Forces in Two Dimensions - Mission F2D4 Detailed Help

A sign that weighs 42.6 N is being hung symmetrically by two cables that make an angle of 17.5 degrees with the horizontal. Draw a free-body diagram and perform a trigonometric analysis to determine the tension in one of the cables.

(Note: Your numbers are selected at random and likely different from the numbers listed here.)


 
Success at a problem in physics is dependent upon a carefully plotted strategy. The strategy below will prove useful in this question:
 
  • Construct a free-body diagram for the sign. Represent each of the three forces by vector arrows that point in the direction of each force; label the forces according to their type.
  • Determine the vertical component of the tension (Fy) in each cable. See Think About It section.
  • Sketch a force triangle and label the sides - Ftens for the hypotenuse and Fy for the vertical side. Label the angle Θ. See Math Magic section.
  • Using a trigonometric function, calculate the tension force from knowledge of the angle Θ and the Fy value. See Math Magic section.


 
All the individual forces acting upon the sign must balance. The cables are at an angle, so each cable has a vertical and a horizontal component of tension. Since the sign is hung symmetrically, the weight of the sign is distributed equally to each cable. Thus, the vertical component of tension is the same in each cable and equal to one-half the weight of the sign.


 
The tension in the cable is a force vector. Vectors are represented by vector arrows. Vectors such as this one have horizontal and vertical components. The components are often represented by constructing a right triangle about the vector such that the vector is the hypotenuse of the right triangle. The components are then the legs of the right triangle.
 
Trigonometric functions can be used to relate the values of the components to the value of the vector. The legendary SOH CAH TOA is applied to the force triangle in this question to give the following results.
 
Fx= Ftens• cosine Θ
Fy= Ftens• sine Θ                  
 
where Θ = angle between the cable and the horizontal


 

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