Inclined Planes Summary 06

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Inclined Planes

The concept of inclined planes is poorly presented, if at all, in high school textbooks. Few situations are presented andvariations in problems are limited. The following pages are an attempt to fill in the missing gaps.

In the early seventeenth century, Galileo make a claim concerning the natural motion of a freely falling body. Galileoclaimed that a falling object will gain equal amounts of velocity in equal amounts of time. This means that the object’sspeed increases as it falls and the rate at which it picks up speed does not change during the fall. However, practicalproblems made it difficult for Galileo to put such a claim to any experimental test. Freely falling objects move tooquickly to study or record the motion directly.

Because seventeenth century clocks could not record the short times involved, Galileo slowed down the motion byreplacing the falling object with a ball rolling down a gently inclined plane.

"In order to make use of motions as slow as possible ... I also thought of making movables descend along an inclined plane not much raised above the horizontal" (Galileo, Two New Sciences).

Galileo guessed that objects descending an incline speed up in exactly the same way falling objects do. He reasoned

as follows. A ball rolling down a steep incline will pick up speed faster than a ball rolling down a gentle incline, but theway in which its speed increases will be the same. Freefall, he reasoned, is simply equivalent to a vertical ramp. Thecharacter of the ball’s motion in freefall should be the same as the character of the motion of a ball “falling” down theinclined plane.

Another technological problem arose in measuring the velocity. The velocity of the freely falling object changescontinuously so Galileo could not measure changing velocities directly, but he could measure distances and times.Galileo, therefore, used math to transform his claim about times and velocities. His mathematical argument issummarized below:

If an object gains speed at a steady rate andIf the object is released from rest,then the total distance traveled by the object willbe proportional to the square of the time needed

for that travel.

Inclined Planes Explained

An object placed on a tilted surface will often slide down the surface. The rate at which theobject 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 tiltedsurface is called an inclined plane. Objects are known to accelerate down inclined planesbecause of an unbalanced force. To understand this type of motion, it is important to analyzethe forces acting upon an object on an inclined plane. The diagram at the right depicts the twoforces acting upon a crate which 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 a ny obje ct tis positioned on an inclined plane - the force of gravity and the normal force. The

hatforce of

gravity (also known as weight) acts in a downward direction; yet the normal force acts in adirection perpendicular to the surface (in fact, normal means "perpendicular").

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The first peculiarity of inclined plane problems is that the normal force is not directed in the direction which we areaccustomed 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 thatthey are always directed perpendicular to the surface that the object is on.

Determining the net force acting upon an object on an inclined plane is difficult because two (or more) forces are not

directed

llel

The perpendicular component of the force of gravity is directed opposite the normal force and as such balances the

raphic.

directed in opposite directions. Thus, one (or more) of the forces will have to be resolved into perpendicular components in order to facilitate their addition to the other forces acting upon the object. Normally, any forceat an angle to the horizontal is resolved into horizontal and vertical components ; this, however, is not the process that

we will use with inclined planes. Instead, the process of analyzing the forces acting upon objects on inclined planes willinvolve resolving the weight vector (F grav ) into two perpendicular components. The force of gravity will be resolved intotwo components of force - one directed parallel to the inclined surface and the other directed perpendicular to theinclined surface. The diagram below shows how the force of gravity has been replaced by two components - a paraand a perpendicular component of force.

normal force; the parallel component of the force of gravity is not balanced by any other force. This object willsubsequently accelerate down the inclined plane due to the presence of an unbalanced force. It is the parallelcomponent of the force of gravity which causes this acceleration. This principle is summarized in the following g

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

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

nce the force of gravity has been resolved into its two components and the inclined plane has been tilted, theproblem should look very familiar. Merely ignore the force of gravity (since it has been replaced by its two components)

represent the simplest examples of constrained fall. Unlike a parabolic fall, the inclined planeprovides partial support for an object falling down it, preventing it from falling as fast as it ordinarily would. The basic

O

and solve for the net force and acceleration.

Constrained Fall

Galileo's inclined planes

rule of falling down an inclined plane is that the falling body will accelerate faster the steeper the plane is. If the plane iscompletely vertical, the body will be in free fall, which represents the fastest gravitational acceleration possible. If theplane is completely horizontal, the body will not fall at all. This principle applies to coasters, even though the track ondrops is usually not straight, like an inclined plane. The steeper a drop, the faster the train accelerates down it. Thus,the train accelerates very quickly down the steep first drop. On the other hand, the train accelerates very slowly downthe shallow drop. However, if a steep and a shallow drop are of the same height, the train will eventually reach thesame speed at the bottom, if we ignore friction. Though the train accelerates more slowly on a shallow drop, it has alonger time in which to accelerate. The factors balance out exactly. This makes sense when we think of energy-- thepotential energy difference from the top to the bottom of a drop depends only on its height, not on its steepness.

Example Problems – Inclined Planes

1. A block is launched up a frictionless incline, as shown to the right, with aninitial speed of 5.5 m/s. What is the maximum displacement, d, of the blockup the incline?

2ov

D = =2

2

(5.5 / )2 sin 2(9.80 / )(sin 40 )o

m sg m sθ

2.4m=

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2. The block shown to the right remains at rest. What is the friction force acting onthe block?

2(0.50 )(9.80 / )F ma F kg m s= = 4.9 N =

N

3. A 4 kg mass initially moving at 10 m/s at the bottom of a 37 o inclinest makes it to the top before coming to a stop. The work done byction is?

4.9 sin 40 3.1o

fr F N = =

jufri

21 E mgh mvΔ = −

2

21

(40 )(3 ) (4 )(10 / ) 802

N m kg m s J = − = −

What is the normal force on the block to the right? 4.

2(2.5 )(9.80 / )F ma F kg m s= = 24.5 N =

lane that makes a 30 o ngle with the horizontal. Find the acceleration of this block.

24.5 cos 25 22.2oF N = =

5. A block slides down a frictionless, inclined pa

sin

sinF mg

a gm m

θ θ = = =

4.2 2sin (9.80 / )(0.5) 9 / g m s m sθ = =

6. A block slides down a 30 o incline which has a coefficient of kineticfriction 0.3μ = . What is the acceleration of the block?

(sin cos )a g θ μ θ = −

2 o o (9.80 / )(sin 30 0.3cos30 ) 2.4 / m s m s= − = 2

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7. A forcewith the ho

of 20 N acts horizontally on a mass of 10 kg being pushed up a frictionless incline that makes a 30 o anglerizontal. What is the acceleration of the box up the incline?

cos sinF mg maθ θ − =

cos sinF mga m

θ θ −=

2(20 )cos30 (10 )(9.8 / )sin 3010

o o N kg m sa

kg

−= s

8. A 40 kg box slides down an incline with a coefficient of kinetic friction of

23.17 / a m=

μ =0.3. The length of the ramp is 10 m.

a. How much work is done by gravity?

d ( sin )gravW mg θ =

)(sin 40 )(10 ) 2520o m J

uch work is done by the normal force?

force is perpendicular to the motion. The work done by this force is 0.

)

(40 )(9.8 / kg m s= 2 = b. How m

Normal

c. How much work is done by friction?

( cos )( fr fr k W F d mg d μ θ = − = − 2 (0.3)(40 )(9.kg= − 8 / )(cos 40 )(10 ) 901om s m J = −

the total work done?

W W W W = ∑ = + +

+ − =

roblems – Inclined Planes & Pulley’s

9. A 1.0 kg block of wood is attached via a pulley to a hanging 0.5 kglculate the

d. What is

W N Total grav F fr

2520= +0 ( 901) 1619 J

Example P

mass. Assuming no friction between the block and the table, caacceleration of the system. ( 210.0 / g m s= )

2(10.0 / )(0.5 ) (1.5 )F ma m s kg kg a= =

s 23.3 / a m=

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10. Two masses are connected by a light cord. The horizontal mass of 2 kg is pulled to the right with a force of 30 N along a frictionlesssurface. What is the tension on the cord? What is the acceleration?

In the x-direction:

9.80 / T m s−

ve two equations: = ration substitute 16.5 N and find a: a = 6.75 m/s 2

) What is the force of friction acting on the 8 kgss?

2F T ma a− = =

In the y-direction: 29.8 / T Mg T m s Ma a− = − = =

30 2 N T a a− = = 2

Combine the abo − 230 2( 9.8 / ) 16.5 N T T m s T N − =To find the accele

11. An 8-kg mass moves up the incline where thecoefficient of kinetic friction is 0.4. Assume an idealpulley.(ama

For the 8 kg mass

x k N F T mg F θ μ = − −

80sin37 0.4 x N F T F = − −

For the 20 kg mass

mg64 0 64

48 0.4(64) 8

N N F F

T a

− = ⇒ =− − =

F F cos y N θ = − 80cos37 y N F F − =

Force of friction

ss?

Find for the 20 kg mass a

0.4(64) 25.6= =

(b) What is the acceleration of each ma

200 200 20F T T = − − =

Add the equations:

d the tension on the connecting cord:

Substitute to find T: N

248 25.6 200 28 4.5 / a a m s− − + = ⇒ =

(c) Fin

200 10(4.5) 110T T − = ⇒ =

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12.Two boxes are attached b light cord that runs over a frictionlesspulley. The mass of the block on the 60 o incline is M= 8 kg, and themass of the block on the 45 o incline is m=2.0 kg.(a) What is the acceleration of the blocks once released?

g

First draw the free body diagrams to visualize the forces involved.

sin 60

sin 45

sin 60 sin 45

oT

oT

o o

M F Ma

F mg ma

Mg mg Ma ma

− =− =

− = +

sin 60 sin 45o o M m−a g

M m=

+

2 2(8 )sin 60 (2 )sin 45(9.80 / ) 5.4 /

8 2

o okg kgm s m s

kg kg

−= =+

(b) What is the total force on the cord? We can use either of the above equations:

sin 45 oT F mg ma= + s kg m s N = + =

From the following diagram:

)i ( cos30 ) j ( cos 45 )jo o oT T F F

2 2(2 )(9.80 / ) sin 45 (2 )(5.4 / ) 24.7okg m

ˆ( sin 30 )i ( sin 45oT T F F F ˆ ˆ⎡ ⎤ ⎡ + − + −= − +⎣ ⎦ ⎣ ˆ ˆ( sin 30 sin 45 )i ( cos30 cos 45 ) jo o o o

T F ⎡ ⎤= − + + − −⎣ ⎦

2 2( sin 30 sin 45 ) ( cos30 cos 45 )o o oT F F = − + + − − o

(24.7 ) 2.518 39 N N = =

Example Problems – Roller-coaster

13. A roller coaster lifts a 700 kg car to point A 90 m high aspoint A, rolls with negligible friction down the incline thr loop is 50 m high.

shown on the diagram below. The cart starts from rest atough a loop of radius 20 m. Point B, the highest point on the

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(a) Locate the point S where the car reaches its maximum speed?(b) What is the maximum speed of the car

max

12 Amv mgy= max 2 Av g= y 2

max 2(9.80 / )(90 )v m s m 42 / m s= =

(c) What is the speed of the car at point B.

12 B B Amv mgy mgy+ = 2 ( ) B A B

22(9.80 / )(90 50 )v g y y= − Bv m s m m= −

hen the car is upside down in the loop what are the forces acting on it. Draw the vectors showing all vectors.

28 / m s=

(d) W

(e) What are the force of N F and mgF ?

N

2(700 )(9.80 / ) 6860mgF kg m s= =

2 B

N

mvF mg

r + =

2 2(700 )(28 / )6860

20 B

N

mv kg m sF mg

r m= − = − N

20,580 N =

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Inclined Planes Summary 06

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