Solving Physics Problem: Moment = F × d = 60X20 = 50d, d = 24 cm

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The discussion centers on solving a physics problem involving moments, where the equation used is Moment = F × d. The calculation presented is 60 × 20 = 50d, leading to the conclusion that d equals 24 cm. It is noted that the weight of the rod was overlooked in the initial calculations. The total torque is set to zero, indicating that the sum of counterclockwise and clockwise torques around the pivot at 40 cm must balance. The conversation emphasizes the importance of considering all forces, including the rod's weight, in static equilibrium problems.
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Homework Statement


http://www.xtremepapers.com/CIE/International%20A%20And%20AS%20Level/9702%20-%20Physics/9702_s11_qp_13.pdf


question 12


Homework Equations


moments=Fd


The Attempt at a Solution


60X20=50d
d=24cm
 
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Yo forgot the weight of the rod. ehild
 


Static of Rigid Bodies.

Total sum of Torque=0
\sum \tau=0

find ccw(counterclockwise) and cw(clockwise) torgue about pivot(40cm).
 
Thread 'Collision of a bullet on a rod-string system: query'

Thread 'Collision of a bullet on a rod-string system: query'

In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
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