U3AOS1 Topic 6: Inclined Planes

When there is a mass sitting on an incline plane there are 3 forces acting on the object.


1. The Normal Force which is perpendicular (90 degrees) to the plane. (green)

2. The force due to gravity which acts directly down from the center of the mass. (blue)

3. Friction force on the mass. (orange)

There are 2 cases that can occur.


1. The block will slide down the slope. This means that the friction is too small to stop the acceleration of the block down the incline. Therefore, there will be a net force (purple) that goes down the slope.

2. The block will not slide down the slope. This will happen if the friction is either equal to or greater than the net force of the gravity and normal components.


To calculate the forces themselves:

Normal force:

\[ \displaystyle \Large F_N = mg\cos(\theta) \]

FN=mgcos(θ)F_N=mgcos(\theta)

Force due to gravity:

\[ \displaystyle \Large F_g = mg \]

Fg=mgF_g=mNet force with no friction:

\[ \displaystyle \Large F_{\text{net}} = mg\sin(\theta) \]

Fnet=mgsin(θ)F_{net}=mgsin(\theta)

Net force with friction:

\[ \displaystyle \Large F_{\text{net}} = mg\sin(\theta) - F_{\text{friction}} \]

Fnet=mgsin(θ)FfrictionF_{net}=mgsin(\theta)-F_{friction}If there are pulleys on the end of inclined planes you must remember that pulleys only redirect forces. Therefore you can treat the entire system as a connected bodies questions after finding the net forces on each block and then solve for the system.


Created with GeoGebra®, by Tom Walsh, Link

Exercise &&1&& (&&2&& Questions)

Calculate the normal force of a 5kg mass on a 30 degree slope.

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Calculate the normal force of a 14kg mass on a 60 degree slope.

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Exercise &&2&& (&&1&& Question)

Calculate the gravitational force of a 15kg mass on a 45 degree slope.

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Exercise &&3&& (&&1&& Question)

Calculate the frictional force on an object with a mass of 5kg on a 25 degree slope and acceleration of 5m/s^2.

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