Question 1195922
.
A 1 m meter stick {{{highlight(cross(of))}}} {{{highlight(cross(negligible))}}} {{{highlight(cross(mass))}}} has center of rotation at point O 
at one end of the stick and two force acted on the other end of the stick. 
Given F1 = 16 N with 30° and F2 = 8 N with -40°, what is the magnitude 
of the net torque on the meter stick about the axis of rotation O?
~~~~~~~~~~~~~~~~~


<pre>
You can imagine that the point O (the rotation center) is the origin 
of the coordinate system and that the stick is placed
along the positive direction of x-axis.


You are given forces F1 and F2 acting at the other end of the stick at
angle 30° and angle -40° to this axis.


Since they ask about the torque, all you need is to know VERTICAL components 
of the forces. They are

    {{{F1[y]}}} = 16*sin(30°) = 16*0.5 = 8 newtons

and

    {{{F2[y]}}} = 8*sin(-40°) = -8*0.6428 = -5.1424 newtons.


So, the net force acting at the end of the stick has y-component of 8 - 5.1424 = 2.8576 newtons,

and therefore the magnitude of the net torque is

    M = 2.8576 N * 1 m = 2.8576 N*m.    <U>ANSWER</U>
</pre>

Solved.


As simple as a cucumber is.



///////////////



The words &nbsp;" of negligible mass " &nbsp;in the problem's formulation are excessive: &nbsp;this condition is not used in the solution 

and does not make any influence on &nbsp;(as well as does not make any contribution to) &nbsp;a physics of this problem.