Page 234 - Fisika Terapan for Engineers and Scientists

434 CHAPTER 14 Statics and Elasticity

acts at right angles to the (horizontal) line from P to the point of application of
2
the force, the magnitude of the torque rF sin 90 for each force is simply the
product of the distance and the force, rF. According to the equilibrium con-
dition, we must set the sum of the three torques equal to zero:

T T T 0 (14.3)
bridge loc pier
4
5
45 m 9.0 10 kg g 30 m 9.0 10 kg g 90 m F 0
1
(14.4)
Here, we have chosen to reckon the first two torques as positive, since they tend
to produce counterclockwise rotation about P , and the last torque must then be
2
reckoned as negative, since it tends to produce clockwise rotation. Equation (14.4)
contains only the single unknown force F . Note that we were able to isolate this
1
unknown force by evaluating the torques about P : the other unknown force F is
2 2
absent because it produces no torque about P . Solving this equation for the
2
unknown F , we find
1
5 4
(45m 9.0 10 kg 30 m 9.0 10 kg) g
F
1
90 m
5
4.8 10 kg g
5 2 6
4.8 10 kg 9.81 m/s 4.7 10 N
Next, consider the torques about the point P .These torques are generated by
1
the weight of the bridge, the weight of the locomotive, and the upward thrust F
2
at point P (the upward thrust of F has zero moment arm and generates no torque
2 1
about P ). Setting the sum of these three torques about the point P equal to zero,
1 1
we obtain
4
5
45 m 9.0 10 kg g 60 m 9.0 10 kg g 90 m F 0
2
This equation contains only the single unknown force F (the force F is absent
2 1
because it produces no torque about P ). Solving for the unknown F , we find
1 2
6
F 5.0 10 N
2
The loads on the piers (the downward pushes of the bridge on the piers) are oppo-
site to the forces F and F (these downward pushes of the bridge on the piers are
1 2
the reaction forces corresponding to the upward thrusts of the piers on the bridge).
6
6
Thus, the magnitudes of the loads are 4.7 10 N and 5.0 10 N, respectively.
COMMENT: Note that the net vertical upward force exerted by the piers is F
1
6
F 9.7 10 N. It is easy to check that this matches the sum of the weights
2
of the bridge and the locomotive; thus, the condition for zero net vertical force,
as required for translational static equilibrium, is automatically satisfied. This
automatic result for the equilibrium of vertical forces came about because we
used the condition for rotational equilibrium twice. Instead, we could have used
the condition for rotational equilibrium once [Eq. (14.4)] and then evaluated F
2
by means of the condition for translational equilibrium [Eq. (14.1)]. The result
for zero net torque about the point P would then have emerged automatically.
1
Also note that instead of taking the bridge as the body whose equilibrium is to
be investigated, we could have taken the bridge plus locomotive as a combined
body. The downward push of the locomotive on the bridge would then not be an
external force, and would not be included in the “free-body” diagram. Instead, the