**Problem 2B-01: Lever**

A lever is shown in Fig. 1. A rigid massless rod AB is supported at point O. The length
of AO is *a*, and the length of BO is *b*. A weight of mass *m*
is placed at point A. A vertically downward force of magnitude F is applied at
point B to lift the weight. What is the minimum value of F needed to lift the weight?

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solution

**Solution:**

Let the rod forms an angle θ with the horizontal line as shown in Fig. 1. When
the weight at point A is touching the ground, the vertically downward gravitational
pull of magnitude *m**g* is cancelled out by the vertically upward
resistance of equal magnitude from the ground acting on point A so there is no
torque created abound point O. However, once the weight is lifted up from the
ground, even by a minuscule amount, the resistance force from the ground
disappears, and the gravitational pull on the weight will generate a toque of
the magnitude τ_{A} to rotate the rod clockwise around point O.
The downward force at the end of point B creates a torque of the magnitude
τ_{B} that will try to rotate the rod counter-clockwise around
point O. Thus the minimum F required is to satisfy the equation
τ_{A} = τ_{B}. According
to Eq. (2) of Topic 2B-01, we have

τ_{A} = *m**g*·*a*·cosθ ,

and

τ_{B} = *m**g*·*b*·cosθ .

Thus τ_{A} = τ_{B} leads to

F = *m**g*·(*a*/*b*) .

To make the use of lever meaningful we must have *a* < *b* so that a
smaller force F can overcome a larger gravitational pull *m**g*.