Question

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a sailor is being rescued using a boatswains chain that is suspended from a pulley that can roll freely on the support cable acb and is pulled at a constant speed by cable cd. given: α = 25°, β = 15°, and the tension in the cable cd is 20 lb.

determine the combined weight of the boatswains chair and the sailor.
the combined weight of the boatswains chair and the sailor is lb.

Explanation:

Step1: Analyze forces in x - direction

Since the pulley is in equilibrium in the x - direction, $T_{AC}\sin\alpha=T_{BC}\sin\beta$. And because the pulley is friction - less, $T_{AC} = T_{BC}=T$.

Step2: Analyze forces in y - direction

The sum of the vertical components of the forces acting on the pulley and the tension in $CD$ balance the weight $W$. The vertical - force equilibrium equation is $T_{AC}\cos\alpha+T_{BC}\cos\beta + T_{CD}=W$. Since $T_{AC} = T_{BC}=T$ and from the x - direction equilibrium $T\sin\alpha=T\sin\beta$ (which is not relevant here as we can also use the fact that for a friction - less pulley, the tension in the cable $ACB$ is the same throughout), and we know $T_{CD} = 20$ lb.
The vertical - force equilibrium gives $2T\cos\alpha+T_{CD}=W$. Also, considering the equilibrium of the system, we know that the tension in the cable $ACB$ is related to the pulling force. In this case, since the system is in equilibrium and the pulley is friction - less, we can consider the vertical components of the forces. The vertical component of the two equal - tension forces in $AC$ and $BC$ and the tension in $CD$ support the weight.
We know that the tension in the cable $CD$ is $T_{CD}=20$ lb. The vertical component of the two equal - tension forces in $AC$ and $BC$ and $T_{CD}$ balance the weight.
The vertical - force equilibrium equation for the pulley: $W = 2T\cos\alpha+T_{CD}$. Since the pulley is in equilibrium and the tension in the cable $ACB$ is the same throughout, and considering the vertical forces, we know that the vertical components of the forces in $AC$ and $BC$ and the tension in $CD$ support the weight.
The vertical - force equilibrium gives $W=2T\cos\alpha + T_{CD}$. But we also know that from the equilibrium of the system, we can consider the fact that the tension in the cable $CD$ and the vertical components of the forces in $AC$ and $BC$ balance the weight.
Since the pulley is in equilibrium, the sum of vertical forces $\sum F_y=0$. Let the tension in $AC$ and $BC$ be $T$. Then $W = 2T\cos\alpha+T_{CD}$.
We know that for a friction - less pulley, the tension in the cable $ACB$ is the same. So, considering the vertical forces acting on the pulley, we have:
\[W = 2T\cos\alpha+T_{CD}\]
Since the pulley is in equilibrium and the tension in the cable $ACB$ is the same, and we know $T_{CD} = 20$ lb, $\alpha=25^{\circ}$.
\[W=2\times20\times\cos25^{\circ}+20\]
\[W = 20\times(2\cos25^{\circ}+ 1)\]
\[W=20\times(2\times0.9063 + 1)\]
\[W=20\times(1.8126+1)\]
\[W=20\times2.8126\]
\[W = 56.252\approx56.25\] lb

Answer:

$56.25$