Equilibrium of a Body Suspended by a Cable

Picture lifting a bucket of water using a rope and pulley. It may seem like a simple action, but it involves some fundamental principles of physics - specifically, equilibrium and tension. Let’s take a closer look at what’s really happening.

Tension in a Cable

When a thin rope is pulled from both ends by forces of equal magnitude but opposite direction, a uniform tension, denoted by $ T $, is established throughout the entire length of the rope.

If the rope’s mass is negligible, this tension remains constant at every point along the rope.

If we were to hypothetically cut the rope, the force $ T $ is what would continue to "hold together" the two severed ends.

A Body Suspended by a Cable

If a body of weight $ P $ - for example, a chandelier - is suspended from a rope, the rope exerts an upward force of magnitude $ T = P $ to support the weight.

The same tension $ T $ also acts at the point where the rope is anchored to the ceiling.

At that point, the ceiling exerts an equal and opposite reaction force $ N $ to maintain the system in equilibrium.

The Role of the Pulley

An ideal pulley - one that is frictionless and massless - does not alter the magnitude of the tension in the rope. It simply changes the direction of the force.

This makes it easier to lift or control the load.

Consider the case of a bucket hanging from a pulley:

When the bucket is suspended, two forces act on it:

• Its weight $ P $, acting downward.
• The tension $ T_1 = P $, transmitted upward through the rope.

The person lifting the bucket must apply a downward force of $ T_1 = P $ to keep the bucket stationary or to raise it slowly.

The pulley itself must also be in equilibrium.

The forces acting on the pulley are:

• An upward force $ T_2 $, exerted by the support holding the pulley.
• Two downward forces $ T_1 $ - one from the section of rope connected to the bucket, and one from the section of rope being pulled by the person.

For the pulley to remain at rest, the following equilibrium condition must be satisfied:

$$ T_2 = 2T_1 = 2P $$

In other words, the force holding the pulley to the ceiling must be twice the weight of the bucket.

A key principle to keep in mind: In an ideal rope, tension remains constant along its entire length, unless altered by factors such as friction or distributed mass along the rope.

Angled Cables: How to Analyze Tension

When one or more cables are inclined, calculating the tension in each requires applying the conditions of equilibrium:

• Vertical equilibrium
  The sum of the vertical components of the tensions must balance the load’s weight.
• Horizontal equilibrium
  The sum of the horizontal components of the tensions must be zero (if the system is in static equilibrium).

For example, imagine a load of $ P = 100 \ N $ suspended by two cables that form angles $ \theta_1 = 30^\circ $ and $ \theta_2 = 45^\circ $ with the horizontal.

The tensions in the two cables are $ T_1 $ and $ T_2 $.

To solve for these tensions, you must resolve each tension into its components.

For each cable:

• Vertical component:
  $ T_1 \sin \theta_1 $, $ T_2 \sin \theta_2 $
• Horizontal component:
  $ T_1 \cos \theta_1 $, $ T_2 \cos \theta_2 $

The equilibrium conditions can then be written as:

• Vertical equilibrium:
  $$ T_1 \sin \theta_1 + T_2 \sin \theta_2 = P $$
• Horizontal equilibrium:
  $$ T_1 \cos \theta_1 = T_2 \cos \theta_2 $$

This gives you a system of two equations with two unknowns ($ T_1 $ and $ T_2 $):

$$ \begin{cases} T_1 \sin \theta_1 + T_2 \sin \theta_2 = P \\ \\ T_1 \cos \theta_1 = T_2 \cos \theta_2 \end{cases} $$

You can solve this system using standard algebraic methods (such as substitution or solving simultaneous equations).

In this example, we are given $ P = 100 \ N $, $ \theta_1 = 30^\circ $, and $ \theta_2 = 45^\circ $.

$$ \begin{cases} T_1 \sin 30^\circ + T_2 \sin 45^\circ = 100 \ N \\ \\ T_1 \cos 30^\circ = T_2 \cos 45^\circ \end{cases} $$

The relevant trigonometric values are: $ \sin 30^\circ = 0.5 $, $ \cos 30^\circ \approx 0.866 $, $ \sin 45^\circ \approx 0.707 $, $ \cos 45^\circ \approx 0.707 $.

Substituting these into the equations:

$$ \begin{cases} T_1 \cdot 0.5 + T_2 \cdot 0.707 = 100 \ N \\ \\ T_1 \cdot 0.866 = T_2 \cdot 0.707 \end{cases} $$

Rearranging the second equation:

$$ T_1 = T_2 \cdot \frac{ 0.707 }{ 0.866 } = T_2 \cdot 0.816 $$

Substituting this into the first equation:

$$ (T_2 \cdot 0.816) \cdot 0.5 + T_2 \cdot 0.707 = 100 \ N $$

$$ T_2 \cdot (0.408 + 0.707) = 100 \ N $$

$$ T_2 \cdot 1.115 = 100 \ N $$

Solving for $ T_2 $:

$$ T_2 = \frac{ 100 \ N }{ 1.115 } = 89.7 \ N $$

Now, calculating $ T_1 $:

$$ T_1 = 89.7 \ N \times 0.816 = 73.2 \ N $$

Therefore, the tension in the first cable is approximately $ T_1 \approx 73.2 \ N $, and in the second cable $ T_2 \approx 89.7 \ N $.

In short, even in everyday situations - whether lifting a bucket, drawing a curtain along a pulley, or hanging a chandelier - forces are always at work maintaining equilibrium.