10 STATICS 10.5 Ladders and walls
about the pivot point to zero, we obtain
l
M g − T l sin θ = 0. (10.23)
2
Note that there is a minus sign in front of the second torque, since this torque
clearly attempts to twist the rod in the opposite sense to the first.
Equations (10.21) and (10.22) can be solved to give
cos φ
T =
sin(θ + φ)M g, (10.24)cos θ
R =
sin(θ + φ)M g. (10.25)Substituting Eq. (10.24) into Eq. (10.23), we obtain
sin(θ + φ) = 2 sin θ cos φ. (10.26)The physical solution of this equation is φ = θ (recall that sin 2 θ = 2 sin θ cos θ),
which determines the direction of the reaction at the pivot. Finally, Eqs. (10.24)
and (10.25) yield
T = R =M g
2 sin θ ,^ (10.27)^
which determines both the magnitude of the tension in the cable and that of the
reaction at the pivot.
One important point to note about the above solution is that if φ = θ then thelines of action of the three forces—R, M g, and T —intersect at the same point,
as shown in Fig. 92. This is an illustration of a general rule. Namely, whenever a
rigid body is in equilibrium under the action of three forces, then these forces are
either mutually parallel, as shown in Fig. 91 , or their lines of action pass through
the same point, as shown in Fig. 92.
10.5 Ladders and walls
Suppose that a ladder of length l and negligible mass is leaning against a vertical
wall, making an angle θ with the horizontal. A workman of mass M climbs