Force Acting on an Inclined Plane


 
 
Concept Explanation
 

Force Acting on an Inclined Plane

The normal forces in an inclined plane are not always directed upward opposite to the direction of gravity but rather they are always directed perpendicular to the surface that the object is on. To understand the direction of the force acting on an inclined  plane the force acting on the object will has to be resolved in two components.The force that is directed to an angle to the horizontal is resolved into horizontal and vertical components. This can be explained with the help of an example.

For example:

In the figure the object is moving down an inclined plane which is making an angle ;theta with the plane, with Force = f

                                                      

 

The normal force  = N and it is perpendicular to the plane

F_w=mg  is the weight of the load.

The component that balances the Normal force =mg ;Cos;theta

The component that is responsible for the sliding motion is =mg ;Sin;theta

In the absence of the friction and the other forces, the acceleration of an object is the value of the parallel component divided by the mass.It results in the equation as 

    a:=:gsintheta

In the presence of the friction and other forces such as applied force and tensional force, it gets slightly complicated.

Q 1. A five kg mass, initially at rest, slide down a friction less 30^{circ} incline.Calculate the force parallel to the incline and the acceleration of the mass . If the incline is 0.8 m long, calculate the velocity of the mass when it reaches the bottom of the incline.

Solution: We have to calculate force parallel to the incline , acceleration and the velocity of the mass

   The force parallel to the incline , F =mg ;Sin;theta   = (5)(9.8)(sin30) = 24.5 N downwards       (sin 30  =  0.5)

  Acceleration of the mass is  ,  a:=:gsintheta  = (9.8)(sin 30) =  4.9;m:/s^2

  Velocity of the mass on reaching the bottom  can be find out using equation

              v^2:=:u^2:+:2as

Here, u = 0 , a =  4.9;m:/s^2   and s =  0.8 m

 So, v:=:(2as)^{1/2}

      v:=:(2as)^{1/2}:=:(2times4.8times0.8)^{1/2};=:2.8;m/s           

Q. A 2 kg mass is incline down an inclined plane with an unknown angle , as shown.The incline is assumed to be frictionless and is 1.2 m long. The mass start from the rest and is observed to take 2.3s to reach the bottom . What is the angle of incline?

Solution: A of the mass along the incline,   a:=:gsintheta     = (9.8)(sintheta )          .....................(i)

                We need to use this formula ,s:=:ut:+:frac{1}{2}at^2

                                                                a:=:frac{2s}{t^2}

                                                               a:=:frac{2times1.2}{(2.3)^2}:=:0.4.5:m/s^2                   .................(ii)

Now, from equation (i) and (ii)                 0.45;=;(9.8)(sintheta ):=:0.046

                                                                 theta :=;2.63^{circ}  

 

Sample Questions
(More Questions for each concept available in Login)
Question : 1

The normal forces in an inclined plane are always directed ___________________ to the surface that the object is on.

Right Option : B
View Explanation
Explanation
Question : 2

Which of the following are correct :

(a) In the presence of the friction and other forces such as applied force and tensional force, it gets slightly complicated.

(b) The force that is directed to an angle to the horizontal is resolved into horizontal and vertical components.

Right Option : C
View Explanation
Explanation
Question : 3

Which of the following are correct :

(a) The normal forces in an inclined plane are not always directed upward opposite to the direction of gravity

(b) The normal forces in an inclined plane are always directed perpendicular to the surface that the object is on.

(c) The force that is directed to an angle to the horizontal is resolved into horizontal and vertical components.

Right Option : D
View Explanation
Explanation
 
 
 


Students / Parents Reviews [10]