Satellites and "Weightlessness"

A satellite is put into orbit by accelerating it to a sufficient high tangential speed with the use of rockets.

If the speed is too high, the spacecraft will not be confined by the Earth's gravity and will escape, never to return.

If the speed is too low, it will return to Earth.

Satellite are usually put into circular (or nearly circular) orbits, because such orbits require the least takeoff speed.






this image is from www.howstuffworks.com

If a satellite stopped moving, it would fall directly to Earth.

But at the very high speed a satellite has, it would quickly fly out into space, if it weren't for the gravitational force of the Earth pulling it into orbit.

In fact, a satellite is falling (accelerating toward Earth), but its high tangential speed keeps it from hitting Earth.

Gravity Near the Earth's Surface: Geophysical Application

When we apply the gravitaional force to the gravitational force between the Earth and an object at its surface,

m(1) becomes the mass of the Earth m(E) .

m(2) becomes tge mass of the object m.

r becomes the distance of the object from the Earth's centre = radius of the earth

The force of gravity due to the Earth is the weight of the object mg.

Newton's Law of Universal Gravitation

G = 6.67 x 10-11 N m^2/kg^2

http://www.physicsclassroom.com/Class/circles/U6L3c.html

Centrifugation

image from http://www.steve.gb.com

Nonuniform Circular Motion

Any object moving in a circle has a centripetal acceleration,

directed toward the center of that circle.

It has a centripetal force acting on it, also directed toward the center of that circle.

"Directed toward the center"="radially".

Any object moving in a circle has a radial acceleration.

That means it has a radial force acting on it.

In addition, the object may be accelerating tangentially , along the direction tangential to the circle or perpendicular to the radius.

Then there is also a tangential force or the net force has a tangential component as well as the radial component.

The total or net acceleration is the vector sum of the radial and tangential components.

The total or net force is the vector sum of the radial and tangential components.

banked curve

highway curves

horizontal circular motion & vertical circular motion

Force on revolving ball

T= mv^2/r

object revolving in a circle

circular motion

Uniform Circular Motion

An object moving in a circle is accelerating.

Accelerating objects are objects which are changing their velocity -

either the speed (i.e., magnitude of the velocity vector) or the direction.

An object undergoing uniform circular motion is moving with constant speed; nonetheless, it is accelerating due to its change in direction.

The direction of the acceleration is inwards.

Circular Motion ; Gravitation (Chapter 5)

Objectives

1. the centripetal acceleration of a point mass in uniform circular motion given the radius of the circle and either the linear speed or the period of the motion.

2. the force that is the cause of the centripetal acceleration and determine the direction of the acceleration vector.

3. Highway Curves, Banked and UnBanked Curves.

4. Nonuniform Circular Motion

5. Centrifugation

6. Newton's universal law of gravitation and explain the meaning of each symbol in the equation.

7.Gravity Near the Earth's Surface; Geophusical Applications

8. Satellites and "Weightlessness".

9. Kepler's Laws and Newton's Synthesis

10.Types Forces in Nature

Practice Questions (Chapter 4)

1 .A packing crate slides down an inclined ramp at constant velocity. Thus we can deduce that
a frictional force is acting on it.

(A) a net force is acting on it.
( B) it may be accelerating.
( C) Gravity is not acting on it.


2. You fall while skiing and one ski (of weight W) loosens and slides down an icy slope (assume no friction), which makes an angle q with the horizontal. The force that pushes it down along the hill is

(A) zero; it moves with constant velocity.
(B) W cos q
(C) W sin q
(D) it is not acted on by a gravitational force.


3. A decoration, of mass M, is suspended by a string from the ceiling inside an elevator. The elevator is traveling upward with a constant speed. The tension in the string is

(A) equal to Mg.
(B) less than Mg.
(C) greater than Mg.
(D) impossible to tell without knowing the speed.


4. An object is placed on an inclined plane. The angle of incline is gradually increased until the object begins to slide. The angle at which this occurs is q. What is the coefficient of static friction between the object and the plane?

(A) sin q
(B) cos q
(C) tan q
(D) Cannot determine without knowing the mass of the object.


5. A mass on a string attached to a ceiling is pulled to the side and let go. When the string makes an angle q with the vertical, which of the following is not true?

(A) The weight is down.
(B) The tension equals mg*cos(q).
(C) The tension equals mg*sin(q).
(D) The tension and the weight are never in the same direction.


6. A roller coaster car does a loop-the-loop. When it is upside down at the very top, which of the following is true?

(A) The normal force and the weight are in opposite directions.
(B) The normal force and the weight are perpendicular to each other.
(C) The weight is zero.
(D) The normal force and the weight are in the same direction.


7. A roller coaster car does a loop-the-loop. When it is right-side up at the very bottom, which of the following is true?

(A) The normal force and the weight are in opposite directions.
(B) The normal force and the weight are perpendicular.
(C) The weight is zero.
(D) The normal force and the weight are in the same direction.


8. A 2 kg box sits on a 3 kg box which sits on a 5 kg box. The 5 kg box rests on a table top. What is the normal force exerted on the 5 kg box by the table top?

(A) 19.6 N
(B) 29.4 N
(C) 49 N
(D) 98 N


9. A 2 kg box sits on a 3 kg box which sits on a 5 kg box. The 5 kg box rests on a table top. What is the normal force exerted by the 5 kg box on the 3 kg box?

(A) 19.6 N
(B) 29.4 N
(C) 49 N
(D) 98 N


10. A 5 kg (4.9 N) book rests on your desk. Which of the following answers is not correct?

(A) The normal force on the book equals its weight because of Newton's First Law.
(B) The normal force on the book equals its weight because of Newton's Second Law.
(C) The normal force on the book equals the weight because of Newton's Third Law.


11. 5 kg (4.9 N) book rests on your desk. What is the reaction force to the 4.9 N weight (force) of the book?

(A) The 4.9 N normal force from the table on the book.
(B) The 4.9 N reaction force from the book on the table.
(C) The weight of the earth.
(D) A 4.9 N force on the Earth.


12. You stand on a bathroom scale in an elevator accelerating upwards and it reads 200 Newtons. Which of the following is true?

(A) You weigh 200 Newtons.
(B) The scale exert a 200 N force on you.
(C) The acceleration of the elevator is 20.4 m/s^2.
(D) None of the above are correct.


13. During the investigation of a traffic accident, police find skid marks 90 m long. They determine the coefficient of friction between the car's tires and the roadway to be 0.5 for the prevailing conditions. To find the speed of the car you must

(A) know the mass of the car.
(B) find the acceleration of the car.
( C) know the mass of the car and find the acceleration of the car.


14. An object moves left to right (right is positive) with speed decreasing at a constant rate and
its acceleration is positive.

(A) the net force on it is decreasing.
(B) the net force on it is increasing.
(C) its acceleration is negative.


15. A car pulls on a rope tied to a tree with a force of 1000 Newtons. Which of the following is not true.

(A) The rope pulls on the tree with a force of 1000 Newtons.
(B) The tree pulls on the rope with a force of 1000 Newtons.
(C) The tension in the rope is 2000 Newtons.
(D) The tension in the rope is 1000 Newtons.



16. A 5 kg block and a 10 kg block slide down a frictionless incline and
they both have the same acceleration.

(A) the 5 kg block has an acceleration of twice that of the 10 kg block.
(B) the 10 kg block has an acceleration of twice that of the 5 kg block.
(C) their acceleration depends on the normal force.

inclines

Fig. Forces on an object sliding down an incline.

Coefficients of Friction

Kinetic Friction

When an object slides along a rough surface, the force of kinetic friction acts opposite to the direction of the object's velocity.

The magnitude of the force kinetic friction depends on the nature of the two sliding surfaces.

The friction force is approximately proportional to the normal force between the two surfaces, which is the force that either object exerts on the other, perpendicular to their common surface of contact.

We write the proportionality between the friction force F (fr) and the normal force F (N) as an equation by inserting a constant proportionality,

Friction

When an object is pulled by an applied force , F(A) along a surface, the force of friction F (fr) opposes the motion. The magnitude of F(fr) is proportional to the magnitude of the normal force , F(N).

Two boxes connected by a cord

Two boxes, A and B , are connected by a lightweight cord and are resting on a smooth (frictionless) table.

Tension in a Flexible Cord

Tension is a force.

When a flexible cord pulls on an object, the cord is said to be under tension, and the force it exerts on the object is the tension F (T).


Two boxes , A and B , are connected by a cord. A person pulls horizontally on the box A with force F (p) = 40.0 N.

T (upper) = T (lower)

By the 2nd law and the fact that the system is at rest,

T (upper) = T(lower) = W

Pulling the Mystery Box

Suppose a friend asks to examine the 10.0 kg box , hoping to guess what is inside; and you respond, "Sure, pull the box by the attached cord, as shown in figure, along the smooth surface of the table.

The magnitude of the force exerted by the person is

F (p) = 40.0 N.

It is exerted at a 30 degree angle as shown in figure.

Calculate (a) the acceleration of the box, and (b) the magnitude of the upward force F(N) exerted by the table on the box. Assume that friction can be neglected.

Solution :

1. Draw a sketch : The situation is shown in Fig. (a) , it shows the box and the force applied by the person, F (p).

2. free-body diagram : Fig. (b) shows the free-body diagram of the box. To draw it correctly, we show all the forces acting on the box and only the forces acting on the box.

They are : the force of gravity mg ; the normal force exerted by the table F (N) ; and the force exerted by the person F(p).

3. Choose axes and resolve vectors : We expect the motion to be horizontal , so we choose the x axis horizontal and the y axis vertical . The pull of 40.0 N has components

F (px) = (40.0 N) ( cos 30 degree) = (40.0 N) ( 0.866) = 34.6 N

F (py) = (40.0 N) ( sin 30 degree) = (40.0 N) (0.500) = 20.0 N.

In the horizontal (x) direction, F (N) and mg have zero components.

The horizontal component of the net force is F(px) .

4. (a) Apply Newton's second law to determine the x component of the acceleration;

F (px) = ma (x).

5. (a) Solve :

a (x) = F(px) /m = (34.6 N)/ (10.0 kg) = 3.46 m/s^2.

The acceleration of the box is 3.46 m/s^2 to the right.

4. (b) Apply Newton's second law to the vertical (y) direction , with upward as positive:

F (y) = ma(y)

F (N) - mg + F (py) = ma (y).

5. (b) Solve: We have mg = (10.0 kg ) ( 9.80 m/s^2) = 98.0 N .

F (py) = 20.0 N.

F (py ) is less then mg, the box does not move vertically, so

a (y) = 0.

Thus, F (N) - 98.0 N + 20 N = 0.

F (N) = 78.0 N.

The Honky Puck

A honky puck is sliding at constant velocity across a flat horizontal ice surface that is assumed to be frictionless. Which of the sketches in the figure is the correct free-body diagram for this punk? What would your answer be if the puck slowed down?

The Resultant Force

By Newton's Second Law,

the acceleration of an object is proportional to the net force acting on the object.

The net force is the vector sum of all forces acting on the object.

The object will start moving at a 45 degree angle and thus the net force acts at a 45 degree angle.

From the theorem of Pythagoras,

F(R) = ( ( 100 N )^2 + (100 N) ^2 ) ^ 1/2 = 141 N.

Accelerating the Box

Example :

What happens when a person pulls upward on the box with a force equal to, or greater than , the box's weight , say F (p) = 100.0 N rather than the 40.0 N shown in figure.

Solution: The net force on the box is

F (y) = F (N) - mg + F (p)

= F (N) - 98.0 N + 100.0 N

The box accelerates upward, so the net force is not zero.

Normal force is zero.

The net force = F (y) = F (p) - mg = 100.0 N - 98.0 N

= 2.0 N

a (y) = F (y)/ m = 2.0 N/10.0 kg = 0.20 m/s^2.

Weight, Normal Force, and a Box

A friend has given you a special gift, a box of mass 10.0 kg with a mystery surprise inside. The box is resting on the smooth (frictionless) horizontal surface of a table.

(a) Determine the weight of the box and the normal force exerted on it by the table.

(b) Now your friend pushes down on the box with a force of 40.0 N , as in fig (b) .
Again determine the normal force exerted on the box by the table.

(c) If your friend pulls upward on the box with a force of 40.0 N (fig. (c)), what now is the normal force exerted on the box by the table.

Approach : The box is at rest on the table, so the net force on the box in each case is zero. (Newton's Second Law).
The weight of the box equals mg in all three cases.

Solution:

(a) The weight of the box is mg = (10.0 kg) (8.80 m/s^2) = 98.0 N

This force acts downward.

The normal force exerted upward on it by the table. Fig.(a)

The upward direction as the positive y direction;

The net force along y direction on the box = F (y) = F (N) - mg .

The box is at rest, so the net force on it = 0.

F(y) = F (N) - mg = 0

F (N) = mg

The normal force on the box, exerted by the table , is 98.0 N upward. It has magnitude equal to the box's weight.


(b) Your friend is pushing down on the box with a force of 40.0 N .

Now there are three forces acting on th ebox. Fig. (b)

The weight of the box = mg = 98.0 N

The net force = F (y) = F (N) - mg - 40.0 N = 0 ( a = 0)

For Normal Force ,

F(N) = mg + 40.0 N = 98.0 N + 40.0 N = 138.0 N


(c) The box's weight is still 98.0 N and acts downward.

The force exerted by your friend and the normal force
both act upward ( positive direction). Fig. (c)

The box does not move, since your friend's upward force is less than the weight.
The net force = 0 (a =0)

F (y) = F (N) - mg + 40.0 N = 0

F (N) = mg - 40.0 N = 98.0 N - 40.0 N = 58.0 N.

The table does not push against the full weight of the box because of the upeard pull exerted by your friend.

Free-Body Diagrams

1. A girl is suspended motionless from a bar which hangs from the ceiling by two ropes.









2. An egg is free-falling from a nest in a tree. Neglect air resistance.



3. A flying squirrel is gliding (no wing flaps) from a tree to the ground at constant velocity. Consider air resistance.







4. A rightward force is applied to a book in order to move it across a desk with a rightward acceleration. Consider frictional forces. Neglect air resistance.







5. A rightward force is applied to a book in order to move it across a desk at constant velocity. Consider frictional forces. Neglect air resistance.







6. A college student rests a backpack upon his shoulder. The pack is suspended motionless by one strap from one shoulder.







7. A skydiver is descending with a constant velocity. Consider air resistance.

8. A force is applied to the right to drag a sled across loosely-packed snow with a rightward acceleration.

9. A football is moving upwards towards its peak after having been booted by the punter. Neglect air resistance.

10. A car is coasting to the right and slowing down.

Normal Force

From Newton's Second Law,

the net force on an object that remains at rest is zero.

There must be another force on the object to balance the gravitational force.

For an object is resting on a table,

the table exerts this upward force.

The table is compressed slightly beneath the object, and due to its elasticity, it pushes up on the object .

The force exerted by the table is often called a contact force, since it occurs when two objects are in contact.

When a contact force acts perpendicular to the common surface of contact, it is referred to as the normal force .

A book is at rest on a table top.

F (G) and F( N) must be equal magnitude and in opposite directions.

Weight-the force of Gravity

All objects accelerate towards the earth.
Thus the earth exerts a force on these objects - this is called the force of gravity.

From Newton's Second Law,
an object of mass m falling due to gravity;
for the acceleration, a, we use the downward acceleration due to gravity, g.
The gravitational force on an object, F(G) = F = mg

The direction of this force is down toward the centre of the Earth.

The magnitude of the force of gravity on an object is called the object's weight.

In SI units, g= 9.8 m/s^2 = 9.80 N/kg.