Why is it the tension BC smallest when BC is perpendicular to AC???? For #2.89,

Question

Why is it the tension BC smallest when BC is perpendicular to AC????

For #2.89, the cable tension is said to be 385N, but when solving for the problem, why is it that the tension in DB is 385N also??  I figured that 385N would be the resultant vector of DB+ BE. What is the theorem responsible for this ?

Fig. P2.46 Knowing that the tension in cable AC is 2130 N, determine the components of the force exerted on the plate at C. A frame ABC is supported in part by cable DBE that passes through a frictionless ring at B. Knowing that the tension in the cable is 385 N, determine the components of the force exerted by the cable on the support at D.

Explanation / Answer

It seems that you are not completely comfortable with the whole concept of tension. The doubts that you have raised here are surelly because of the difficulty in grasping the concept rather than the mathematical skills.

The first problem that you mentioned can be solved if you assume the angle at B as "x" and then solve for the tension in BC in terms of "x". Thus, you will arrive at the expression of tension in BC in terms of "X". Using the concept of maxima & minima, you can easily determine the angle where tension in BC is max.

Now, Let me explain to you the concept of tension through a very straigh-forward example that would surely help you in future. The tension in a massless string or a weightless truss member is uniform throughout the length of the string/member. This is the most important thing that you need to keep in mind. Remember that it is valid only for a weightless string/member. SInce, it isnt mentioned, it can be safely assumed that the string/member is weightless.

Further, in problem 2, you seem to be thinking that since the string passes through a ring, the tension in the two portions od the string should be different. However, if you look at the question, it says that the ring is frictionless. Thus the ring doesnot restrict the movement of the string in anyway. So the tension in both the parts of the ring will be same and equal to the overall tension in the string.

Now let me tell you a very easy approach to tackle such problems. Whenever there is tension involved in a string/member and you have to include the tension force in some free body diagram, imagine that the string has been cut in the middle. Then there will be two parts of the string, each pulling the other towards itself with a force equal to the tension in the string. This tension is infact what holds the two parts together. (note that this is true only when string is weighless and the joints etc, are frictionless)

If you still have doubt, you can comment here..!!!

Cheers

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