Unit 2 - Forces - Tension in Strings FRQ

Please enter your responses in paragraph form, using equations as necessary to support your claim. I’ll go through on Friday (4/10) and critique your responses.

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None of them

Attached are my FBD illustrations.

In both cases, the y component of the diagonal string is equal to mg. However, as theta decreases, the tension becomes more and more horizontal and less vertical, meaning that a greater total tension in the string is necessary in Case B. Because the sine of a small angle is smaller than the sine of a larger one, which gives a vertical force component, the value for T1sin() must increase. Therefore, T1 is greater in Case B. This is true because mg = T1sin(), and () is smaller in Case B, necessitating a greater total T1 tension. As for the horizontal string (on the left) tension, it is also greater because, in both cases, T2 = T1cos(). () is smaller in case B, and the cosine of a small angle is greater than the cosine of a larger one. Therefore T is greater in both strings in case B.

John,
Nice Job! You got it exactly!
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