A block is resting over an incline plane as shown in the
figure. The angle of
the incline can be changed by pulling on the string. If the block only
starts sliding when the angle is greater than qc.
What is the value of the static frictional force when
the inclination angle is q smaller
None of the above.
and the coefficient of static friction between the
None of the above.
When the angle of the inclined is q, smaller
than qc, the block is at rest,
first picture of sequence on the left, . The forces acting on the block are
the weight of the block, the normal force, and the static friction.
These forces are shown on the next figure of the sequence . Based on the
direction of the forces,
again it is convenient to select a coordinate system with the x-axis
parallel to the inclined plane and the y-axis perpendicular to
. Thus, only
the vector representing the weight of the block need to be broken down
into components. Just as in the case of the
studied before, the angle of the incline appears between the bottom part
of the y-axis and the vector representing the weight of the block
. Accordingly, the resulting components
of the weight are
, along the x-axis pointing to the right; and, along the
y-axis pointing down,
. Finally, the corresponding FBD for the block is the last
figure of the sequence
. In the present case, the block is at rest.
From the diagram, the net force in the y-direction is
Applying Newton's second law to this component (no motion present)
From the two relations for the net force, the following equation and
solution is obtained,
This result allows you to calculate the normal force at
Also the net force in the x-direction is
In this case, the block is not sliding down the inclined
Equating the two relations, the following equation is
The previous results indicates how the static frictional
force adjust to mach the force pulling the block down depending on the
angle of the inclined.
equals to qc the static frictional
force is maximum, .
In the maximum case, the maximum static friction is related to the
normal force by the relation,
Substituting the value of the normal force at the critical angle,
As calculated above, at any angle the static frictional
force is given by
evaluated at the critical angle is
From the previous two relations an equation for the
coefficient of static friction is obtained,