5 kg*40cm = m kg *30 cm
m = 5(4/3) = 20/3 kg
a meter rod is pivoted at the center 5kg body is attched 10cm from the one end 30cm from the pivot at the other side how much mass be connected to keep the meter in equilibrium
2 answers
A uniform rod of mass 5kg is an lenghth 50cm is pivoted at the 30cm mark. what force would be applied at the 50cm mark to keep the rod in equilibrium?
1 answer