Initially, the wrecking ball A of mass m is held stationary by the horizontal ca
ID: 2114981 • Letter: I
Question
Initially, the wrecking ball A of mass m is held stationary by the horizontal cable AB . The cable AB is then released so that the wrecking ball A starts swinging about the %uFB01xed point O . Determine the tension in the cable OA before the cable AB is released and immediately after it is released. What is the percent change in cable tension if %u03F4=30?
Explanation / Answer
before the cable AB is released , force equation in vertical direction would give =Tsin theta = mg =>T = mg / cos (theta) = 2mg/sqrt(3) (theta =30 degrees) after the cable AB is released , net force will be 0 in centrifugal (cable ) direction => T = mgcos theta = mgsqrt(3) / 2 => percent change = 25%