Question

What would be the total torque of two "children" on a teetertotter if the one on the right has a mass of 150 kg located 2.5 m for the fulcrum, and the one on the left has a mass of 80 kg located 3.5 m from the fulcrum? The board has a mass/unit length of 10 kg/m

Answer 1

M_r = 150kg mass of right side children.

d_r = 2.5m distance of right children.

M_l = 80kg mass of right side children.

d_l = 3.5m distance of right children.

Board mass/length u=10kg/m

Mg force is acting on children and board downwards.

Now caculating torque in right side

Torque for children

T = f(force) ×distance

= m_rg×d_r

=150×9.8×2.5

=3675 Nm

Now torque for board.

As distance is variable we have to integrate it.

Take same strip dx at d=x

Hence mass of it m= udx

Force on it mg = 10dx×9.8

Now for torque in board in right side = fd

As x is variable from 0 to 2.5

F = 98×x×dx

F = 306.25N

So total Torque on right side 3675+306.25 =3981.25Nm

Now for left side

Torque for children

T = f(force) ×distance

= m_lg×d_l

= 80×9.8×3.5

=2744

Now Torque on board

As x is variable from 0 to 3.5

T = 98×x×dx

T = 600.25Nm

Total torque on left side = 2744+600.25

=3344.25Nm

As both torque is opposite to each other as one children is move downward other move upward

So resultant torque

T= 3981.25 - 3344.25

T = 637Nm