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## AP®︎/College Physics 1

### Course: AP®︎/College Physics 1>Unit 2

Lesson 4: Applications of Newton's second law

# Ice accelerating down an incline

Explore the physics of an ice block sliding down an icy incline. Understand the forces at play, including gravity and the normal force, and how they contribute to the block's acceleration. Learn to calculate these forces using trigonometry and Newton's second law. Created by Sal Khan.

## Want to join the conversation?

• Maybe I missed it in the video, but why do the Fg(parallel) and the Fg (perpendicular) not ad up to 98 N? shouldn't they? Because all we did was split up the gravity vector into 2 vectors, so if we add them it should be the same, no?.. Thank you.
• Actually they do add up to 98! :-) These are vectors, so adding them depends on the angle between them, not just their values. If they were pointed the same way, then they would add up to the original value, but they are at 90 degrees to each other.
Explanation:
Look at the screenshot of the video (you don't have to replay it to see). There is a right triangle with the hypotenuse as the force from gravity (98) and the two component values for the perpendicular force and the parallel force. If you use the Pythagorean theorem with the components, you get 98 for the magnitude of the force due to gravity vector.
Details:
Fgparallel = 98 sin (30)
Fgperpendicular = 98 cos (30)
These are the legs of a right triangle.
The hypotenuse then is Ftotal so...
c^2 = a^2 + b^2
Ftotal^2 = Fgparallel ^2 + Fgperpendicular ^2
Ftotal^2 = (98sin(30)) ^2 + (98cos(30)) ^2
Ftotal^2 = (98 * 1/2) ^2 + 98 * (sqrt(3)/2) ^2
Ftotal^2 = 2401 + 7203
Ftotal^2 = 9604 (take the square root of both sides)
Ftotal = 98!
• Hey there Sal, it was a great video and i looove your website but after watching this, I had a question in my mind.
It the force acting on the block parallel to the slope is 49 N and the force due to gravity is 98 N, where does the remaining 49 N go?
Is the remaining 49 N the force that is being counteracted by the normal force on the block?
thx
• Hi Tushar, you are on the right lines but remember that forces are vectors so they have direction and cannot be simply added or subtracted like that unless they are acting in the same or opposite directions. What Sal did is to decompose the 98N into two components: the 49N acting parallel to the surface of the slope and the 49(3^0.5)N which is the force that the block exerts perpendicular to the slope. And as you say that component is exactly equal and opposite to the normal force which the slope exerts on the block. But you cannot do 98-49N because they are vectors in different directions. Please remember to be very wary of doing that with vectors.

All the best,
Sheridan
• Does ice itself actually have negligible friction? Isn't it because melting ice creates a surface layer of water that enables it to move smoothly?
(Sorry, this isn't really related to the video in itself, but I've been having this doubt for a while)
• the reasons for ice being slippery are not well understood. The explanation that relies on the application of pressure to melt a thin layer of water doesn't hold up to careful experimentation and calculations.
• Hello!
This video has helped me understand the basics of forces on the inclined plane (THANK YOU!) but I have a question:
From to , why is the Cossin calculated on the y component and the sin calculated on the x component? Please explain.

Udyant Aggarwal
• Understandably, you seem to have gotten turned around! The cosine is used in this case because what we have is an angle on a right triangle. Based on this angle, we want to know the the side beside it, which we will then compare to the hypotenuse.

Basically, cos and sin should not be thought of as x and y (even though in the unit circle they can be used this way). To keep yourself less confused, the sine of an angle is the ratio of the leg directly opposite to it over the hypotenuse, while the cosine is the leg beside the angle over the hypotenuse. Trying to rotate and reflect everything into terms of x and y will just get you doing extra work.

If you still have questions, I suggest you draw a bunch of triangles and turn them around to see what I mean on your own.
• If a body is held stationary on an incline by more than just friction, would the frictional forcre be regarded as static or dynamic?
• Hi, I know I'm really late :) but it depends on whether the body is accelerating or not. Since it's at rest and not sliding, there is static friction but because of the incline the body is trying to overcome static friction.

Hope this helps.. others as well.
(1 vote)
• How does the surface area of an object affect the sliding frictional force present while pulling it across a surface?
• It also depends on what you mean by "surface area." An increase in surface area in the sense that the surface is bumpier or rougher would increase the frictional force (since it increases the coefficient of friction) but just increasing the surface area in the sense that the surface of contact is greater would generally have minimal impact on the frictional force.
(1 vote)
• What would happen if kinetic friction was involved in the problem but it was still accelerating (rather than being stationary in the next video)?
• You subtract the force of kinetic friction from the parallel to the surface of the slope component of the force of gravity.
(1 vote)
• can you explain why dropping things of different mass reach the ground at the same time but sliding things of different mass down an incline do not?