Question

A see-saw made from 4 meter board sits on a pivot at its center. A child weighing 280 N sits 1.75 m away from the center. A second child who weighs 330 N sits on the other side such that the board remains horizontal and steady.

Now you come and push on the end of the seesaw nearest the second child with a force of 110 N. You don’t push straight up, but at an angle of 25° from the vertical – tilted a bit toward the center of the seesaw. The first child moves to maintain equilibrium.

(d) Based on your intuition, which way should the child move? Towards or away from the center?

(e) Calculate how far she is now from the center?

(f) What is the force now (magnitude and direction) that the pivot applies to the board?

Answer 1

initial condition (before applying force):

let the second child sits at a distance of d from the center.

then for torque balancing about the pivot,

280*1.75=330*d

==>d=280*1.75/330=1.4848 m from the center

final condition:(after the force is applied)

as the vertical component of the force applied by you in vertically upward direction, direction of torque due to this component (as torque due to horizontal component of force applied =0) will be same as direction of torque applied by 280 N (first child)

hence to make the total torque equal to torque due to 330 N, the 280 N child has to move closer to the center in order to balance the torque

let her new position be d1.

then 280*d1+110*cos(25)*2=330*1.4848

==>d1=(330*1.4848-110*cos(25)*2)/280=1.0378 m

part c:

net vertical force on the board=280+330-110*cos(25)=510.31 N,in downward direction

net horizontal force on the board=110*sin(25)=46.488 N, to the right

force due to pivot will be equal and opposite to the force applied on the board in order to maintain equilibrium

hence vertical force by the pivot=510.31 N, in vertically upward direction

horizontal force by pivot=46.488 N, to the left

net force=sqrt(46.488^2+510.31^2)=512.42 N

direction with left side horizontal=arctan(510.31/46.488)=84.795 degrees

then angle with right side =180-84.795=95.205 degrees