*Notes to a video lecture on http://www.unizor.com*

__Motion on Inclined Plane__

*Problem A*

Consider a slope fixed on the ground that makes an angle

*φ*with horizon. Let's analyze the motion of an object, as it slides down a

slope under its own weight. Our task is to determine its acceleration

along the slope.

First of all, we have to choose a reference frame - a system of coordinates suitable for this problem.

The main force acting on this object is its

*weight*

**- a vector of**

*W**gravity*

directed vertically down to the ground. That prompts us to choose a

frame of reference with horizontal X-axis and vertical Y-axis. So, the

vector of weight has non-zero Y-component and zero X-component:

**= {**

*W***,**

*0***},**

*W***is a magnitude of vector of weight**

*W***. However, this obvious choice is not the best in a case like this.**

*W*There are usually numerous forces involved in an experiment and only few

objects. In this case there is only one object. So, instead of catering

to one particular force, like

*weight*, it's better to simplify the motion of a single object.

Much more convenient frame of reference would be the one, where our object in motion has only one non-zero coordinate.

Recall that there must be another force acting on an object -

*reaction force*of the slope, that prevents an object to go vertically down to the ground through a slope and forces it, in cooperation with

*gravity force*, to slide along a slope. This reaction force

*R*is always perpendicular to the surface, where an object is on (in this

case, perpendicular to a slope), and its magnitude is such that the

*resultant*of the weight

**and reaction**

*W***is directed along a slope.**

*R*Consider a frame of reference with X-axis going along the slope, where the object slides down and Y-axis perpendicular to it.

Granted, the weight now has both coordinates non-zero:

(a) perpendicular to a surface of a slope and (on the picture above) directed along negative Y-coordinates vector

**with magnitude**

*W*_{R}**, that causes reaction force**

*W*_{R}=W·cos(φ)*, that is equal in magnitude and opposite in direction to*

**R****, and**

*W*_{R}(b) parallel to a slope, directed (on the picture) towards negative X-coordinate, vector

**with magnitude**

*W*_{F}**, that is the cause of motion of our object down a slope.**

*W*_{F}=W·sin(φ)As it is pictured, both components of weight are negative since they are

directed towards negative direction of the X- and Y-axes:

**= {**

*W***,**

*−W·sin(φ)***}**

*−W·cos(φ)*Force

**, as opposite and equal in magnitude to**

*R***, is**

*W*_{R}**= {**

*R***,**

*0***}.**

*W·cos(φ)*The resultant of three vectors

**,**

*W*_{R}**and**

*W*_{F}**is**

*R***, directed down a slope, equaled in magnitude to**

*W*_{F}**.**

*W·sin(φ)*Therefore, an object of mass

*will slide down a slope with acceleration equaled in magnitude to*

**M**

**a = W**_{F}/M = W·cos(φ)/MSince weight and mass of an object are related as

*, where*

**W=M·g***is a known acceleration of free falling (9.8 m/sec² on the Earth ground), the resulting acceleration equals in magnitude to*

**g**

**a = M·g·sin(φ)/M = g·sin(φ)**In vector form in the chosen reference frame:

**= {**

*a**}*

**−g·sin(φ), 0***Problem B*

Consider an object A of mass

*, sliding without friction*

**m**on a slope of slide B, which itself lies on a horizontal surface and can

slide on it without friction. An angle of a slope of slide B is

**with horizon, its mass is**

*φ**.*

**M**Our task is to determine acceleration

*of slide B.*

**a**Let's analyze the motion of object A on a slope and the motion of a

slide B, as it moves horizontally, as a result of the weight of object A

on it.

As object A presses down with its weight, it slides downhill along a

slope of a slide B. At the same time slide B moves to the right (on the

picture above).

The horizontal component of the pressure

*of*

**W**_{R}an object A on slide B perpendicularly to its slope is the cause of the

motion of a slide B. However, this pressure is not the same as in the

problem A above. It will be less. Its opposite reaction force

*, that is equal in magnitude to*

**R***, but acting on the object A, will also be less than in the problem A above.*

**W**_{R}The

*resultant*of weight

*and reaction force*

**W***is force*

**R***that is not parallel to a slope, but tilted downwards.*

**W**_{F}So, the combination of object A sliding downhill on a slope and slide B

movement to the right produces the resultant move of object A that is

not parallel to a slope, neither it is directed vertically down, but

will be somewhere in-between.

Let's consider the same reference frame as in the problem A above. Now

both object A and slide B, as they move, have both X- and Y-components

not equal to zero.

Consider only Y-coordinate of the A object now and Y-components of forces acting on it.

In the direction of Y-axis the force

**=**

*W*_{F}**+**

*W*

*R***=**

*W*_{Fy}**+**

*W*_{y}**=**

*R*_{y}

**R − W·cos(φ)**This force

**R − W·cos(φ)***, multiplied by a Y-component of its acceleration, which so far is unknown.*

**m**Let

*be an acceleration of slide B in the direction of*

**a**its horizontal movement. Since displacement of object A in the direction

of the Y-axis (perpendicularly to a slope) equals to horizontal

displacement of a slide B multiplied by

*, the*

**sin(φ)**acceleration of slide B in the horizontal direction and acceleration of

the object A in a direction perpendicular to a slope, maintain the same

factor.

Therefore, the acceleration of object A perpendicularly to a slope of slide B equals to

*, where*

**a·sin(φ)***is the acceleration of the slide B that we have to determine in this problem.*

**a**The Newton's Second Law for object A in the direction of the Y-axis (perpendicular to a slope of slide B) is

**R − W·cos(φ) = −m·a·sin(φ)**(minus on the right because the acceleration of object A relative to Y-axis is negative).

This equation is the first in a system of two equations that include two unknowns

*and*

**R***.*

**a**On the other hand, a horizontal component of vector

**is the cause of horizontal acceleration**

*−R**of slide B that has mass*

**a***.*

**M**Therefore,

**R·sin(φ) = M·a**This is the second equation in a system of two equations that include two unknowns

*and*

**R***.*

**a**Solving this system as follows.

From the second equation:

**R = M·a/sin(φ)**Substitute it in the first equation:

**M·a/sin(φ) − W·cos(φ) =**

= −a·m·sin(φ)= −a·m·sin(φ)

The solution for horizontal acceleration of slide B is

*[*

**a = W·cos(φ) /**

//

*]*

**m·sin(φ) + M/sin(φ)***[*

**=**

= W·sin(φ)·cos(φ) /

/= W·sin(φ)·cos(φ) /

/

*]*

**m·sin²(φ)+M**So, our final result for an acceleration

*of slide B, as it moves horizontally, is*

**a***[*

**W·sin(φ)·cos(φ)/***]*

**m·sin²(φ)+M**
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