Question

In the movie “Young Frankenstein”, when the monster sits down on a see-saw, the little girl on the other end is launched into the air. Suppose the see-saw is 4 meters long, has a mass of 30kg, and pivots around its mid point.

a.) Suppose the girl weighed 40kg, and at the exact moment the see-saw was level, the girl was being accelerated at 20.2 m/s^2. What would be the magnitude of the torque of the little girl on the see-saw?

b.) At the same moment in the previous question, what was the angular acceleration of the see-saw?

c.) At the same moment as in the previous two questions, what was the net torque on the see-saw?

d.) At the same moment as the previous questions, how much torque was “the monster” exerting on the see-saw?

Answer 1

a)

F = force on the seesaw by the girl = ?

a = acceleration upward = ?

m = mass of the girl = 40 kg

force equation for the motion of girl in up direction is given as

F - mg = ma

F = m (g + a)

F = (40) (9.8 + 20.2)

F = 1200 N

L = length of the see=saw = 4 m

r = distance of the gril from the pivot = L/2 = 4/2 = 2 m

Torque is given as

= F r Sin90

= 1200 x 2 = 2400 Nm

b)

angular acceleration is given as

= a/r = 20.2/2 = 10.1 rad/s²

c)

M = mass of see-saw = 30 kg

I = moment of inertia of the see-saw = ML²/12 = (30) (4)²/12 = 40 kgm²

Net torque is given as

_net = I

_net = 40 x 10.1

_net = 404 Nm

d)

_net = _girl -  _monster

400 = 2400 - _monster

_monster = 2000 Nm