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The body in Fig. 10-40 is pivoted at O. Three forces act on FA = 10N it: at point A,8.0m from O; FB = 16N at B,4.0m from O ; FC = 19Nandat C,3.0m from O. What is the net torque about O ?

Short Answer

Expert verified

The net torque about O is, 12N.m .

Step by step solution

01

Understanding the given information

  1. Magnitude of position vector 1 is, rA=8.0m.
  2. Magnitude of position vector 2 is, rB=4.0m.
  3. Magnitude of position vector 3 is, rC=3.0m.
  4. FA=10N
  5. FB=16N
  6. FC=19N
  7. A=135.0o
  8. B=90.0o
  9. C=160.0o
02

Concept and formula used in the given question

Torqueis a turning action on a body about a rotation axis due to a force. If force is applied at a point, then total torque is the cross product of radial vector and force exerted on the body. The magnitude of torque is=rFsin

=rF=rFsin

03

Calculation for the net torque about O 

If torque causes counterclockwise rotation, the torque is considered as positive and if torque causes clockwise rotation, the torque is considered as negative.

=rF=rFsin=A+B+C=rAFAsinA-rBFBsinB+rCFCsinC

Substitute all the value in the above equation.

=8m10Nsin135-4m16Nsin90+3m19Nsin160=12.06N.m12N.m

Hence the torque is, 12N.m .

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