SOLUTION: A 1 m meter stick of negligible 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 =
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Question 1195922: A 1 m meter stick of negligible 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? Answer by ikleyn(52771) (Show Source):
You can put this solution on YOUR website! .
A 1 m meter stick 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?
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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
= 16*sin(30°) = 16*0.5 = 8 newtons
and
= 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. ANSWER
Solved.
As simple as a cucumber is.
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The words " of negligible mass " in the problem's formulation are excessive: this condition is not used in the solution
and does not make any influence on (as well as does not make any contribution to) a physics of this problem.
