Solved Examples on Inclined Plane
Example 1: A block of mass 5 kg rests on an inclined plane inclined at an angle of 30 degrees to the horizontal. If the coefficient of friction between the block and the inclined plane is 0.2, calculate:
a) The normal force acting on the block.
b) The frictional force acting on the block if it is on the verge of sliding down.
c) The acceleration of the block if it is released from rest.
Solution:
a) The normal force (N) can be calculated using the formula:
N = mg cos(θ)
where m 5 kg, g =9.81 m/s (acceleration due to gravity), and θ = 30
N (5 kg)(9.81 m/s2) cos(30°)
N = (5)(9.81)(√3/2)
N = 42.73 N
b) The frictional force (f) can be calculated using the formula:
f = uN;
where u= 0.2 (coefficient of friction).
f = (0.2) (42.73)
f = 8.55 N
c) The net force acting on the block (Fnet) when it is released from rest is the component of the
gravitational force parallel to the inclined plane minus the frictional force:
Fnet = mg sin(θ) — f
Fnet = (5) (9.81) sin(30°) – 8.55
Fnet = 24.52 – 8.55
Fnet = 15.97 N
The acceleration (a) of the block can be calculated using Newton’s second law (Fnet = ma)
15.97 = (5)a
a = 15.97/5
a = 3.194 m/s
Example 2: A 10 kg box is placed on an inclined plane inclined at an angle of 45 degrees to the horizontal. If the coefficient of friction between the box and the inclined plane is 0.3, determine the force parallel to the incline required to move the box up the incline at constant velocity.
Solution:
The force parallel to the incline required to move the box up the incline at constant velocity is
equal to the force of friction acting down the incline.
f=uN
Where u = 0.3 (coefficient of friction).
N = mg cos(θ)
N = (I0 kg)(9.81 m/s2) cos(45°)
N = 69.3N
f = 20.79 N
Example 3: A block of mass 12 kg is placed on an inclined plane inclined at an angle of 30 degrees to the horizontal. If the coefficient of friction between the block and the inclined plane is 0.25, calculate the acceleration of the block when it is released from rest.
Solution:
The net force acting on the block( net) when it is released from rest is the component of the
gravitational force parallel to the inclined plane
Fnet = mg sin(θ) — f
f = uN
N = mg cos(θ)
Fnet = mg sin(θ) — umg cos(θ)
Fnet =m(g sin(0) — ug cos(θ))
Fnet = 12(9.81 × sin(30°) – 0.25 × 9.81 × cos(30°))
Fnet = 12(4.905 – 0.25 × 8.484)
Fnet =12(4.905 – 2.121)
Fnet = 12 × 2.784
Fnet = 33.408 N
The acceleration (a) of the block can be calculated using Newton’s second law (Fnet=m*a)
33.408 = 12 × a
a = 33.408/12
a = 2.784m s
Inclined Plane
Inclined Plane is the most fundamental forms of mechanical devices used in physics. In order to get around physical obstacles and simplify tasks, inclined planes have been used for centuries in both ancient and recent construction projects. A flat surface that is angled with respect to the horizontal plane is the fundamental component of an inclined plane. It is a basic mechanism that works by extending the force over a greater distance in order to decrease the force required to move an object vertically.
In this article, we will learn in detail about inclined plane, the mechanics behind it, the resolution of forces into horizontal perpendicular component acting on inclined plane and solve examples based on it.