Movement on an inclined plane formula. How do inclined planes work? Basic formula of dynamics
On the surface of the Earth gravity (gravity) is constant and equal to the product of the mass of the falling body and the acceleration of gravity: F g = mg
It should be noted that the acceleration of free fall is a constant value: g=9.8 m/s 2 , and is directed towards the center of the Earth. Based on this, we can say that bodies with different masses will fall to Earth equally quickly. How so? If you throw a piece of cotton wool and a brick from the same height, the latter will make its way to the ground faster. Don't forget about air resistance! For cotton wool it will be significant, since its density is very low. In an airless space, brick and wool will fall simultaneously.
The ball moves along an inclined plane 10 meters long, the angle of inclination of the plane is 30°. What will be the speed of the ball at the end of the plane?
The ball is affected only by the force of gravity Fg, directed downward perpendicular to the base of the plane. Under the influence of this force (component directed along the surface of the plane), the ball will move. What will be the component of gravity acting along the inclined plane?
To determine the component, it is necessary to know the angle between the force vector F g and the inclined plane.
Determining the angle is quite simple:
- the sum of the angles of any triangle is 180°;
- the angle between the force vector F g and the base of the inclined plane is 90°;
- the angle between the inclined plane and its base is α
Based on the above, the desired angle will be equal to: 180° - 90° - α = 90° - α
From trigonometry:
F g slope = F g cos(90°-α)
Sinα = cos(90°-α)
F g slope = F g sinα
It really is like this:
- at α=90° (vertical plane) F g tilt = F g
- at α=0° (horizontal plane) F g tilt = 0
Let's determine the acceleration of the ball from the well-known formula:
F g sinα = m a
A = F g sinα/m
A = m g sinα/m = g sinα
The acceleration of a ball along an inclined plane does not depend on the mass of the ball, but only on the angle of inclination of the plane.
Determine the speed of the ball at the end of the plane:
V 1 2 - V 0 2 = 2 a s
(V 0 =0) - the ball begins to move from place
V 1 2 = √2·a·s
V = 2 g sinα S = √2 9.8 0.5 10 = √98 = 10 m/s
Pay attention to the formula! The speed of the body at the end of the inclined plane will depend only on the angle of inclination of the plane and its length.
In our case, a billiard ball, a passenger car, a dump truck, and a schoolboy on a sled will have a speed of 10 m/s at the end of the plane. Of course, we don't take friction into account.
Dynamics and kinematics are two important branches of physics that study the laws of movement of objects in space. The first considers 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 branches of physics must be used to successfully solve problems involving motion on an inclined plane. Let's look at this issue in the article.
Basic formula of dynamics
Of course, we are talking about the second law, which was postulated by Isaac Newton in the 17th century while studying the mechanical motion of solid bodies. Let's write it in mathematical form:
The action of an external force F¯ causes the appearance of linear acceleration a¯ in 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 rotational motion, Newton's second law is written as:
Here M and I are inertia, respectively, α is angular acceleration.
Kinematics formulas
Solving problems involving motion 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 distance traveled into equalities. For uniformly accelerated (uniformly decelerated) rectilinear motion, the following formulas are used:
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 along a straight path during time t. A "+" sign should be added if the speed of the body increases over time. Otherwise (uniformly slow motion), the “-” sign should be used 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:
ω = ω 0 ± α*t;
θ = ω 0 *t ± α*t 2 /2
Here α and ω are the speed, respectively, θ is the angle of rotation of the rotating body during time t.
Linear and angular characteristics are related to each other by the formulas:
Here r is the radius of rotation.
Movement on an inclined plane: forces
This movement is understood as the movement of an object along a flat surface that is inclined at a certain angle to the horizon. Examples include a block sliding on a board or a cylinder rolling on an inclined sheet of metal.
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 can be the following forces:
- heaviness;
- support reactions;
- and/or slipping;
- thread tension;
- external traction force.
The first three of them are always present. The existence of the last two depends on the specific system of physical bodies.
To solve problems involving movement along an inclined plane, it is necessary to know not only the force modules, but also their directions of action. If a body rolls down a plane, the friction force is unknown. However, it is determined from the corresponding system of equations of motion.
Solution method
Solving problems of this type begins with determining the forces and their directions of action. To do this, the force of gravity is first considered. It should be decomposed into two component vectors. 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 downwards, provides its linear acceleration. This happens anyway. The second is equal to All these indicators can have different parameters.
The friction force when moving along an inclined plane is always directed against the movement of the body. When it comes to sliding, the calculations are quite simple. To do this, use the formula:
Where N is the support reaction, µ is the friction coefficient, which has no dimension.
If only these three forces are present in the system, then their resultant along the 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, we can use Newton's law to determine the linear acceleration a. The latter, in turn, is used to determine the speed of movement along an inclined plane after a known period of time and the distance traveled by the body. If you look into it, you can understand that everything is not so complicated.
In the case when a 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 - It is unknown. When a body rolls, the force of gravity does not create a moment, since it is applied to the axis of rotation. In turn, F r creates the following moment:
Considering that we have two equations and two unknowns (α and a are related to each other), we can easily solve this system, and therefore the problem.
Now let's look at how to use the described technique to solve specific problems.
Problem involving the movement of a block on an inclined plane
The wooden block is at the top 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 it will take for the block to descend along this plane as a result of sliding. Take the friction coefficient equal to 0.4.
We write 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 block must travel, we can write the following formula for the path during uniformly accelerated motion without an initial speed:
Where should we express time and substitute known values:
t = √(2*S/a) = √(2*1/4.162) ≈ 0.7 s
Thus, the time it takes to move along the inclined plane of the block will be less than a second. Note that the result obtained does not depend on body weight.
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. You should calculate its maximum linear speed that it will gain when rolling down a plane if its length is 1.5 meters.
Let's 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:
Let's substitute this value into the second formula, express the friction force F r from it and replace it with the resulting expression in the first equation, we have:
F r *r = 1/2*m*r 2 *a/r = >
m*g*sin(φ) - 1/2*m*a = m*a =>
a = 2/3*g*sin(φ)
We 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:
Then the maximum speed of movement along 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 conditions into the final formula, and we get the answer: v ≈ 3.132 m/s.
This article talks about how to solve problems about moving along an inclined plane. A detailed solution to the problem of the motion of coupled bodies on an inclined plane from the Unified State Examination in Physics is considered.
Solving the problem of motion on an inclined plane
Before moving directly to solving the problem, as a tutor in mathematics and physics, I recommend carefully analyzing its condition. You need to start by depicting the forces that act on connected bodies:
Here and are the thread tension forces acting on the left and right bodies, respectively, are the support reaction force acting on the left body, and are the gravity forces acting on the left and right bodies, respectively. Everything is clear about the direction of these forces. The tension force is directed along the thread, the gravity force is vertically downward, and the support reaction force is perpendicular to the inclined plane.
But the direction of the friction force will have to be dealt with separately. Therefore, in the figure it is shown as a dotted line and signed with a question mark. It is intuitively clear that if the right load “outweighs” the left one, then the friction force will be directed opposite to the vector. On the contrary, if the left load “outweighs” the right one, then the friction force will be co-directed with the vector.
The right weight is pulled down by force N. Here we took the acceleration of gravity m/s 2. The left load is also pulled down by gravity, but not all of it, but only a “part” of it, since the load lies on an inclined plane. This “part” is equal to the projection of gravity onto an inclined plane, that is, a leg in a right triangle shown in the figure, that is, equal to N.
That is, the right load still “outweighs”. Consequently, the friction force is directed as shown in the figure (we drew it from the center of mass of the body, which is possible in the case when the body can be modeled by a material point):
The second important question that needs to be addressed is whether this coupled system will move at all? What if it turns out that the friction force between the left load and the inclined plane will be so great that it will not allow it to move?
This situation will be possible in the case when the maximum friction force, the modulus of which is determined by the formula (here - the coefficient of friction between the load and the inclined plane - the support reaction force acting on the load from the inclined plane), turns out to be greater than the force that is trying to bring the system into motion. That is, that very “outweighing” force that is equal to N.
The modulus of the support reaction force is equal to the length of the leg in the triangle according to Newton’s 3rd law (with the same magnitude of force the load presses on the inclined plane, with the same magnitude of force the inclined plane acts on the load). That is, the support reaction force is equal to N. Then the maximum value of the friction force is N, which is less than the value of the “overweighing force”.
Consequently, the system will move, and move with acceleration. Let us depict in the figure these accelerations and coordinate axes, which we will need later when solving the problem:
Now, after a thorough analysis of the problem conditions, we are ready to begin solving it.
Let's write down Newton's 2nd law for the left body:
And in the projection onto the axes of the coordinate system we get:
Here, projections are taken with a minus, the vectors of which are directed opposite the direction of the corresponding coordinate axis. Projections whose vectors are aligned with the corresponding coordinate axis are taken with a plus.
Once again we will explain in detail how to find projections and . To do this, consider the right triangle shown in the figure. In this triangle 
And 
. It is also known that in this right triangle . Then and.
The acceleration vector lies entirely on the axis, and therefore . As we already mentioned above, by definition, the modulus of the friction force is equal to the product of the friction coefficient and the modulus of the support reaction force. Hence, . Then the original system of equations takes the form:
Let us now write down Newton’s 2nd law for the right body:
In projection onto the axis we get.
A mass of 26 kg lies on an inclined plane 13 m long and 5 m high. The friction coefficient is 0.5. What force must be applied to the load along the plane in order to pull the load? to steal the load
SOLUTION
What force must be applied to lift a trolley weighing 600 kg along an overpass with an inclination angle of 20°, if the coefficient of resistance to movement is 0.05
SOLUTION
During laboratory work, the following data were obtained: the length of the inclined plane is 1 m, the height is 20 cm, the mass of the wooden block is 200 g, the traction force when the block moves upward is 1 N. Find the coefficient of friction
SOLUTION
A block of mass 2 kg rests on an inclined plane 50 cm long and 10 cm high. Using a dynamometer located parallel to the plane, the block was first pulled up the inclined plane and then pulled down. Find the difference in dynamometer readings
SOLUTION
To hold the cart on an inclined plane with an angle of inclination α, it is necessary to apply a force F1 directed upward along the inclined plane, and to lift it upward, it is necessary to apply a force F2. Find drag coefficient
SOLUTION
The inclined plane is located at an angle α = 30° to the horizontal. At what values of the friction coefficient μ is it more difficult to pull a load along it than to lift it vertically?
SOLUTION
There is a mass of 50 kg on an inclined plane 5 m long and 3 m high. What force directed along the plane must be applied to hold this load? pull up evenly? pull with an acceleration of 1 m/s2? Friction coefficient 0.2
SOLUTION
A car weighing 4 tons moves uphill with an acceleration of 0.2 m/s2. Find the traction force if the slope is 0.02 and the drag coefficient is 0.04
SOLUTION
A train weighing 3000 tons moves down a slope of 0.003. The coefficient of resistance to movement is 0.008. With what acceleration does the train move if the traction force of the locomotive is: a) 300 kN; b) 150 kN; c) 90 kN
SOLUTION
A motorcycle weighing 300 kg began to move from rest on a horizontal section of road. Then the road went downhill, equal to 0.02. What speed did the motorcycle acquire 10 seconds after it started moving, if it covered a horizontal section of the road in half this time? The traction force and the coefficient of resistance to movement are constant throughout the entire path and are respectively equal to 180 N and 0.04
SOLUTION
A block of mass 2 kg is placed on an inclined plane with an angle of inclination of 30°. What force, directed horizontally (Fig. 39), must be applied to the block so that it moves uniformly along the inclined plane? The coefficient of friction between the block and the inclined plane is 0.3
SOLUTION
Place a small object (rubber band, coin, etc.) on the ruler. Gradually lift the end of the ruler until the object begins to slide. Measure the height h and base b of the resulting inclined plane and calculate the coefficient of friction
SOLUTION
With what acceleration a does a block slide along an inclined plane with an inclination angle α = 30° with a friction coefficient μ = 0.2
SOLUTION
At the moment the first body began to fall freely from a certain height h, the second body began to slide without friction from an inclined plane having the same height h and length l = nh. Compare the final velocities of the bodies at the base of the inclined plane and the time of their movement.