- #1

Gustav3497

m = 5[kg]

v = 2.8 [m/s]

g = 10 [m/(s^2)]

String in massless and air friction is negligible

The question is what the tension in the string is when the ball is in the bottom position. My reasoning is that the string tension only should consist of the weaker force, the gravitational force (F= mg). However, the solution says that the string tension is the sum of both the centripetal and gravitational force:

F_tension=m(v^2)/r + mg = 150 [N]

I'm having trouble buying that logic, because in a tug of war the rope tension is limited by the force with which the weaker team pulls. So why is that not the case here? Why is the tension the sum of both forces in opposite directions, and not equal to the weaker force, the gravitational one?

Thanks in advance!