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Course: Class 11 Physics (India)>Unit 9

Lesson 8: Inclined planes

Inclined planes review

Review the key concepts and skills for inclined planes, including how to write Newton's second law for forces parallel and perpendicular to the incline.

Key terms

Term (symbol)Meaning
Inclined planeA tilted surface, sometimes called a ramp or incline.

How to write Newton’s second law for forces on an incline

1) Draw a free body diagram for the object (see Figure 3). Remember to rotate the coordinate axes to align with the incline (see Figure 1 below).
If there is any acceleration, it will typically be along the parallel axis (labeled $\parallel$) of the incline.
The perpendicular axis (labeled $\perp$) typically has no acceleration and ${a}_{\perp }=0$.
2) Write the Newton’s second law statement for the direction of interest.
$m{a}_{\perp }=\Sigma {F}_{\perp }$ OR $m{a}_{\parallel }=\mathrm{\Sigma }{F}_{\parallel }$
The perpendicular direction’s equation simplifies because ${a}_{\perp }=0$:
$\begin{array}{rl}m\left(0\right)& =\mathrm{\Sigma }{F}_{\perp }\\ \\ 0& =\mathrm{\Sigma }{F}_{\perp }\end{array}$
3) Substitute the sum of all the forces acting in the direction of interest ($\perp$ or $\parallel$) for $\mathrm{\Sigma }F$. Use your free body diagram to identify which forces are acting in the direction of interest.
Sometimes a force is completely aligned in the parallel or perpendicular direction like normal force and friction.
Some forces have components in both the parallel and perpendicular direction, such as the force of gravity. In that case, the force should be broken down into the parallel and perpendicular components (see Figure 2 below) for substitution in the net force equations.
The parallel and perpendicular components of weight are (Figure 2):
$\begin{array}{rl}{F}_{g\perp }& ={F}_{g}\mathrm{cos}\theta \\ \\ {F}_{g\parallel }& ={F}_{g}\mathrm{sin}\theta \end{array}$

Common mistakes and misconceptions

People forget the force directions. The diagram below shows the forces on an object resting on an incline.
• Weight ${F}_{g}$ is straight down.
• Normal force ${F}_{N}$ pushes perpendicular to the incline.
• Friction ${F}_{f}$ acts parallel to the incline.

To view a worked example of an object sliding down a ramp, watch our video of ice accelerating down a ramp.
To check your understanding and work toward mastering these concepts, check out the exercises on forces and inclined planes.

Want to join the conversation?

• Is there a video or a question that explains how friction is applied to an object in motion on an incline? Say if you wish to find the force applied to an object being pushed up an incline.
• I believe it would be in the friction unit, the 2nd video isn't specific to incline planes but I think it can be pretty easily aplied
• Why is acceleration ZERO on an inclined plane with no friction? You didn't explain that.
• Can a non contact force like mg be ever considered to be a contact force?
• Gravity is a non-contact force that can act on objects within the gravitational field of the exerting body, whereas in the case of contact forces the exerting body has to be in direct contact with the object the force is applied on. So it is a non contact force.
• is this in radians or degrees
• It doesn't matter. Just make sure your calculator is set in degree mode if you're given angles in degrees.
• Is the angle of the incline always equal to the angle between Fg and Fgcos?
• Yes- that's because of the Alternate interior angle theorem! You can revisit this theorem from Sal's videos in the Geometry section. It is intuitive.
(1 vote)
• How does it change when a force is applied horizontally to prevent the object (box) from sliding down the ramp? Does the magnitude of the normal force change while the box is being held in place on a frictionless slope? Thanks.
• An inclined plan makes an angle of 30° with the horizontal. Find the constant force, applied parallel to the inclined plan, required to cause a 20 kg box to slide I. down the incline with acceleration 1,00 m/s2?
II. up the incline with acceleration 1,00 m/s2?
(1 vote)
• I don't understand why and how sine and cosine come into play. I know sine is a ratio of the opposite side to hypotenuse and cosine is adjacent to hypotenuse. But how does this give us our force of gravity? What about this ratio gives it to us? What about them makes it helpful in finding it and how come one of them uses sine and the other is cosine?
(1 vote)
• If you have an inclined force (call that F1, and you know how big that force is and also the angle), horizontal force (Fx), and vertical force (Fy) (both vertical and horizontal force have unknown magnitude), try to make a rectangle with the horizontal and vertical force, you will see that the inclined force split the rectangle and form 2 triangle.You will see that :
sin(theta) = Fy/F1 => rearrange : Fy = F1.sin(theta)
cos(theta) = Fx/F1 => Fx = F1.cos(theta)
using sin and cos help you to decompose an inclined force into an easier one which is horizontal and vertical force