Question

A uniform beam of length= 1.0 ${\rm m}$ and mass 10 ${\rm kg}$ is attached to a wall by a cable at angle $\theta = 30^\circ$ to the horizontal, as shown in the figure. The beam is free to pivot at the point where it attaches to the wall.What is the tension in the cable?

Answer 1

Concepts and reason

The concept of moments in rotational equilibrium is used to solve the problem. The tension in the cable can be determined by applying the principle of moments to the beam.

Fundamentals

The beam would be in equilibrium when the total forces acting on it add up to zero and when the algebraic sum of moments of the forces acting on the body about any point is zero.

The beam is under the action of three forces- Its Weight W acting downwards at its center of mass, the tension T acting along the length of the cable and the normal force N acting perpendicular to the wall and along the length of the beam.

This is shown in the diagram below.

Moment of a force is defined as the product of the force and the perpendicular distance between the pivot and the line of action of the force.

The rod AB is uniform, which implies that the center of mass of the beam lies at its geometric center O. The weight W is the beam lies at its midpoint. Resolve the tension into two components.

This is shown in the diagram below:

Calculate the value of the horizontal and the vertical components ${T_h}$ and $T$ of Tension T.

$\begin{array}{c}\\{T_h} = T\cos 30^\circ \\\\ = T\left( {\frac{{\sqrt 3 }}{2}} \right)\\\end{array}$

$\begin{array}{c}\\{T_v} = T\cos 30^\circ \\\\ = T\left( {\frac{1}{2}} \right)\\\end{array}$

The weight of the beam W is given by,

$W = mg$

Here, m is the mass of the beam and g is the acceleration of free fall.

Substitute 10 kg for m and 9.8 m/s² for g.

$\begin{array}{c}\\W = mg\\\\ = \left( {10{\rm{ kg}}} \right)\left( {9.8{\rm{ m/}}{{\rm{s}}^2}} \right)\\\\ = 98{\rm{ N}}\\\end{array}$

The weight W acting on the beam at C produces a clockwise moment about point A.

Write the expression for the moment ${m_W}$ of the force W about A.

${m_W} = \left( {AO} \right) \times W$

The vertical component ${T_v}$ of the tension produces an anticlockwise moment about A.

Write the expression for the moment ${m_v}$ of the force ${T_v}$ about A.

${m_v} = \left( {AB} \right) \times {T_v}$

The forces N and ${T_h}$ , have their lines of action passing through the point A. Hence they produce no rotation about A.

For equilibrium, the clockwise moment ${m_v}$ should be equal to the anticlockwise moment ${m_W}$ .

Therefore,

${m_v} = {m_W}$

Or

$\left( {AB} \right) \times {T_v} = \left( {AO} \right) \times W$

Substitute 1.0 m for AB, $\frac{T}{2}$ for ${T_v}$ , 0.5 m for AO and 98 N for W. Solve for T.

$\begin{array}{c}\\\left( {AB} \right) \times {T_v} = \left( {AO} \right) \times W\\\\\left( {1.0{\rm{ m}}} \right)\left( {\frac{T}{2}} \right) = \left( {0.5{\rm{ m}}} \right)\left( {98{\rm{ N}}} \right)\\\\T = 98{\rm{ N}}\\\end{array}$

Ans:

The tension in the cable is 98 N.