Question

consider a sign suspended on a boom that is supported by a cable, as shown. what is the proper equation to use for finding the net force in the y direction? $f_{nety} = (f_t)(sin 32^circ) - f_g$; $f_{nety} = (f_t)(cos 32^circ) - f_g$; $f_{nety} = (f_t)(cos 32^circ) + f_g$; $f_{nety} = (f_t)(sin 32^circ) + f_g$

Explanation:

Step1: Analyze Force Components

The tension force \( F_T \) in the cable has a vertical component. Using trigonometry, the vertical component of \( F_T \) is \( F_T \sin(32^\circ) \) (since the angle with the horizontal is \( 32^\circ \), the vertical component is opposite the angle, so sine is used). The weight force \( F_g \) acts downward. For net force in the \( y \)-direction, we subtract the downward force from the upward component.

Step2: Formulate Net Force Equation

The upward vertical component of \( F_T \) is \( (F_T)(\sin 32^\circ) \), and the downward force is \( F_g \). So the net force in the \( y \)-direction is \( F_{\text{net}y} = (F_T)(\sin 32^\circ) - F_g \).

Answer:

\( F_{\text{net}y} = (F_T)(\sin 32^\circ) - F_g \) (the first option among the given choices)