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Physics LibreTexts

18.3: Generalized Lever Law

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We can extend the Lever Law to the case in which two external forces F1 and F2 are acting on the pivoted beam at angles θ1 and θ2 with respect to the horizontal as shown in the Figure 18.4. Throughout this discussion the angles will be limited to the range [0θ1,θ2π]. We shall again neglect the thickness of the beam and take the pivot point to be the center of mass.

clipboard_e1aa4feef68e554bf6942e8276fe3203e.png
Figure 18.4 Forces acting at angles to a pivoted beam.

The forces F1 and F2 can be decomposed into separate vectors components respectively (F1,,F1,) and (F2,,F2,), where F1, and F2, are the horizontal vector projections of the two forces with respect to the direction formed by the length of the beam, and F1, and F2, are the perpendicular vector projections respectively to the beam (Figure 18.5), with

F1=F1,+F1,

F2=F2,+F2,

clipboard_e7bed16456e1c198297db730118780ae9.png
Figure 18.5 Vector decomposition of forces.

The horizontal components of the forces are

F1,=F1cosθ1

F2,=F2cosθ2

where our choice of positive horizontal direction is to the right. Neither horizontal force component contributes to possible rotational motion of the beam. The sum of these horizontal forces must be zero,

F1cosθ1F2cosθ2=0

The perpendicular component forces are

F1,=F1sinθ1

F2,=F2sinθ2

where the positive vertical direction is upwards. The perpendicular components of the forces must also sum to zero,

Fpivot mbg+F1sinθ1+F2sinθ2=0

Only the vertical components F1, and F2, of the external forces are involved in the lever law (but the horizontal components must balance, as in Equation (18.3.5), for equilibrium). Then the Lever Law can be extended as follows.

Generalized Lever Law A beam of length l is balanced on a pivot point that is placed directly beneath the center of mass of the beam. Suppose a force F1 acts on the beam a distance d1 to the right of the pivot point. A second force F2 acts on the beam a distance d2 to the left of the pivot point. The beam will remain in static equilibrium if the following two conditions are satisfied:

1) The total force on the beam is zero,

2) The product of the magnitude of the perpendicular component of the force with the distance to the pivot is the same for each force,

d1|F1,|=d2|F2,|

The Generalized Lever Law can be stated in an equivalent form,

d1F1sinθ1=d2F2sinθ2

We shall now show that the generalized lever law can be reinterpreted as the statement that the vector sum of the torques about the pivot point S is zero when there are just two forces F1 and F2 acting on our beam as shown in Figure 18.6.

clipboard_e69d291e8f6652a3130da241dcf610325.png
Figure 18.6 Force and torque diagram.

Let’s choose the positive z -direction to point out of the plane of the page then torque pointing out of the page will have a positive z -component of torque (counterclockwise rotations are positive). From our definition of torque about the pivot point, the magnitude of torque due to force F1 is given by

τS,1=d1F1sinθ1

From the right hand rule this is out of the page (in the counterclockwise direction) so the component of the torque is positive, hence,

(τS,1)z=d1F1sinθ1

The torque due to F2 about the pivot point is into the page (the clockwise direction) and the component of the torque is negative and given by

(τS,2)z=d2F2sinθ2

The z -component of the torque is the sum of the z -components of the individual torques and is zero,

(τS, total )z=(τS,1)z+(τS,2)z=d1F1sinθ1d2F2sinθ2=0

which is equivalent to the Generalized Lever Law, Equation (18.3.10),

d1F1sinθ1=d2F2sinθ2


This page titled 18.3: Generalized Lever Law is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Peter Dourmashkin (MIT OpenCourseWare) via source content that was edited to the style and standards of the LibreTexts platform.

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