Question

3. A uniform light beam is pivoted halfway along its length. At one end it supports a load of 5 kN while the other end is tethered to a fixed point by a rope inclined at 45° to the horizontal. If the beam is in equilibrium, what is the tension in the rope?​

Answer 1

Approximately
`7.1\; {\rm kN}` (
`5 √(2)\; {\rm kN}`.)

Step-by-step explanation:

Let
`F_(1)` and
`F_(2)` denote the two forces that act on this beam. Let
`s_(1)`,
`s_(2)`,
`\theta_(1)`, and
`\theta_(2)` denote the distance from pivot and angle relative to the beam of the two forces, respectively. The magnitude of the torques that the two forces exert on this beam will be
`F_(1)\, s_(1)\, \sin(\theta_(1))` and
`F_(2)\, s_(2)\, \sin(\theta_(2))`, respectively.

The two forces in this question act on the beam from opposite sides of the pivot. Hence, for the beam to be in equilibrium, the torque from the two forces need to be equal in magnitude. In other words:

`F_(1)\, s_(1)\, \sin(\theta_(1)) = F_(2)\, s_(2)\, \sin(\theta_(2))`.

Let
`F_(1)` denote the
`5\; {\rm kN}` force that the load exerts on this beam;
`\theta = 90^(\circ)` since this load is placed directly on the beam. The normal force from the load will be perpendicular to the beam.

Let
`F_(2)` denote the force that the rope exerts on this beam;
`\theta = 45^(\circ)`.

Note that
`s_(1) = s_(2)` since the pivot is exactly halfway between the two forces.

Rearrange the equation
`F_(1)\, s_(1)\, \sin(\theta_(1)) = F_(2)\, s_(2)\, \sin(\theta_(2))` to find the unknown
`F_(2)`:

`\begin{aligned}F_(2) &= (F_(1)\, s_(1)\, \sin(\theta_(1)))/(s_(2)\, \sin(\theta_(2))) \\ &= (F_(1)\, \sin(\theta_(1)))/(\sin(\theta_(2))) && (\text{since $s_(1) = s_(2)$}) \\ &= \frac{5\; {\rm kN}\, \sin(90^(\circ))}{\sin(45^(\circ))} \\ &= \frac{5\; {\rm kN}}{(1 / √(2))} \\ &= 5 √(2)\; {\rm kN} \\ &\approx 7.1\; {\rm kN} \end{aligned}`.

The tension in the rope will be equal in magnitude to the force exerted on the beam: approximately
`7.1\; {\rm kN}` (
`5 √(2)\; {\rm kN}`.)