Dynamics and kinematics are two important branches of physics that study the laws of movement of objects in space. The first examines the forces acting on the body, while the second deals directly with the characteristics of the dynamic process, without delving into the reasons for what caused it. Knowledge of these sections of physics must be applied to successfully solve problems of motion along an inclined plane. Consider this issue in the article.
The basic formula of dynamics
Of course, we are talking about the second law that Isaac Newton postulated in the XVII century, studying the mechanical motion of solids. We write it in mathematical form:
F¯ = m * a¯
The action of an external force F¯ causes the appearance of linear acceleration a¯ of a body with mass m. Both vector quantities (F¯ and a¯) are directed in the same direction. The force in the formula is the result of the action on the body of all the forces that are present in the system.
In the case of rotation motion, Newton’s second law is written as:
M = I * α
Here M and I are the moments of force and inertia, respectively, α is the angular acceleration.
Kinematics formulas
Solving problems on movement on an inclined plane requires knowledge of not only the main formula of dynamics, but also the corresponding expressions of kinematics. They connect acceleration, speed and the distance traveled in the equal. For uniformly accelerated (equally slow) rectilinear motion, the following formulas apply:
a = Δv / Δt;
v = v 0 ± a * t;
S = v 0 * t ± a * t 2/2
Here v 0 is the value of the initial velocity of the body, S is the path traveled over time t along a straight path. The sign “+” should be put if the speed of the body increases over time. Otherwise (equidistant motion), use the "-" sign in the formulas. This is an important point.
If the movement is carried out along a circular path (rotation around an axis), then the following formulas should be used:
α = Δω / Δt;
ω = ω 0 ± α * t;
θ = ω 0 * t ± α * t 2/2
Here, α and ω are the angular acceleration and velocity, respectively, θ is the angle of rotation of the rotating body over time t.
Linear and angular characteristics are interconnected by formulas:
a = α * r;
v = ω * r
Here r is the radius of rotation.
Inclined movement: forces
By this movement is meant the movement of an object along a flat surface that is inclined at a certain angle to the horizon. Examples include sliding a bar over a board or rolling a cylinder over a tilted metal sheet.
To determine the characteristics of the type of movement under consideration, it is necessary first of all to find all the forces that act on the body (bar, cylinder). They may be different. In general, these may be the following forces:
- severity
- support reactions;
- rolling and / or sliding friction ;
- thread tension;
- force of external traction.
The first three of them are always present. The existence of the latter two depends on the specific system of physical bodies.
To solve the problem of moving along an inclined plane, it is necessary to know not only the force modules, but also their direction of action. If the body rolls along the plane, the friction force is unknown. However, it is determined from the corresponding system of equations of motion.
Solution technique
Solving problems of this type begins with determining the forces and their directions of action. For this, gravity is primarily considered. It should be decomposed into two components of the vector. One of them should be directed along the surface of the inclined plane, and the second should be perpendicular to it. The first component of gravity, in the case of a body moving downward, provides its linear acceleration. It happens anyway. The second is equal to the reaction force of the support. All these indicators can have various parameters.
The force of friction when moving along an inclined plane is always directed against the movement of the body. When it comes to gliding, the calculations are pretty simple. To do this, use the formula:
F f = µ * N
Where N is the reaction of the support, µ is the coefficient of friction, which does not have a dimension.
If the system contains only these three forces, then their resulting along an inclined plane will be equal to:
F = m * g * sin (φ) - μ * m * g * cos (φ) = m * g * (sin (φ) - μ * cos (φ)) = m * a
Here φ is the angle of inclination of the plane to the horizon.
Knowing the force F, one can determine the linear acceleration a according to Newton’s law. The latter, in turn, is used to determine the speed of movement along an inclined plane through a known period of time and the distance traveled by the body. If you delve into, you can understand that everything is not so complicated.
In the case when the body rolls down an inclined plane without slipping, the total force F will be equal to:
F = m * g * sin (φ) - F r = m * a
Where F r - rolling friction force. She is unknown. When the body rolls, gravity does not create a moment, since it is applied to the axis of rotation. In turn, F r creates the following moment:
M = F r * r = I * α
Given that we have two equations and two unknowns (α and a are related to each other), we can easily solve this system, and hence the problem.
Now we will consider how to use the described technique in solving specific problems.
The task of moving the bar along an inclined plane
A wooden block is located in the upper part of the inclined plane. It is known that it has a length of 1 meter and is located at an angle of 45 o . It is necessary to calculate how long the block will lower along this plane as a result of sliding. The coefficient of friction should be taken equal to 0.4.
We write down Newton's law for a given physical system and calculate the value of linear acceleration:
m * g * (sin (φ) - μ * cos (φ)) = m * a =>
a = g * (sin (φ) - μ * cos (φ)) ≈ 4.162 m / s 2
Since we know the distance that the bar must pass, we can write the following formula for the path with uniformly accelerated movement without an initial speed:
S = a * t 2/2
Where should the time be expressed, and substitute known values:
t = √ (2 * S / a) = √ (2 * 1 / 4,162) ≈ 0.7 s
Thus, the travel time on the inclined plane of the bar will be less than a second. Note that the result obtained does not depend on body weight.
The problem with a cylinder rolling down a plane
A cylinder with a radius of 20 cm and a mass of 1 kg is placed on a plane inclined at an angle of 30 o . It is necessary to calculate its maximum linear speed, which he will gain when rolling from a plane, if its length is 1.5 meters.
We write the corresponding equations:
m * g * sin (φ) - F r = m * a;
F r * r = I * α = I * a / r
The moment of inertia of cylinder I is calculated by the formula:
I = 1/2 * m * r 2
We substitute this value into the second formula, express from it the friction force F r and replace it with the expression obtained in the first equation, we have:
F r * r = 1/2 * m * r 2 * a / r =>
F r = 1/2 * m * a;
m * g * sin (φ) - 1/2 * m * a = m * a =>
a = 2/3 * g * sin (φ)
We have found that linear acceleration does not depend on the radius and mass of the body rolling off the plane.
Knowing that the length of the plane is 1.5 meters, we find the time of movement of the body:
S = a * t 2/2 =>
t = √ (2 * S / a)
Then the maximum speed on the inclined plane of the cylinder will be equal to:
v = a * t = a * √ (2 * S / a) = √ (2 * S * a) = √ (4/3 * S * g * sin (φ))
We substitute all the quantities known from the problem condition into the final formula, we get the answer: v ≈ 3.132 m / s.