Question

required information
cable ac exerts on beam ab a force p directed along line ac. the force p must have a 330 - lb vertical component.
determine the horizontal component of the force.
the horizontal component of the force is 471 lb.

Explanation:

Step1: Recall force component formula

The vertical component of a force \( P \) at an angle \( \theta \) (with respect to the vertical) is given by \( P_{\text{vertical}} = P \cos\theta \), and the horizontal component is \( P_{\text{horizontal}} = P \sin\theta \). We can also relate the horizontal and vertical components using \( \tan\theta=\frac{P_{\text{horizontal}}}{P_{\text{vertical}}} \), so \( P_{\text{horizontal}} = P_{\text{vertical}} \tan(90^\circ - \theta) \) or since \( \theta = 55^\circ \) (with respect to vertical), the angle with respect to horizontal is \( 35^\circ \), but easier to use \( \tan\theta=\frac{\text{opposite}}{\text{adjacent}} \) where \( \theta = 55^\circ \), vertical is adjacent, horizontal is opposite. So \( \tan(55^\circ)=\frac{P_{\text{horizontal}}}{P_{\text{vertical}}} \).

Step2: Solve for horizontal component

Given \( P_{\text{vertical}} = 330 \) lb, \( \theta = 55^\circ \). Then \( P_{\text{horizontal}} = P_{\text{vertical}} \tan(55^\circ) \). Calculate \( \tan(55^\circ) \approx 1.4281 \). So \( P_{\text{horizontal}} = 330 \times 1.4281 \approx 471 \) lb. Alternatively, using \( \sin \) and \( \cos \): first find \( P \) from \( P_{\text{vertical}} = P \cos(55^\circ) \), so \( P=\frac{330}{\cos(55^\circ)} \approx \frac{330}{0.5736} \approx 575.3 \) lb. Then horizontal component \( P_{\text{horizontal}} = P \sin(55^\circ) \approx 575.3 \times 0.8192 \approx 471 \) lb.

Answer:

The horizontal component of the force is \(\boxed{471}\) lb.