2. Balancing forces on a modified Seesaw:
Consider the same seesaw (M = 40 kg and length L = 5.0 m), but now the pivot P is at the far
end, with a child of $m_2$ = 30 kg sitting a distance $r_2$ = 4.0 m to the left of the pivot. To balance
the net torque on the beam, a perpendicular rope that can exert a torque via upwards tension
force is attached to the beam at a distance r = 1.0 m from the pivot. Assume the torque from
the child and the beam balances the torque from the rope's tension, so that the system is in
rotational equilibrium, at rest.
a) Calculate the upward force F (tension in the rope) in terms of M, $m_2$, $r_2$, r, L, and g, and
then plug in the values above to find the force F that is needed for equilibrium.
b) Now, consider that the rope tension balancing the torque is at an angle $\theta$ = 120°, rather
than being vertical. Calculate the new force F' in terms of M, $m_2$, $r_2$, r, L, and g, and
then plug in the values above to find the force F' that is needed for equilibrium.
