A brick is initially at rest on top
of a rough horizontal surface, as shown in the diagram. The brick is pushed by a force 𝐹
parallel to the surface. The weight of the brick equals 30
newtons. The coefficient of static friction
of the brick with the surface is 0.75. What must the magnitude of 𝐹
exceed to start the brick moving across the surface?
Okay, so we have this brick
initially at rest on a rough horizontal surface. And we see that this brick is being
pushed by a force 𝐹 parallel to the surface. For small enough values of 𝐹, the
brick actually won’t move. And that’s due to frictional
effects between the brick and the surface, leading to a frictional force on the
brick. This frictional force is in
opposition to the applied force 𝐹. But the frictional force has some
maximum value it can achieve. If the applied force 𝐹 exceeds
that maximum value, then the brick will start to move.
To figure out the greatest possible
static frictional force that acts on the brick, we can recall that the magnitude of
the frictional force acting on some object on a flat surface is equal to the
object’s weight multiplied by the coefficient of static friction between the object
and the surface it rests on. We can say then that the maximum
static frictional force acting on the brick is equal to its weight, here we have the
magnitude of its weight, multiplied by the coefficient of static friction between
the brick and the rough surface.
Just as a side note, we say that
this is the maximum static frictional force because the force actually could be less
than this. When the force 𝐹 that we apply to
the brick is very small, yet increasing, the static frictional force on the brick
keeping it in place increases with 𝐹. But as we’ve said, this happens
only up to the point where that applied force is either less than or equal to the
weight of this brick multiplied by the coefficient of static friction.
When 𝐹 exceeds this value, it’s at
that exact moment that the brick will start to move. And that’s why we’re solving for 𝐹
sub f max. Substituting in our given values,
this is equal to 30 newtons times 0.75. And if we keep just two significant
figures from our answer, that ends up as 23 newtons. Once our applied force 𝐹 exceeds
this value, the static frictional force will no longer be able to keep up, so to
speak, and the brick will start to move.