10 STATICS 10.6 Jointed rods
→thermore, the torques associated with these two forces act in opposite directions.
Hence, setting the net torque about the bottom of the ladder to zero, we obtain
M g x cos θ − S l sin θ = 0. (10.30)The above three equations can be solved to give
R = M g, (10.31)and
f = S =x (^)
l tan θ
M g. (10.32)
Now, the condition for the ladder not to slip with respect to the ground is
f < μ R. (10.33)
This condition reduces to
x < l μ tan θ. (10.34)
Thus, the furthest distance that the workman can climb along the ladder before
it slips is
xmax = l μ tan θ. (10.35)
Note that if tan θ > 1/μ then the workman can climb all the way along the ladder
without any slippage occurring. This result suggests that ladders leaning against
walls are less likely to slip when they are almost vertical (i.e., when θ 90 ◦).
10.6 Jointed rods
Suppose that three identical uniform rods of mass M and length l are joined
together to form an equilateral triangle, and are then suspended from a cable, as
shown in Fig. 94. What is the tension in the cable, and what are the reactions at
the joints?
Let X 1 , X 2 , and X 3 be the horizontal reactions at the three joints, and let Y 1 , Y 2 ,
and Y 3 be the corresponding vertical reactions, as shown in Fig. 94. In drawing
this diagram, we have made use of the fact that the rods exert equal and opposite