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.
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
When conducting 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.
The movement of a body along an inclined plane is a classic example of the movement of a body under the influence of several non-directional forces. Standard method solving problems of this kind of motion consists in decomposing the vectors of all forces into components directed along the coordinate axes. Such components are linearly independent. This allows us to write Newton's second law for components along each axis separately. Thus, Newton's second law, which is a vector equation, turns into a system of two (three in the three-dimensional case) algebraic equations.
The forces acting on the block are
case of accelerated downward movement
Consider a body that is sliding down an inclined plane. In this case, the following forces act on it:
- Gravity m g , directed vertically downwards;
- Ground reaction force N , directed perpendicular to the plane;
- Sliding friction force F tr, directed opposite to the speed (up along the inclined plane when the body slides)
When solving problems in which an inclined plane appears, it is often convenient to introduce an inclined coordinate system, the OX axis of which is directed downward along the plane. This is convenient, because in this case you will have to decompose only one vector into components - the gravity vector m g
, and the friction force vector F
tr and ground reaction forces N
already directed along the axes. With this expansion, the x-component of gravity is equal to mg sin( α
) and corresponds to the “pulling force” responsible for the accelerated downward movement, and the y-component is mg cos( α
) = N balances the ground reaction force, since there is no body movement along the OY axis.
Sliding friction force F tr = µN proportional to the ground reaction force. This allows us to obtain the following expression for the friction force: F tr = µmg cos( α
). This force is opposite to the "pulling" component of gravity. Therefore for body sliding down
, we obtain expressions for the total resultant force and acceleration:
F x = mg(sin( α
) – µ
cos( α
));
a x = g(sin( α
) – µ
cos( α
)).
It's not hard to see what if µ < tg(α ), then the expression has positive sign and we are dealing with uniformly accelerated motion down an inclined plane. If µ >tg( α ), then the acceleration will have negative sign and the movement will be equally slow. Such movement is possible only if the body is given an initial speed down the slope. In this case, the body will gradually stop. If provided µ >tg( α ) the object is initially at rest, it will not begin to slide down. Here the static friction force will completely compensate for the “pulling” component of gravity.
When the friction coefficient is exactly equal to the tangent of the angle of inclination of the plane: µ = tg( α ), we are dealing with mutual compensation of all three forces. In this case, according to Newton's first law, the body can either be at rest or move with constant speed(Wherein uniform motion only possible downwards).
The forces acting on the block are
sliding on an inclined plane:
case of slow motion upward
However, the body can also drive up an inclined plane. An example of such motion is the movement of a hockey puck up an ice slide. When a body moves upward, both the frictional force and the “pulling” component of gravity are directed downward along the inclined plane. In this case, we are always dealing with uniformly slow motion, since the total force is directed in the direction opposite to the speed. The expression for acceleration for this situation is obtained in a similar way and differs only in sign. So for body sliding up an inclined plane , we have.
Dynamics is one of the important branches of physics, which studies the reasons for the movement of bodies in space. In this article, we will consider from a theoretical point of view one of the typical problems of dynamics - the movement of a body along an inclined plane, and also give examples of solutions to some practical problems.
Basic formula of dynamics
Before moving on to studying the physics of body motion along an inclined plane, we present the necessary theoretical information for solving this problem.
In the 17th century, Isaac Newton, thanks to practical observations of the motion of macroscopic surrounding bodies, derived three laws that currently bear his name. All classical mechanics is based on these laws. We are interested in this article only in the second law. Its mathematical form is given below:
The formula says that the action external force F¯ will give acceleration a¯ to a body of mass m. We will further use this simple expression to solve problems of body motion along an inclined plane.
Note that force and acceleration are vector quantities directed in the same direction. In addition, force is an additive characteristic, that is, in the above formula, F¯ can be considered as the resulting effect on the body.
Inclined plane and forces acting on the body located on it
The key point on which the success of solving problems of body motion along an inclined plane depends is the determination of the forces acting on the body. The definition of forces is understood as knowledge of their modules and directions of action.
Below is a drawing that shows that a body (car) is at rest on a plane inclined at an angle to the horizontal. What forces are acting on it?
The list below lists these forces:
- heaviness;
- support reactions;
- friction;
- thread tension (if present).
Gravity
First of all, this is the force of gravity (F g). It is directed vertically downwards. Since the body has the ability to move only along the surface of the plane, when solving problems, the force of gravity is decomposed into two mutually perpendicular components. One of the components is directed along the plane, the other is perpendicular to it. Only the first of them leads to the appearance of acceleration in the body and, in fact, is the only driving factor for the body in question. The second component determines the occurrence of the support reaction force.
Ground reaction
The second force acting on the body is the ground reaction (N). The reason for its appearance is related to Newton's third law. The value N shows the force with which the plane acts on the body. It is directed upward perpendicular to the inclined plane. If the body were on a horizontal surface, then N would be equal to its weight. In the case under consideration, N is equal only to the second component obtained from the expansion of gravity (see paragraph above).
The support reaction does not provide direct impact on the nature of the movement of the body, since it is perpendicular to the plane of inclination. Nevertheless, it causes friction between the body and the surface of the plane.
Friction force
The third force that should be taken into account when studying the movement of a body on an inclined plane is friction (F f). The physical nature of friction is complex. Its appearance is associated with microscopic interactions of contacting bodies having inhomogeneous contact surfaces. There are three types of this force:
- peace;
- slip;
- rolling.
Static and sliding friction are described by the same formula:
where µ is a dimensionless coefficient, the value of which is determined by the materials of the rubbing bodies. So, with sliding friction of wood on wood, µ = 0.4, and ice on ice - 0.03. The coefficient for static friction is always greater than that for sliding.
Rolling friction is described using a formula different from the previous one. It looks like:
Here r is the radius of the wheel, f is a coefficient having the dimension of the inverse length. This frictional force is usually much less than the previous ones. Note that its value is affected by the radius of the wheel.
The force F f, whatever its type, is always directed against the movement of the body, that is, F f tends to stop the body.
Thread tension
When solving problems of body motion on an inclined plane, this force is not always present. Its appearance is determined by the fact that a body located on an inclined plane is connected with another body using an inextensible thread. Often the second body hangs by a thread through a block outside the plane.
On an object located on a plane, the tension force of the thread acts either accelerating it or slowing it down. Everything depends on the magnitude of the forces acting in the physical system.
The appearance of this force in the problem significantly complicates the solution process, since it is necessary to simultaneously consider the movement of two bodies (on the plane and hanging).
Problem of determining the critical angle
Now the time has come to apply the described theory to solve real problems of movement along an inclined plane of a body.
Let's assume that a wooden beam has a mass of 2 kg. It is on a wooden plane. It is necessary to determine at what critical angle of inclination of the plane the beam will begin to slide along it.
Sliding of the beam will occur only when the total force acting downward along the plane on it is greater than zero. Thus, to solve this problem, it is enough to determine the resulting force and find the angle at which it becomes greater than zero. According to the conditions of the problem, only two forces will act on the beam along the plane:
- gravity component F g1 ;
- static friction F f .
For a body to begin sliding, the following condition must be met:
Note that if the component of gravity exceeds the static friction, then it will also be greater than the sliding friction force, that is, the movement that has begun will continue with constant acceleration.
The figure below shows the directions of all the acting forces.
Let us denote the critical angle by the symbol θ. It is easy to show that the forces F g1 and F f will be equal:
F g1 = m × g × sin(θ);
F f = µ × m × g × cos(θ).
Here m × g is the weight of the body, µ is the coefficient of static friction force for the wood-wood pair of materials. From the corresponding table of coefficients you can find that it is equal to 0.7.
Substituting the found values into the inequality, we get:
m × g × sin(θ) ≥ µ × m × g × cos(θ).
Transforming this equality, we arrive at the condition for body motion:
tan(θ) ≥ µ =>
θ ≥ arctan(µ).
We got a very interesting result. It turns out that the meaning critical angleθ does not depend on the mass of the body on the inclined plane, but is uniquely determined by the coefficient of static friction µ. Substituting its value into the inequality, we obtain the value of the critical angle:
θ ≥ arctan(0.7) ≈ 35 o .
The task of determining acceleration when moving along an inclined plane of a body
Now let's solve a slightly different problem. Let there be a wooden beam on a glass inclined plane. The plane is inclined at an angle of 45 o to the horizon. It is necessary to determine with what acceleration the body will move if its mass is 1 kg.
Let us write down the main equation of dynamics for this case. Since the force F g1 will be directed along the movement, and F f against it, the equation will take the form:
F g1 - F f = m × a.
We substitute the formulas obtained in the previous problem for the forces F g1 and F f, we have:
m × g × sin(θ) - µ × m × g × cos(θ) = m × a.
Where do we get the formula for acceleration:
a = g × (sin(θ) - µ × cos(θ)).
Once again we have a formula that does not include body weight. This fact means that blocks of any mass will slide down an inclined plane at the same time.
Considering that the coefficient µ for rubbing materials wood-glass is 0.2, we substitute all the parameters into the equality and get the answer:
Thus, the technique for solving problems with an inclined plane is to determine the resultant force acting on the body and then apply Newton's second law.
Physics: body movement on an inclined plane. Examples of solutions and problems - all interesting facts and achievements of science and education on the site
Similar to a lever, inclined planes reduce the force required to lift bodies. For example, it is quite difficult to lift a concrete block weighing 45 kilograms with your hands, but dragging it up an inclined plane is quite possible. The weight of a body placed on an inclined plane is decomposed into two components, one of which is parallel and the other perpendicular to its surface. To move a block up an inclined plane, a person must overcome only the parallel component, the magnitude of which increases with increasing angle of inclination of the plane.
Inclined planes are very diverse in design. For example, a screw consists of an inclined plane (thread) that spirals around its cylindrical part. When a screw is screwed into a part, its thread penetrates into the body of the part, forming a very strong connection due to the high friction between the part and the threads. The vice transforms the action of the lever and rotational movement screw into a linear compressive force. The jack used to lift heavy loads works on the same principle.
Forces on an inclined plane
For a body located on an inclined plane, the force of gravity acts parallel and perpendicular to its surface. To move a body up an inclined plane, a force is required equal in magnitude to the component of gravity parallel to the surface of the plane.
Inclined planes and screws
The relationship between the screw and the inclined plane can be easily traced if you wrap a diagonally cut sheet of paper around the cylinder. The resulting spiral is identical in location to the screw thread.
Forces acting on the propeller
When a screw is turned, its thread creates a very large force applied to the material of the part into which it is screwed. This force pulls the propeller forward if it is turned clockwise and backward if it is turned counterclockwise.
Weight Lifting Screw
The rotating screws of jacks generate enormous force, allowing them to lift objects as heavy as cars or trucks. By turning the central screw with a lever, the two ends of the jack are pulled together, producing the necessary lift.
Inclined planes for splitting
The wedge consists of two inclined planes connected by their bases. When driving a wedge into a tree, the inclined planes develop lateral forces sufficient to split the strongest lumber.
Strength and work
Although an inclined plane may make a task easier, it does not reduce the amount of work required to complete it. Lifting a concrete block weighing 45 kg (W) 9 meters vertically upward (far picture on the right) requires 45 x 9 kilograms of work, which corresponds to the product of the weight of the block and the amount of movement. When the block is on a 44.5° inclined plane, the force (F) required to pull the block in is reduced to 70 percent of its weight. Although this makes it easier to move the block, now, in order to raise the block to a height of 9 meters, it must be dragged along a plane of 13 meters. In other words, the gain in strength is equal to the height of the lift (9 meters) divided by the length of movement along the inclined plane (13 meters).
