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Chapter 12: Problem 42

A 160 -kg highway sign of uniform density is \(2.3 \mathrm{m}\) wide and \(1.4 \mathrm{m}\) high. At one side it's secured to a pole with a single bolt, mounted a distance \(d\) from the top of the sign. The only other place where the sign contacts the pole is at its bottom corner. If the bolt can sustain a horizontal tension of \(2.1 \mathrm{kN},\) what's the maximum permissible value for the distance \(d\) ?

Short Answer

Expert verified
The maximum permissible value for the distance \(d\) is approximately 0.52 meters.

Step by step solution

01

Identify the forces

Firstly, identify all the forces acting on the sign: the weight, acting in the middle of the sign, and the tension in the bolt. The weight is the force due to gravity, which can be calculated as mass times gravitational acceleration \(Weight (W) = mass (m) \times gravity (g)\). The mass is 160 kg and gravity is \(9.8 m/s^2\). So \(W = 160kg \times 9.8m/s^2 = 1568N\). The tension is given as \(2.1kN = 2100N\).
02

Calculate the center of gravity

The center of gravity of the sign is the point about which all the weight is evenly distributed. For a sign of uniform density, it is located at the geometric center. In this case, the height of the sign is 1.4 m, so the center of the gravity is located at the height of \(1.4m/2 = 0.7m\) from the base of the sign.
03

Apply the equilibrium condition for torques

For the sign to remain in equilibrium, the sum of the torques from the center of gravity and from the bolt must be equal. Torque (τ) is calculated as the force multiplied by the distance from the rotation point. The torque provided by the weight is \( W \times 0.7m = 1568N \times 0.7m = 1097.6Nm\). The torque provided by the tension in the bolt is \( T \times d = 2100N \times d\). Setting these equal gives the equation \(2100N \times d = 1097.6Nm\). Solving for \(d\) gives \(d = 1097.6Nm / 2100N ~ 0.52m\).
04

Interpret the result

The maximum value permissible for \(d\) is 0.52 meters. This is the maximum distance \(d\) that the bolt can be located from the top of the sign such that the sign is in equilibrium without the bolt snapping from excessive tension.

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