Newton's second law: **F**_{net} = m**a**

To find the net force on an object, all the vector forces that act
on the object have to be added. We use **free-body diagrams** to help us with this task. Free-body
diagrams are diagrams used to show the relative magnitude and direction
of all forces acting on an object in a given situation. The object is
represented by a box or some other simple shape. The forces are represented by arrows. The length
of the arrow in a free-body diagram is proportional the magnitude of the
force. The direction of the arrow gives the direction of the force.
Each force arrow in the diagram is labeled. All forces, which act on
the object, must be represented in the free-body diagram. A free body
diagram only includes the forces that act on the object, not the forces
the object itself exerts on other objects.

A free-body diagram for a freely falling ball:

Neglecting air friction, the only force acting on the ball is gravity.

A free-body diagram for a ball resting on the ground:

Gravity is acting downward. The ball is at rest. The ground must exert
a force equal in magnitude and opposite in direction on the ball. This
force is called the **normal force, n**, since it is normal to the surface.

A free-body diagram for a mass on an inclined plane:

Gravity acts downward. The component of
**F**_{g}
perpendicular to the surface is cancelled out by the normal force the
surface exerts on the mass. The mass does not accelerate in the
direction perpendicular to the surface. The component of **F**_{g}
parallel to the surface causes the mass to accelerate in that direction.

Solid objects are incompressible. In general, we cannot push an object into another solid object.

- A table rests on the kitchen floor. Gravity is acting on the
table, pulling it downward. But it does not accelerate into the ground. No
acceleration implies no net force. The net force on the table is zero. The
ground exerts a support force on the table which is equal in magnitude and
opposite in direction to the table's weight. The support force direction is
perpendicular or
**normal**to the ground surface. - A book rests on the table. Gravity is pulling it downward,
but the table exerts a support force on the book which is equal in magnitude
and opposite in direction to the book's weight. The support force direction
is perpendicular or
**normal**to the table surface.

Suppose that you want to move a heavy file cabinet,
which is standing in the middle of your office, into a corner. You push
on it, but nothing happens. What is going on?

You exert a force, but there is no acceleration. The net force
must be zero.

Which force of equal magnitude points in a direction opposite to the
direction of the force you are applying?

The force of **static friction**
(**f**_{s}) cancels the applied force when the cabinet is
at rest while you are pushing on it.

You push harder. Eventually the cabinet breaks away and starts
accelerating. But you have to keep on pushing just to keep it moving
with a constant velocity. When you stop pushing, it quickly slows down
and comes to rest. Why?

While the cabinet is moving the force of
**kinetic friction** (**f**_{k}) opposes the
applied force. When it is moving with constant velocity, the two
forces exactly cancel.

Where do these frictional forces come from? Frictional forces are intermolecular forces. These forces act between the molecules of two different surfaces that are in close contact with each other. On a microscopic scale, most surfaces are rough. Even surfaces that look perfectly smooth to the naked eye show many projections and dents under a microscope. The intermolecular forces are strongest where these projections and dents interlock resulting in close contact. The component of the intermolecular force normal to the surfaces provides the normal force which prevents objects from passing through each other and the component parallel to the surface is responsible for the frictional force.

Assume a cabinet is resting on the floor. Nobody is pushing on it. The net intermolecular force between the molecules of two different surfaces is normal to the surface. The force of gravity acting on the cabinet (red arrow) is balanced by the normal force from the floor acting on it (black arrow).

Now assume that you are pushing against the cabinet. The cabinet is not moving, but the surface molecules are displaced by microscopic amounts. This results in a net intermolecular force which has a component tangential to the surface (the force of static friction). This tangential component opposes the applied force. The net force on the cabinet is zero. The harder you push the greater is the microscopic displacement of the surface molecules and the greater is the tangential component of the net intermolecular force.

When you push hard enough, some of the projections on the surfaces will break off, i.e. some of the surface molecules will be completely displaced. The horizontal component of the net intermolecular force diminishes and no longer completely opposes the applied force. The cabinet accelerates. But while the horizontal component has diminished, it has not vanished. It is now called the force of kinetic friction. For the cabinet to keep accelerating, you have to push with a force greater in magnitude than the force of kinetic friction. To keep it going with constant velocity you have to push with a force equal in magnitude to the force of kinetic friction. If you stop pushing, the force of kinetic friction will produce an acceleration in the opposite direction of the velocity, and the cabinet will slow down and stop.

The frictional force always acts between two surfaces, and opposes the relative motion of the two surfaces.

The **maximum force of static friction**
between two surfaces is roughly proportional to the magnitude of the
force pressing the two surfaces together. The proportional constant
is called the **coefficient of static
friction μ**** _{s}**.
The
magnitude of the force of static friction is always smaller than or
equal to μ

As long as the applied force has a magnitude
smaller than μ_{s}N, the force of static friction f_{s}
has the same magnitude as the applied force, but points in the
opposite direction.

The magnitude of the force of kinetic friction acting on an
object is f_{k }= μ_{k}N, where **μ**_{k}**
is the coefficient of kinetic friction**. For most
surfaces, μ_{k} is less than μ_{s}.

A woman at an airport is towing her 20 kg suitcase
at constant speed by pulling on a strap at an angle of θ above the horizontal.
She pulls on the strap with a 35 N force, and the
frictional force on the suitcase is 20 N.

(a) Draw a free body diagram of the suitcase.

(b) What angle does the strap make with the horizontal?

(c) What normal force does the ground exert on the suitcase?

Solution:

- Reasoning:

constant velocity <--> no acceleration <--> no net force

The x- and y-components of all forces have to add up to zero.

To keep track of all the forces, it helps to draw a free-body diagram. - Details of the calculation:

(a) See the drawing on the right.

(b) The suitcase moves with constant velocity, the net force on the suitcase is zero.

F_{x }= 0 implies (35 N)cosθ = 20 N, or cosθ = 20/35, θ = cos^{-1}(20/35) = 55.15^{o}.

(c) F_{y }= 0 implies mg = n + (35 N)sinθ, (20 kg)(9.8 m/s^{2}) = n + (35 N)0.82,

n = 167.3 N.

Determine the stopping distance for a skier with a speed of 20 m/s on a slope that makes an angle θ with the horizontal. Assume μ_{k
}= 0.18 and θ = 5^{o}.

Solution:

- Reasoning:

Draw a free-body diagram.

The component of the net force perpendicular to the slope must be zero.

The bormal force must cancel the perpendicular component of gravity.

Therefore the magnitude of the normal force n = mg cosθ.

The magnitude of the frictional force is proportional to the normal force, f_{k }= μ_{k}mg cosθ.

The component of the net force down the slope is F = mg sinθ - μ_{k}mg cosθ.

It is the vector sum of the frictional force and the tangential component of gravity.

The acceleration down the slope produced by this net force is a F/m = g sinθ - μ_{k}g cosθ. - Details of the calculation:

μ_{k}g cosθ = 0.18 (9.8 m/s^{2}) cos(5^{o}) = 1.76 m/s^{2}.

g sinθ = (9.8 m/s^{2}) sin(5^{o}) = 0.85 m/s^{2}.^{ }a = g sinθ - μ_{k}g cosθ = -0.91 m/s^{2}

The acceleration points up the slope, in a direction opposite to the velocity.

The acceleration is constant. We now can kinematic equations to solve for the stopping distance.

We have 0 - v_{0 }= -at, t = v_{0}/a = (20/0.91) s = 21.98 s for the time it takes the skier to come to a stop.

In this time he travels a distance s = v_{0}t - ½at^{2}. s = 219.8 m.

Link: Solving inclined plane problems (Youtube)

Suppose you are driving a car along a highway at a high speed. Why should you avoid slamming on your brakes if you want to stop in the shortest distance? That is, why should you keep the wheels turning as you brake?

Solution:

- Reasoning:

When the wheels are turning, friction between the surfaces of the brake pads and the disks or drums of the wheels is responsible for decelerating the car. When the wheels are locked, friction between the tires and the road decelerates the car. The brake pads are designed for the job, and the coefficient of kinetic friction between the brake pads and the disk or drum is large.