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  5. Examples of static equilibrium

12.2 Examples of static equilibrium  (Page 8/9)

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We select the pivot at point P (upper hinge, per the free-body diagram) and write the second equilibrium condition for torques in rotation about point P :

pivot at P : τ w + τ B x + τ B y = 0 .

We use the free-body diagram to find all the terms in this equation:

τ w = d w sin ( − β ) = − d w sin β = − d w b / 2 d = − w b 2 τ B x = a B x sin 90 ° = + a B x τ B y = a B y sin 180 ° = 0 .

In evaluating sin β , we use the geometry of the triangle shown in part (a) of the figure. Now we substitute these torques into [link] and compute B x :

pivot at P : − w b 2 + a B x = 0 ⇒ B x = w b 2 a = ( 400.0 N ) 1 2 · 2 = 100.0 N.

Therefore the magnitudes of the horizontal component forces are A x = B x = 100.0 N . The forces on the door are

at the upper hinge: F → A on door = −100.0 N i ^ + 200.0 N j ^ at the lower hinge: F → B on door = + 100.0 N i ^ + 200.0 N j ^ .

The forces on the hinges are found from Newton’s third law as

on the upper hinge: F → door on A = 100.0 N i ^ − 200.0 N j ^ on the lower hinge: F → door on B = −100.0 N i ^ − 200.0 N j ^ .

Significance

Note that if the problem were formulated without the assumption of the weight being equally distributed between the two hinges, we wouldn’t be able to solve it because the number of the unknowns would be greater than the number of equations expressing equilibrium conditions.

Check Your Understanding Solve the problem in [link] by taking the pivot position at the center of mass.

F → door on A = 100.0 N i ^ − 200.0 N j ^ ; F → door on B = −100.0 N i ^ − 200.0 N j ^

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Check Your Understanding A 50-kg person stands 1.5 m away from one end of a uniform 6.0-m-long scaffold of mass 70.0 kg. Find the tensions in the two vertical ropes supporting the scaffold.

Figure is a schematic drawing of a woman standing 1.5 m away from one end and 4.5 m away from another end of a scaffold.

711.0 N; 466.0 N

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Check Your Understanding A 400.0-N sign hangs from the end of a uniform strut. The strut is 4.0 m long and weighs 600.0 N. The strut is supported by a hinge at the wall and by a cable whose other end is tied to the wall at a point 3.0 m above the left end of the strut. Find the tension in the supporting cable and the force of the hinge on the strut.

Figure is a schematic drawing of a sign which hangs from the end of a uniform strut. The strut is 4.0 m long and is supported by a 5.0 m long cable tied to the wall at a point 3.0 m above the left end of the strut.

1167 N; 980 N directed upward at 18 ° above the horizontal

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Summary

  • A variety of engineering problems can be solved by applying equilibrium conditions for rigid bodies.
  • In applications, identify all forces that act on a rigid body and note their lever arms in rotation about a chosen rotation axis. Construct a free-body diagram for the body. Net external forces and torques can be clearly identified from a correctly constructed free-body diagram. In this way, you can set up the first equilibrium condition for forces and the second equilibrium condition for torques.
  • In setting up equilibrium conditions, we are free to adopt any inertial frame of reference and any position of the pivot point. All choices lead to one answer. However, some choices can make the process of finding the solution unduly complicated. We reach the same answer no matter what choices we make. The only way to master this skill is to practice.

Conceptual questions

Is it possible to rest a ladder against a rough wall when the floor is frictionless?

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Show how a spring scale and a simple fulcrum can be used to weigh an object whose weight is larger than the maximum reading on the scale.

(Proof)

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Source:  OpenStax, University physics volume 1. OpenStax CNX. Sep 19, 2016 Download for free at http://cnx.org/content/col12031/1.5
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