Question 2.SP.6: For a new sailboat, a designer wants to determine the drag f......
For a new sailboat, a designer wants to determine the drag force that may be expected at a given speed. To do so, she places a model of the proposed hull in a test channel and uses three cables to keep its bow on the centerline of the channel. Dynamometer readings indicate that for a given speed, the tension is 40 lb in cable AB and 60 lb in cable AE. Determine the drag force exerted on the hull and the tension in cable AC.
STRATEGY: The cables all connect at point A, so you can treat that as a particle in equilibrium. Because four forces act at A (tensions in three cables and the drag force), you should use the equilibrium conditions and sum forces by components to solve for the unknown forces.
- Tension in cable AB: 40 lb
- Tension in cable AE: 60 lb
Final Answer
MODELING and ANALYSIS:
Determining the Angles. First, determine the angles α and β defining the direction of cables AB and AC:
\tan \alpha=\frac{7 ~\mathrm{ft}}{4 ~\mathrm{ft}}=1.75 \quad \tan \beta=\frac{1.5~ \mathrm{ft}}{4~ \mathrm{ft}}=0.375 \\\alpha=60.26^{\circ} \quad \beta=20.56^{\circ}Free-Body Diagram. Choosing point A as a free body, draw the free-body diagram (Fig. 1). It includes the forces exerted by the three cables on the hull, as well as the drag force \mathbf{F}_D exerted by the flow.
Equilibrium Condition. Because point A is in equilibrium, the resultant of all forces is zero:
\mathbf{R}=\mathbf{T}_{A B}+\mathbf{T}_{A C}+\mathbf{T}_{A E}+\mathbf{F}_D=0 (1)
Because more than three forces are involved, resolve the forces into x and y components (Fig. 2):
\begin{aligned}\mathbf{T}_{A B} & =-(40~ \mathrm{lb}) \sin 60.26^{\circ} \mathbf{i}+(40~ \mathrm{lb}) \cos 60.26^{\circ} \mathbf{j} \\& =-(34.73 ~\mathrm{lb}) \mathbf{i}+(19.84 ~\mathrm{lb}) \mathbf{j} \\\mathbf{T}_{A C} & =T_{A C} \sin 20.56^{\circ} \mathbf{i}+T_{A C} \cos 20.56^{\circ} \mathbf{j} \\& =0.3512 T_{A C} \mathbf{i}+0.9363 T_{A C} \mathbf{j} \\\mathbf{T}_{A E} & =-(60~ \mathrm{lb}) \mathbf{j} \\\mathbf{F}_D & =F_D \mathbf{i}\end{aligned}Substituting these expressions into Eq. (1) and factoring the unit vectors i and j, you have
\left(-34.73~ \mathrm{lb}+0.3512 T_{A C}+F_D\right) \mathbf{i}+\left(19.84 ~\mathrm{lb}+0.9363 T_{A C}-60~ \mathrm{lb}\right) \mathbf{j}=0This equation is satisfied if, and only if, the coefficients of i and j are each equal to zero. You obtain the following two equilibrium equations, which express, respectively, that the sum of the x components and the sum of the y components of the given forces must be zero.
\left(\Sigma F_x=0:\right) \quad-34.73~ \mathrm{lb}+0.3512 T_{A C}+F_D=0 (2)
\left(\Sigma F_y=0:\right) \quad 19.84~ \mathrm{lb}+0.9363 T_{A C}-60~ \mathrm{lb}=0 (3)
From Eq. (3), you find
T_{A C}=+42.9 ~\mathrm{lb}Substituting this value into Eq. (2) yields
F_D=+19.66 ~\mathrm{lb}REFLECT and THINK: In drawing the free-body diagram, you assumed a sense for each unknown force. A positive sign in the answer indicates that the assumed sense is correct. You can draw the complete force polygon (Fig. 3) to check the results.
