Question

this is part of the last problem from the set of practice problems from the previous class on wednesday. if you already did it, congratulations it is exactly the same problem. if not, now you have a chance to try it out and see if you can get the correct answer. a box with a mass of 200 kg is suspended from two cables as shown in the diagram below. the tension force to the left $f_{t1}$ is equal to 2,336.4 newtons. calculate the tension force $f_{t2}$ of the tension force that is at an angle.

Explanation:

Step1: Analyze Vertical and Horizontal Forces

The box is in equilibrium, so vertical and horizontal forces balance. Horizontal: \( F_{T1} = F_{T2}\cos\theta \). Vertical: \( F_{T2}\sin\theta = mg \), where \( m = 200\,\text{kg} \), \( g = 9.8\,\text{m/s}^2 \), so \( mg = 200 \times 9.8 = 1960\,\text{N} \).

Step2: Find \( \cos\theta \) from Horizontal Force

Given \( F_{T1} = 2336.4\,\text{N} \), let \( F_{T2} = T \). Then \( 2336.4 = T\cos\theta \). From vertical, \( T\sin\theta = 1960 \).

Step3: Use Trigonometric Identity

Divide vertical equation by horizontal: \( \tan\theta = \frac{1960}{2336.4} \approx 0.839 \), so \( \theta \approx 40^\circ \) (since \( \tan40^\circ \approx 0.839 \)). Then \( \cos40^\circ \approx 0.766 \).

Step4: Solve for \( F_{T2} \)

From \( 2336.4 = T \times 0.766 \), \( T = \frac{2336.4}{0.766} \approx 3050\,\text{N} \). Alternatively, use \( T = \sqrt{(F_{T1})^2 + (mg)^2} = \sqrt{2336.4^2 + 1960^2} \approx \sqrt{5,468,800 + 3,841,600} = \sqrt{9,310,400} \approx 3051\,\text{N} \) (small difference due to \( \tan\theta \) approximation).

Answer:

The tension force \( F_{T2} \) is approximately \( \boldsymbol{3050\,\text{Newtons}} \) (or \( 3051\,\text{N} \) using Pythagoras directly).