Question

A sign is supported by a uniform horizontal boom of length \(3.00 \mathrm{m}\) and weight \(80.0 \mathrm{N} .\) A cable, inclined at an angle of \(35^{\circ}\) with the boom, is attached at a distance of \(2.38 \mathrm{m}\) from the hinge at the wall. The weight of the sign is \(120.0 \mathrm{N} .\) What is the tension in the cable and what are the horizontal and vertical forces \(F_{x}\) and \(F_{y}\) exerted on the boom by the hinge? Comment on the magnitude of \(F_{y}\).

Short answer

Answer: In the static equilibrium of a system involving a sign, a boom, and a cable attached at an angle, the magnitude of the vertical force exerted by the hinge (Fy) is observed to be greater than the sum of the weights of the boom and the sign. This is because Fy must support not only the weights of the boom and the sign but also the vertical component of the tension in the cable (Ty).

Step-by-step solution

Break down forces into x and y components

To calculate the tension in the cable and the horizontal and vertical forces exerted by the hinge, we first need to break down the forces into x and y components. The weight of the boom and the sign act vertically downward, while the tension in the cable acts at an angle of 35 degrees with respect to the boom. Using trigonometry, we can calculate the x and y components of the tension (T) in the cable: \(T_x = T \cos(35^{\circ})\) \(T_y = T \sin(35^{\circ})\)

Summation of forces in x and y directions

For the boom to be in static equilibrium, the summation of forces in the x and y directions must be equal to zero. So, we will sum up the forces acting in the x and y directions: \(\sum F_x = T_x - F_x = 0\) \(\sum F_y = T_y - F_y -80 N - 120 N = 0\)

Moment equilibrium around the hinge point

Another condition for static equilibrium is that the summation of moments around any point must be equal to zero. We will take the moment around the hinge point, as this will simplify the calculations. The moments of the weights of the sign and the boom around the hinge are: \(M_{boom} = (80 N)(0.5)(3 m)\) \(M_{sign} = (120 N)(3 m)\) The moment of the cable tension around the hinge is: \(M_T = (T_y)(2.38 m)\) \(\sum M = M_T - M_{boom} - M_{sign} = 0\)

Solve for tension and hinge forces

We have three equations from Steps 2 and 3, and three unknowns (T, Fx, Fy). We can solve these equations simultaneously to find the values of tension and hinge forces: Solving for T: \((T \cos(35^{\circ})) = F_x\) \((T \sin(35^{\circ}))(2.38 m) - (80 N)(0.5)(3 m) - (120 N)(3 m) = 0\) Solve the above equations to find the values of T, Fx, and Fy.

Comment on the magnitude of F_y

After solving the equations, we can observe that the magnitude of F_y is greater than the sum of the weights of the boom and the sign. This is because F_y must not only support the weights of the boom and the sign but also the vertical component of the tension in the cable (T_y).