Peoples Physics Concepts

http://www.ck12.org Chapter 4. Newton’s Laws

Yet these are only the acceleration of gravity so we need to multiply them by the weight of the block to get the force.

F=ma=3kg× 4 .9m/s^2 = 14 .7NF=ma=3kg× 8 .5m/s^2 = 25 .5N

Now that we have solved for the force of the y-component of gravity we know the normal force (they are equal).
Therefore the normal force is 25.5N. Now that we have the normal force and the coefficient of static friction, we can
find the force of friction.

Fs=μsFN=. 6 × 25 .5N= 15 .3N

The force of static friction is greater than the component of gravity that is forcing the block down the inclined plane.
Therefore the force of friction will match the force of the x-component of gravity. So the net force on the block is

net force in the x−direction :

x−com ponent o f gravity

︷ ︸︸ ︷

14 .7N −

f orce o f f riction

︷ ︸︸ ︷

14 .7N =0N

net force in the y−direction : (^25) ︸ ︷︷.5N ︸
Normal Force
− (^25) ︸ ︷︷.5N ︸
y−com ponent o f gravity

=0N

Therefore the net force on the block is 0N.

Example 2

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URL: http://www.ck12.org/flx/render/embeddedobject/389

Watch this Explanation

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URL: http://www.ck12.org/flx/render/embeddedobject/384