6.6. Free-Body Diagram Example http://www.ck12.org
6.6 Free-Body Diagram Example
Question: Using the diagram below, find the net force on the block. The block weighs 3kg and the inclined plane
has a coefficient of friction of.6.
Answer:
The first step to solving a Newton’s Laws problem is to identify the object in question. In our case, the block on the
slope is the object of interest.
Next, we need to draw a free-body diagram. To do this, we need to identify all of the forces acting on the block and
their direction. The forces are friction, which acts in the negative x direction, the normal force, which acts in the
positive y direction, and gravity, which acts in a combination of the negative y direction and the positive x direction.
Notice that we have rotated the picture so that the majority of the forces acting on the block are along the y or x axis.
This does not change the answer to the problem because the direction of the forces is still the same relative to each
other. When we have determined our answer, we can simply rotate it back to the original position.
Now we need to break down gravity (the only force not along one of the axises) into its component vectors (which
do follow the axises).
The x component of gravity : 9.8m/s^2 ×cos 60 = 4 .9m/s^2
The y component of gravity : 9.8m/s^2 ×sin 60 = 8 .5m/s^2Yet 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