A meter stick has mass 39 g, and has a pivot placed at the center at the 50 cm mark. A mass of m1 is place at the 5 cm mark. Another mass of 10 g is placed at the 70 cm mark. What mass m in grams has to be placed at the 80 cm mark to create translational and rotational equilibrium? Hint: To solve this problem you have to use the equation from Newton's second law set equal to zero and the rotational equilibrium equation.
Not sure how to do this one. aren't the newtons 2nd law and and rotational equilibrium equations the same?
2 answers
47gram
How did you come to that answer please?
0 answers