Another service from Omega

Seesaw Torque


*****


Problem:


A 40-lb child sits on a seesa, 3 ft from the fulcrum. What torque is exerted when the child is 2 ft above the horizontal? What is the maximum torque exerted by the child?.
picture a picture here


Solution:


Let t be the angle between the horizontal and the seesaw. The torque T of the weight of the child at Q about the fulcrum P is,
T = PQ x F
where F is the force of the weight of the child from Q to the floor. Using a coordinate system with the origin at P and i to the right along the horizontal and j down along the vertical we have,

> PQ := 3*cos(t)*i+2*j;

                            PQ := 3 cos(t) i + 2 j
> F := 2*j;
                                   F := 2 j

and the torque T is,

> T := crossprod(PQ,F);

                             T := [0, 0, 6 cos(t)]

From the picture we have that,

> t := arcsin(2/3);

                               t := arcsin(2/3)

thus,

> T := T[3]*k;

                                         1/2
                                 T := 2 5    k

Clearly the maximum value of the magnitud of the torque is 6 when cos(t)=1 i.e. when t=0 (i.e. horizontal seesaw).


Link to the commands in this file
Carlos Rodriguez <carlos@math.albany.edu>
Last modified: Mon Sep 18 13:25:30 EDT 2000