<h1>Problem</h1>
The diagram represents a system of cables AB, AC, and AD supporting a 50-kg weight W. The system is in static equilibrium. What is the tension in cable AB? Give answer to the nearest Newton.
<h1>Solution</h1>
To get a better sense of the system, we can draw a free-body diagram. The weight pulls down on cable AD giving it a tension <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><msub><mi>T</mi><mrow><mi>A</mi><mi>D</mi></mrow></msub></mrow><annotation encoding="application/x-tex">T_{AD}</annotation></semantics></math>TAD​ = mg = (50kg)(9.81<math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mfrac><mi>m</mi><msup><mi>s</mi><mn>2</mn></msup></mfrac></mrow><annotation encoding="application/x-tex">\frac{m}{s^2}</annotation></semantics></math>s2m​) = 490.5 N. Since we're in equilibrium we know that the forces in the x-axis and y-axis sum to zero.
Therefore, we can write the following equations:
<div class="math math-display"><math xmlns="http://www.w3.org/1998/Math/MathML" display="block"><semantics><mrow><mo>∑</mo><msub><mi>F</mi><mi>x</mi></msub><mo>=</mo><mn>0</mn><mo>⇒</mo><mo stretchy="false">(</mo><msub><mi>T</mi><mrow><mi>A</mi><mi>C</mi></mrow></msub><mo stretchy="false">)</mo><mi>c</mi><mi>o</mi><mi>s</mi><mn>4</mn><msup><mn>5</mn><mo lspace="0em" rspace="0em">∘</mo></msup><mo>−</mo><mo stretchy="false">(</mo><msub><mi>T</mi><mrow><mi>A</mi><mi>B</mi></mrow></msub><mo stretchy="false">)</mo><mi>c</mi><mi>o</mi><mi>s</mi><mn>3</mn><msup><mn>0</mn><mo lspace="0em" rspace="0em">∘</mo></msup><mo>=</mo><mn>0</mn></mrow><annotation encoding="application/x-tex">\sum{F_x} = 0 \Rightarrow (T_{AC})cos45^{\circ} - (T_{AB})cos30^{\circ} = 0</annotation></semantics></math>∑Fx​=0⇒(TAC​)cos45∘−(TAB​)cos30∘=0</div>
<div class="math math-display"><math xmlns="http://www.w3.org/1998/Math/MathML" display="block"><semantics><mrow><mo>∑</mo><msub><mi>F</mi><mi>y</mi></msub><mo>=</mo><mn>0</mn><mo>⇒</mo><mo stretchy="false">(</mo><msub><mi>T</mi><mrow><mi>A</mi><mi>C</mi></mrow></msub><mo stretchy="false">)</mo><mi>s</mi><mi>i</mi><mi>n</mi><mn>4</mn><msup><mn>5</mn><mo lspace="0em" rspace="0em">∘</mo></msup><mo>+</mo><mo stretchy="false">(</mo><msub><mi>T</mi><mrow><mi>A</mi><mi>B</mi></mrow></msub><mo stretchy="false">)</mo><mi>s</mi><mi>i</mi><mi>n</mi><mn>3</mn><msup><mn>0</mn><mo lspace="0em" rspace="0em">∘</mo></msup><mo>−</mo><mn>490.5</mn><mi>N</mi><mo>=</mo><mn>0</mn></mrow><annotation encoding="application/x-tex">\sum{F_y} = 0 \Rightarrow (T_{AC})sin45^{\circ} + (T_{AB})sin30^{\circ} - 490.5N = 0</annotation></semantics></math>∑Fy​=0⇒(TAC​)sin45∘+(TAB​)sin30∘−490.5N=0</div>
We can then solve for <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><msub><mi>T</mi><mrow><mi>A</mi><mi>B</mi></mrow></msub></mrow><annotation encoding="application/x-tex">T_{AB}</annotation></semantics></math>TAB​.

Cables and Weight System