problem 1.5 - find the forces acting along shafts ab and cb, respectively, which are equivalent to the…

Explanation:

Step1: Set up force - equilibrium equations

Let the force in shaft AB be $F_{AB}$ and in shaft CB be $F_{CB}$. At point B, the sum of vertical and horizontal forces is zero. In the vertical - direction: $F_{AB}\sin30^{\circ}+F_{CB}\sin40^{\circ}=200$ In the horizontal - direction: $F_{AB}\cos30^{\circ}=F_{CB}\cos40^{\circ}$, so $F_{AB}=\frac{F_{CB}\cos40^{\circ}}{\cos30^{\circ}}$

Step2: Substitute $F_{AB}$ into the vertical - force equation

Substitute $F_{AB}=\frac{F_{CB}\cos40^{\circ}}{\cos30^{\circ}}$ into $F_{AB}\sin30^{\circ}+F_{CB}\sin40^{\circ}=200$ $\frac{F_{CB}\cos40^{\circ}}{\cos30^{\circ}}\sin30^{\circ}+F_{CB}\sin40^{\circ}=200$ $F_{CB}(\frac{\cos40^{\circ}\sin30^{\circ}}{\cos30^{\circ}}+\sin40^{\circ}) = 200$ We know that $\cos40^{\circ}\approx0.766$, $\sin30^{\circ}=0.5$, $\cos30^{\circ}\approx0.866$, $\sin40^{\circ}\approx0.643$ $\frac{\cos40^{\circ}\sin30^{\circ}}{\cos30^{\circ}}=\frac{0.766\times0.5}{0.866}\approx0.442$ $F_{CB}(0.442 + 0.643)=200$ $F_{CB}=\frac{200}{0.442 + 0.643}=\frac{200}{1.085}\approx184.33$ lb

Step3: Calculate $F_{AB}$

Since $F_{AB}=\frac{F_{CB}\cos40^{\circ}}{\cos30^{\circ}}$ $F_{AB}=\frac{184.33\times0.766}{0.866}\approx162.4$ lb

Answer:

The force in shaft AB is approximately $162.4$ lb and the force in shaft CB is approximately $184.33$ lb.