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

__Resultant Force__

According to the

*Principle of Superposition of Forces*, we can replace actions of many forces acting upon a point-object (each one being a vector) with a

**vector sum of these forces**(called

**resultant**force), getting exactly the same effect on the motion of a point-object.

An immediate consequence from the

*Principle of Superposition of Forces*

is that any single force acting upon a point-object can be replaced by a

vector sum of a few forces, which produce exactly the same effect on

the motion of a point-object, as long as this vector sum of forces

exactly equals to the vector of original force.

We will demonstrate this Principle on a few examples.

*Example 1*

In the frame of reference associated with the ground we have a point-object sliding down a slope.

Given are: mass of an object

**, force of gravitation of this object (its weight, directed vertically down)**

*m***and an angle**

*W***between a slope of a slide and horizontal ground.**

*φ*Our task is to determine the acceleration of this object on its

trajectory down a slope, considering there is no friction that slows

down the object's movement down a slope.

*Solution*

Trajectory of our object is a straight line (the slope), mass

*m*is known, so all we need to find the acceleration from the Newton's

Second Law is the force that pushes this object along its trajectory

down a slope.

Obviously, the slope's angle

**plays an important role here. Consider a slope with**

*φ***, that is the "slope" is not really a slope, but a horizontal plane. The object will not move then because its weight**

*φ=0***, directed vertically down perpendicularly to a horizontal plane, is completely balanced by the reaction force of the plane**

*W*

*R*directed vertically up and equal in magnitude to the weight, according

to the Newton's Third Law. The resultant force is zero-vector and there

is no acceleration.

On the other extreme, if our slide is vertical with

**, an object does not press on it, there is no reaction force, all the object's weight**

*φ=90°***is directed along the vertical trajectory downward, and the acceleration can be determined from the Newton's Second Law as**

*W***=**

*a***.**

*W/m*To meaningfully include the slope's angle

**into our calculation, we will represent its weight**

*φ***as a vector sum of two forces:**

*W*(a) the force

**, directed perpendicularly to a slope, causing slope's reaction force**

*W*_{R}**to balance it, and**

*R*(b) the force

**, directed parallel to a slope, pushing the object forward along its trajectory and causing its acceleration.**

*W*_{F}If we succeed in such representation, we will say that the force

**is the one and only force that pushes the object of mass**

*W*_{F}**down a slope and can be used to determine the object's acceleration from the Newton's Second Law as**

*m***=**

*a***.**

*W*_{F}/mThis representation of the weight of an object as a sum of two vectors

**, knowing the angle**

*W=W*_{R}+W_{F}**, is easy.**

*φ*The magnitude of vectors

**and**

*W*_{R}**, which we will denote as**

*W*_{F}**and**

*W*_{R}**correspondingly, equal to**

*W*_{F}

*W*_{R}= W·cos(φ)

*W*_{F}= W·sin(φ)where

**is the magnitude of a vector of weight**

*W***.**

*W*The force

*W*_{R}is completely balanced by the reaction force of the slope (they both

are perpendicular to a slope, equal in magnitude and oppositely

directed, according to the Newton's Third Law).

So, the only force affecting the motion of an object on its trajectory down a slope and acting along this trajectory is

**, the magnitude of which we have calculated above.**

*W*_{F}Therefore, an acceleration of an object down a slope will be a vector directed parallel to a slope with a magnitude

**=**

*a***.**

*W·sin(φ)/m**Example 2*

A motorcycle stuntman decides to demonstrate his skills by going inside a vertical loop.

His mass is

**.**

*m*His linear speed inside this loop is

**.**

*v*The magnitude of the gravity force (his weight with a motorcycle) is

**.**

*W*The radius of a loop is

**.**

*R*Find the pressure on the loop at its top and at its bottom from a moving motorcycle.

*Solution*

Let's set our frame of reference on the ground with Z-axis directed upward and a loop being in the XZ-plane.

Since an object of mass

**moves with a constant linear speed**

*m***along a circular trajectory of radius**

*v***,**

*R*there is a force that keeps it on this trajectory. This force is always

directed towards the center of a circle and is equal in magnitude to

**, where**

*m·a***is a magnitude of its acceleration towards a center of a circle called**

*a**centripetal acceleration*.

From the Kinematics of rotation we know that

**, where**

*a=R·ω²***is the angular speed of rotation and is related to linear speed of rotation**

*ω***as**

*v***.**

*v=R·ω*From this we derive the

*centripetal acceleration*

**in terms of linear speed and radius:**

*a***.**

*a=v²/R*Knowing the force that keeps an object on a circular trajectory, let's analyze what is the source of this force.

On one hand, there is always a weight directed downward (towards negative direction of Z-axis) and equal in magnitude to

**.**

*W*On the other hand, an object presses onto the loop with the force

*P(t)*perpendicularly to the loop, and the loop reacts back unto object with

the same by magnitude but oppositely directed reaction force. This

reaction force is directed upward, when the object is at the bottom of

the loop (we will denote its magnitude as

**), or downward, when an object is at the top (we will denote its magnitude as**

*P*_{b}**).**

*P*_{t}The combination of weight and the reaction force of the loop is the

resultant force that keeps the object on its circular trajectory.

Therefore, at the bottom of the loop, when weight is directed downward,

reaction is directed upward and the centripetal force is directed

upward,

*−W+P*_{b}= m·R·ω²At the top of the loop, when weight is directed downward, reaction is

directed downward and the centripetal force is directed downward,

*−W−P*_{t}= −m·R·ω²From the first equation we can derive the reaction force at the bottom

that is, by magnitude, equals to the pressure on a loop at this point:

*P*_{b}= m·R·ω² + WFrom the second equation we can derive the reaction force at the top

that is, by magnitude, equals to the pressure on a loop at this point:

*P*_{t}= m·R·ω² − WTheory of gravitation, that will be studied later in this course, states

that weight is a force of gravity proportional to a product of masses

of a planet and an object on its surface and inversely proportional to a

square of a distance between centers of mass. The latter is

approximately equal to a radius of a planet.

For Earth it is expressed as

**,**

*W = m·g***is an acceleration of the free falling on Earth, which is equal to**

*g***.**

*9.8 m/sec²*On Jupiter, by the way, this constant - free fall acceleration - is equal to

**.**

*25 m/sec²*Using this, we can write our formulas for pressure as follows:

*P*_{b}= m·R·ω² + m·g = m·(R·ω² + g)

*P*_{t}= m·R·ω² − m·g = m·(R·ω² − g)It should be noted that, while

**is always positive,**

*P*_{b}

*P*_{t}might be negative, if the speed of an object is not sufficient. If

negative, the movement will not be circular and object will fall down

from its trajectory.

If

**or**

*R·ω²=g***, or**

*ω=√g/R***,**

*v=√g·R*the object will experience weightlessness at the top position, since it

will not press onto loop and, therefore, loop will not press on it. For

loop of, say, 10 meters in diameter

**)**

*R=5 m*

*7 m/sec*will produce an effect of weightlessness at the top of a loop. This

linear speed with a given radius of a loop is equivalent to the angular

speed of

*1.4 rad/sec*Slower speed will result in falling from the circular trajectory.

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