular displacement of very very small rads or approaches 0.

1st characteristicsVA – VB = VVA + (– VB) = V

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By similar triangle.

V/V = s/r

V/V = vt/r

V/t = v2/r

a = v2/r

Centripetal Acceleration in terms of ω

Acp = v2/r but v = ωr

Acp = (ωr)2/r

Acp = ω2 r

Centripetal Acceleration in terms of number of revolutions (n)

n (rev/s, rev/min)

ω nω =k n since k = 2 =360o

ω = 2n

Acp = (2n) 2 r

Acp = 42n2 r

Centripetal Force (Fcp)- net force or a constant pull that deflects a body moving in a rectilinear path and compels it to move in

a circular path.

By Newton’s 2nd law of motionF = ma

Fcp = mAcp

Fcp = m v2/r

Fcp = mω2 r

Fcp = m42n2 r

Where centripetal force is in terms of newtons and dynes8

Centrifugal force (Fcf)- inertial reactive force exerted by the moving body in its attempt to continue its straight line motion. It

is equal to the centripetal force in magnitude but oppositely directed.

By Newtons’ 3rd law of motion

Fcp = - Fcf

Sample Problems1. UCM in vertical circle

A 50g body tied at one end of a 100 cm string is whirled in a vertical circle at a rate of 2 rev/sec. calculate the tension in the string at the ff. position:

a. bottom of VCb. top of VCc. horizontal diameter of VCd. at the point where the string makes an angle 30o above the horizontal.

2. UCM along a circular track (unbanked)What is the maximum safe speed at which an automobile can round a curve of 24 m radius on a level road if a coefficient of friction between the tires and the road is 0.30?

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3. Banking of curves – roads are banked, inclined at a certain angle, in order to eliminate the effect of friction.A curve of radius 30 m is to be banked so that an automobile may make the turn at the speed of 12 m/s, without depending on friction. What must be the slope of the curve?

4. Banked curve with friction.What is the maximum safe speed at which an automobile can round a curve at 24 m radius at an angle of 16o if the coefficient of friction between the road and the tire is 0.30?

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