Force Components Along Two Struts
The 500-lb force is to be resolved into two components acting along the axes of the struts AB and AC. If the component of force along AB is required to be 250 lb, directed from A to B, determine the magnitude of the force acting along AC and the angle θ of the 500-lb force. Strut AB meets the ground at B at 45° and strut AC meets the ground at C at 60°, and θ is measured from the horizontal at A.
- \(F = 500\text{ lb}\) applied at the apex \(A\), directed up and to the left at an angle \(\theta\) above the horizontal
- The component along \(AB\) is \(250\text{ lb}\), directed from \(A\) to \(B\)
- Strut \(AB\) meets the ground at 45°, strut \(AC\) at 60°
The magnitude of the force acting along strut \(AC\), and the angle \(\theta\) of the 500 lb force measured from the horizontal at \(A\).
Write a unit vector along each strut, taken from \(A\). Strut \(AB\) falls away from the apex at 45° below the horizontal toward \(B\) on the left, and \(AC\) falls at 60° below the horizontal toward \(C\) on the right. The 500 lb force pulls up and to the left, so its \(AB\) component runs from \(A\) toward \(B\) while its \(AC\) component runs back from \(C\) toward \(A\):
The two struts themselves meet at \(A\) with 75° between them — that is what is left of a straight angle once 45° and 60° are taken off. One component points away from \(A\) and the other points into it, so the angle between the two component directions is the supplement, and that is the angle the force triangle is built on:
The 500 lb force is the resultant of the two components, so the three vectors close a triangle and the law of cosines ties them together. With \(F_{AB} = 250\text{ lb}\) known, this leaves a quadratic in \(F_{AC}\):
Because \(\cos 105^\circ\) is negative the linear term flips sign. Only the positive root is a physical force:
With both components known, add them in \(x\) and \(y\) and take the arctangent of the result. Both components push the resultant to the left, while the larger \(AC\) component wins in \(y\) and tilts it above the horizontal:
\(F_{AC} = 502.5\text{ lb}, \quad \theta = 31.1^\circ\)
No solution steps shown yet.