A traffic light is hanging from a horizontal pole as
shown in the figure. The two metallic cables holding the light can be
considered massless. The total mass of the traffic light feature is 100
Kg. What is the tension in each of the two cables?

The tension on the cable at 60^{0}with the horizontal

a)

5432 N

b)

1516 N

N

c)

760 N

d)

9800 N

e)

None of the above.

and the tension on the cable at 40^{0}with the horizontal

a)

5432 N

N

b)

494 N

c)

2333 N

d)

9800 N

e)

None of the above.

Solution:

This problem involves only one object, the traffic light. However, there
are two unknown in this problem, the tension in each of the cables. To
solve for the two unknown, two equations are needed it. This two
equations will be derived by using Newton's second law by components.

Over this object, there are three forces acting, the
weight of the traffic light, mg, the tension of the cable at 40^{0}
with the horizontal, T_{1}, and the tension of the cable
at 60^{0} with the horizontal, T_{2}, . Notice that the weight of the traffic light is
along the vertical axis. On the other side, following step 4 for solving
Newton's applications, the tensions must be broken down into their
corresponding components along the horizontal and vertical axis. The
second drawing of the sequence
shows the vertical and horizontal
components of the tension T_{1}. These two components are

The third drawing of the sequence
adds the breaking
down into its components for the tension T_{2}, resulting
in the relations

The fourth drawing of the sequence
shows the FBD for
the traffic light with the vectors decomposed into their components
along the x-axis and the y-axis.

Horizontal Component (x-axis)

From the FBD, the two forces acting in the horizontal
direction are the horizontal component of the tension T_{1}
pulling the traffic light to the right and the horizontal component of
the tension T_{2} pulling the traffic light to the left.
The resulting net force is

The traffic light is supposed to be static, no
acceleration in either direction. Thus, Newton's second law applied to
the horizontal component is

Putting together these two results, the following equation is
derived:

Equation 1

Vertical Component (y-axis)

From the FBD, the three forces acting in the vertical direction are
the vertical components of the tensions T_{1} and T_{2}
pulling the traffic light up; and, the weight of the traffic light
pulling the light down. The resulting net force in this direction is

Again, the traffic light is not accelerating in this direction which
implies that Newton's second law applied to this component is

Combining these two results,

Equation 2

Equations 1 and 2 can be used to solved for
the two unknown. In this case, a direct method for solving these
equations is to solve for T_{1} in equation 1, and
substitute it in equation 2. After T_{2} = 760 N
has been found, this value can be substituted back in order to solve for
T_{1} = 494 N. The applet on the left shows the
different algebraic steps necessary for reaching to the solutions.