**Second Newton's Law Exercises**

**Inclined Plane, Detailed Example With and
Without Friction**

The pulley shown below is weightless and frictionless.
The block of mass m_{1 }is on the plane, inclined at an angle
with the horizontal. The block of mass m_{2} is connected to m_{1}
by a string.

a) Assuming there is no friction, show a formula
for the acceleration of the system in terms of m_{1}, m_{2},
and g.

b) What condition is required for m_{1} to go up the incline?

c) Assume that the coefficient of kinetic friction between m_{1} and
the plane is 0.2, m_{1} = 2 kg, m_{2} = 2.5 kg and the angle
= 30º. Calculate
the acceleration of m_{1} and m_{2}.

d) What is the maximum value of the friction coefficient so the system can
still move.

Before attempting to solve a Second Newton's Law problem like this it is very useful to take into account the following considerations:

**Free Body Diagram.-**
In every problem where the Second Newton's Law applies
it is fundamental to draw what is called the **Free Body Diagram**. This
diagram must show **all** the external forces acting on a body. We **isolate**
the body and the forces due to that strings and surfaces are replaced by arrows;
of course, the friction forces and the force of gravity must be included.
If there are several bodies, a separate diagram should be drawn for each one.

**FREE BODY DIAGRAM**

- The force that m_{1} exerts on m_{2}
through the rope has the same magnitude T. This is so because a rope only
changes the direction of a force, not its magnitude assuming a weightless
rope.

- The magnitude of the acceleration is the same at both ends of the rope assuming an inextensible rope.

**Components of Force.-**
Notice from the diagram
the weight of m_{1} has been split into the components m_{1}gsen
parallel to the incline, and m_{1}gcos
perpendicular to it.

a) Let's assume the direction of the acceleration
makes m_{1} to go upward.

Sum of forces on m_{1} in the direction of the incline: T-m_{1}gsen=m_{1}a

Sum of vertical forces on m_{2} = m_{2}g-T = m_{2}a

And adding both equations we get m_{2}g-m_{1}gsen
= a(m_{1}+m_{2})

Or a = g(m_{2}-m_{1}senß)/(m_{1}+m_{2})

b) For a to be positive (i.e. m_{1}
going up) m_{2} > m_{1}sen

For a to be negative(i.e. m_{1} going down) m_{2} < m_{1}sen

c) Now appears a friction force, always in
an opposite direction to the movement. The magnitude of this friction force
is F_{f} = F_{N}.
F_{N} is obtained using:

Sum of forces perpendicular to the incline = F_{N}-m_{1}gcos
= 0

or F_{N} = m_{1}gcos.
The friction force is then F_{f} = m_{1}gcos

The sum of forces on m_{1} in the sense of the incline is now

(1) T-m_{1}gsen-m_{1}gcos
= m_{1}a

The sum of vertical forces on m_{2} is

(2) m_{2}g-T = m_{2}a

Adding (1) and (2) we get m_{2}g-m_{1}g(sen+µcos)
= a(m_{1}+m_{2})

Replacing values, with
= 0.2, m_{1} = 2 kg, m_{2} = 2.5 kg and
= 30º results a = 2.51 m/s^{2}.

d) As the coefficient of friction increases,
the acceleration decreases until reaching a value near zero or zero. This
condition is obtained with zeros in the right side of equations (1) and (2)
used before in answer c):

T-m_{1}gsen-m_{1}gcos
= 0

m_{2}g-T
= 0

And adding, we get m_{2}g-m_{1}gsen-m_{1}gcos
= 0

Or = (m_{2}-m_{1}sen)/m_{1}cos

Replacing values we get
= 0.87.

**Newton's Second Law Links:**

** Second Newton's Law, Theory and Examples
Second Newton's Law Examples, Part One**

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