Mechanics

Inclined plane

Inclined plane is the part of Dynamics that deals with movement on surfaces that form an angle with the horizontal axis.

Representation of the different inclinations of a plane depending on the angle it makes with the horizontal axis.

inclined plane is the part of Newtonian dynamics that studies the motion of objects inclined on surfaces that form an angle with the horizontal axis, with or without friction . We can find it on ramps, stairs, knives, that is, anything that generates an inclined movement.

To calculate it, we use the Newton’s second law , which determines that the resultant force acting on the block is equal to the product of mass and acceleration. We also use the weight force formula. If there is friction acting on the block, we use the friction force formula.

Inclined plane summary

  • Inclined plane is the term used to designate situations in Dynamic Physics in which movements occur on planes with inclinations.
  • There are two cases of inclined plane exercises: when there is or when there is no friction force acting on the moving object.
  • To calculate the inclined plane, the most used formula is Newton’s second law, which determines that the resultant force is the mass multiplied by the acceleration.
  • The resultant force is the sum or subtraction of the forces acting on the body. To find out its value, we use various formulas, such as the weight force or the friction force.

What is an inclined plane?

An inclined plane is any surface that makes an angle , other than zero, with the horizon. , forming a triangular geometric figure. Some examples of inclined planes are ramps, moving walks, stairs, and anything else that is inclined.

Types of inclined plane

There are two cases in which we work with the inclined plane: when the object is not rubbing against the surface and when it is. We know which case it is according to the information in the statement.

  • frictionless inclined plane

In the case of the inclined plane without the friction forcefat tfat→, the forces present in the block are always the normal force No⃗ N→, which is perpendicular to the surface of the plane, and the weight forceP⃗ P→, which points towards the center of the Earth (in most cases, downwards). If there is something pulling the block, there will also be the pulling force. F⃗ F→. In the image, we can see the forces acting on the plane contained in the frictionless inclined plane.

  • Inclined plane with friction

In the case of the inclined plane with the friction forcefat tfat→, this force acts close to the ground following the movement, as if it were dragging the ground.

In addition, there is the normal forceNo⃗ N→, the weight forceP⃗ P→  and if there is something pulling the block, there will also be the pulling forceF⃗ F→. In the image, we can see the forces acting on the plane contained in the inclined plane with friction.

Decomposition of forces in inclined plane exercises

In inclined plane exercises, it is common for us to use a mathematical device called force decomposition, which is nothing more than “breaking” the vector in two directions: vertically, which is the coordinateandy, and horizontally, which is the coordinatexx. That is, the strengthP⃗ P→ will turnPxPx→ andPandPy→.

Resulting force on the inclined plane

Newton ‘s second law is used to calculate the resultant force on a block, which is the sum or subtraction of forces acting in the same direction on the block .

The sum (forces in the same direction) or subtraction (forces in opposite directions) of the forces will be determined by the exercise model. For example, there may be cases where the weight force is added to the friction force, while in others they may be subtracted.

It is worth mentioning that direction refers to right and left, and direction refers to vertical, horizontal or diagonal.

inclined plane formulas

The formulas used in the inclined plane exercises are Newton’s second law (Fundamental Principle of Dynamics), which states that the resultant force is calculated using the formulas for the weight force , the weight force decomposition formula and, in some cases, the frictional force.

  • Newton’s second law

FRThe⃗ FR→=m∙a→

    • FRFR→ is the resultant force on the block. It is the sum (if the forces are oriented to the same side) or subtraction (if the forces are on opposite sides) of all horizontal or vertical forces. It is measured in newton [N].
    • mm is the mass of the object, measured in kilograms [kg].
    • The⃗ a→ is the acceleration of the object, measured in meters per second squared/stwo][m/s2].
  • strength weight

P⃗  g⃗ P→= m∙g→

    • P⃗ P→ is the weight force, measured in newtons [N].
    • mm is the mass of the object, measured in kilograms [kg].
    • g⃗ g→ is the acceleration due to gravity, measured in meters per second squared/stwo][m/s2].
  • Decomposition of weight force

PxP∙ and θ Px→=P∙sen θ

PandP∙ θ Py→=P∙cos θ

  • PxPx→ is the horizontal coordinate of the weight force, measured in newton [N].
  • PandPy→ is the vertical coordinate of the weight force, measured in newton [N].
  • P  is the magnitude of the weight force, measured in newton [N].
  • and θ sen θ is the sine of the angle of inclination of the plane, measured in degrees.
  • θ cos θ is the cosine of the angle of inclination of the plane, measured in degrees.
  • Frictional force

fat tμ No⃗ fat→=μ∙N→

    • fat tfat→ is the friction force, measured in newtons [N].
    • μμ is the coefficient of friction . It can be static, when the block is on the verge of movement (almost moving), or kinetic, when the block is moving. It has no unit of measurement.
    • No⃗ N→ is the normal force, measured in newtons [N].

Note : If the object’s acceleration is not given, we can calculate it using the MUV formulas .

Exercises on an inclined plane

Question 01

(Unimep-SP) A block of mass 5 kg is dragged along a frictionless inclined plane.

For the block to acquire an acceleration of/stwo3m/s2up, the intensity ofF⃗ F→shall be:

(g10  _ g=10 m/s²,and θ 8 sen θ=0,8 andθ 6 cos θ=0,6)

a) equal to the weight of the block

b) less than the weight of the block

c) equal to the reaction of the plane

d) equal to 55N

e) equal to 10N

Resolution

Alternative D 

First, we will find the forces that are acting on the block. In this case, there is the normal forceNo⃗ N→, which is perpendicular to the surface but not relevant to the calculation, there is the pulling forceF⃗ F→ and the weight forceP⃗ P→. The weight force must be decomposed into two:PxPx→ horizontally andPandPy→ vertically, following the Cartesian plane.

Using Newton’s second law, which says that the sum of all the forces involved is equal to the mass multiplied by the acceleration, we obtain the expression:

F⃗ PxThe⃗ F→−Px→=m∙a→

Using the weight force decomposition formula, we havePxP∙ and θ Px→=P∙sen θ. Then:

F⃗ − P∙ and θ  The⃗ F→−P∙sen θ=m∙a→

F⃗ − ∙ g∙ and θ  The⃗ F→−m∙g∙sen θ=m∙a→

Replacing the data:  

  • gravity: 10 m/s²
  • acceleration: 3 m/s²
  • mass: 5 kg
  • without θ 8θ=0,8

F⃗ − ∙ 10 ∙ ∙ 3F→−5∙10∙0,8=5∙3

F⃗ − 40 15F→−40=15

F⃗ 55 N F→=55 N

question 2

(Uerj — adapted) A block of mass 1.0 kg rests in equilibrium on an inclined plane. This plane has a length equal to 50 cm and reaches a maximum height from the ground equal to 30 cm. Calculate the coefficient of friction between the block and the incline.

Resolution

The inclined plane is formed by a right triangle with a hypotenuse of 50 cm and a side of 30 cm. By means of the Pythagorean theorem , we determine the other side, which must be 40 cm. Thus, we were able to find the value of its sine and cosine.

and α 30 / 50 6 sen α=30/50=0,6

α 40 / 50 8 cos α=40/50=0,8

Since the block is in equilibrium, the net force on it is zero. So:

P⃗ P=fat tP→P=fat→

Breaking down the weight force similarly to the previous exercise and remembering that the normal force is equal toPandPy→, we have:

Pxμ No⃗ Px→=μ∙N→

Pxμ PandPx→=μ∙Py→

Using the weight force decomposition formula,Pand=P⃗ ∙ α Py→=P→∙cos α, we have:

P∙ and α μ  P⃗ ∙ α P∙sen α=μ∙P→∙cos α

∙ g∙ and α μ ∙ ∙ g ∙ α m∙g∙sen α=μ∙m∙g∙cos α

Substituting for the values ​​given in the utterance and using severity as10 / stwo10 m/s2, we get:

∙ 10 ∙ μ ∙ ∙ 10 ∙ 81∙10∙0,6=μ∙1∙10∙0,8

μ ∙ 86=μ∙8

75 μ

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