O inclined plane it is a flat, raised and inclined surface, for example a ramp.
In physics, we study the motion of objects as well as acceleration and acting forces that occur on an inclined plane.
Frictionless Inclined Plane
They exist 2 types of forces that act in this frictionless system: the normal force, which makes 90º in relation to the plane, and the weight force (downward vertical force). Note that they have different directions and senses.
THE normal force acts perpendicular to the contact surface.
To calculate the normal force on a flat horizontal surface, use the formula:
Being,
N: normal force
m: object mass
g: gravity
already the strength weight, acts by virtue of the force of gravity that “pulls” all bodies from the surface towards the center of the Earth. It is calculated by the formula:
Where:
P: strength weight
m: pasta
g: gravity acceleration
Inclined plane with friction
When there is friction between the plane and the object, we have another acting force: the friction force.
To calculate the friction force, use the expression:
Where:
Funtil: frictional force
µ: coefficient of friction
N: normal force
The formula for the normal force N on the inclined plane is:
For, the force N is equal in value to the weight component in this direction.
Note: The coefficient of friction (µ) will depend on the contact material between the bodies and their condition.
Acceleration on the Inclined Plane
On the inclined plane there is a height corresponding to the elevation of the ramp and an angle formed in relation to the horizontal.
In this case, the acceleration of the object is constant due to the acting forces: weight and normal.
To determine the amount of acceleration on an inclined plane, we need to find the net force by decomposing the weight force into two planes (x and y).
Therefore, the components of the weight force:
Px: perpendicular to the plane
Py: parallel to the plane
To find the acceleration on the frictionless inclined plane, use the trigonometric relations of the right triangle:
Px = P. if you are
Py = P. cos θ
According to the Newton's second law:
F = m. The
Where,
F: strength
m: pasta
The: acceleration
Soon,
Px = m.a
P. sin θ = m .a
m. g. sin θ = m .a
a = g. if you are
Thus, we have the formula for acceleration used on the frictionless inclined plane, which will not depend on the mass of the body.
Entrance Exam Exercises with Feedback
question 1
(UNIMEP-SP) A block of mass 5kg is dragged along an inclined plane without friction, as shown in the figure.
For the block to acquire an acceleration of 3m/s² upwards, the intensity of F must be: (g = 10m/s², sin θ = 0.8 and cos θ = 0.6).
a) equal to the block weight
b) less than the weight of the block
c) equal to plan reaction
d) equal to 55N
e) equal to 10N
Alternative d: equal to 55N
Exercise solved
Data:
frictionless
m = 5kg
a = 3m/s²
sin θ = 0.8
cos θ = 0.6
Question: What is the F-force?
Making the organization of the forces and the decomposition of the weight force.
We apply Newton's 2nd law in the direction of motion.
⅀F = resulting F = m.a.
F - mgsen θ= m.a.
F = m.a + mgsen θ
F = 5.3 + 5.10.0.8
F = 55N
question 2
(UNIFOR-CE) A block with a mass of 4.0 kg is abandoned on an inclined plane of 37º with the horizontal with which it has a coefficient of friction of 0.25. The acceleration of the block movement is in m/s². Data: g = 10 m/s²; sin 37° = 0.60; cos 37° = 0.80.
a) 2.0
b) 4.0
c) 6.0
d) 8.0
e) 10
Alternative b: 4.0
Exercise solved
Data:
M = 4kg
g = 10 m/s²
sin 37th = 0.60
cos 37º = 0.80
= 0.25 (coefficient of friction)
Question: What is the acceleration?
We do the decomposition of the weight force.
Since there is friction, let's calculate the friction force, Fat.
Fat = . N
By decomposing the force weight, we have that N = mgcos θ.
So, Fat = . mgcos θ
Applying Newton's 2nd Law in the direction of motion, we have:
⅀F = resulting F = m.a.
mg sin θ - Fat = ma
mgsen θ - mi.mgcos θ = m.a
4.10. 0,6 - 0,25.4.10.0,8 = 4. The
Isolating it, we have:
a = 4 m/s²
question 3
(Vunesp) On the inclined plane in the figure below, the coefficient of friction between block A and the plane is 0.20. The pulley is friction-free and the air effect is neglected.
Blocks A and B have masses equal to m each and the local acceleration of gravity has an intensity equal to g. The intensity of the tension force in the rope, supposedly ideal, is:
a) 0.875 mg
b) 0.67 mg
c) 0.96 mg
d) 0.76 mg
e) 0.88 mg
Alternative e: 0.88 mg
Exercise solved
As there are two blocks, we apply Newton's 2nd Law to each one, in the direction of motion.
Where T is the tension in the string.
Block B (equation 1)
P - T = m.a.
Block A (equation 2)
T - Fat - mgsen θ = ma
Making a system of equations and adding the two equations, we have:
P - T = m.a.
T - Fat - mgsen θ = ma
P - Fat - mgsen θ = ma
To proceed, let's determine Fat, then come back to that point.
Fat = mi. N
Fat = mi. mgcos θ
Now, let's determine the values of sin θ and cos θ.
According to the image and applying the Pythagorean theorem:
Since there is the hypotenuse
h² = 4² + 3²
h = 5
Thus, by the definition of sinθ and cosθ
sin θ = 5/3
cos θ = 4/3
Going back to the equation and replacing the found values:
P - Fat - mgsenθ = ma
mg - mi. mgcosθ - mgsenθ = ma
Putting mg in evidence
mg (1 - mi.cox - senX) = 2ma
mg (1 - 0.2. 0.8 - 0.6) = 2ma
0.24mg =2 ma
ma = 0.12mg
Now, let's substitute this value into equation 1
(equation 1)
P - T = m.a.
Isolating T and replacing ma:
T = P - ma
T = mg - 0.24mg
T = mg (1 - 0.12)
T = 0.88mg
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