8.4. Applications of Equilibrium Conditions http://www.ck12.org
Check Your Understanding
The result of Example 2 shows that the “daredevil” cannot tip the beam. The beam only extends one-third of its
length beyond the building. Even at the very edge of the beam, there is no chance of it tipping, since he would need
to be^34 Lbeyond the point,P.
What weight beam would he need in order to ensure that tipping begins at the end of the beam, thus making this feat
a bit more worthy of a daredevil?
Answer:We wish the beam to be on the verge of tipping when the daredevil reaches the end of the beam, that is,
when he is^13 Lpast the pivot point. At this point, the torque due to the weight of the beam must just balance the
torque due to the weight of the daredevil. The distance between the center of mass of the beam and the pivot^16 Lis
one-half the distance from the pivot to the edge of the beam; note,^16 is one-half of^13. Therefore the beam must weigh
twice that of the daredevil.
Illustrative Example 3 Trucking
A loaded 18-wheeler of massmt= 3. 6 × 104 kg is traveling across a flat bridge of lengthL, 75.0m and massmbof
6. 0 × 106 kg. The Free Body Diagram for the bridge is shown inFigure8.13. At the point the truck is 10.0m from
pointPwhat forces,FAandFBmust the supports provide in order for the system to remain in static equilibrium?
FIGURE 8.13
Answer:There are two unknowns in the problem, so we will need two independent equations in order to solve for
FAandFB. Equation 1 below maintains that the translational equilibrium condition is satisfied, that is, that the net
force on the bridge is zero.
Equation 1:∑F= 0 →FA+FB−mtg−mbg= 0
Equation 2 provides that the rotational equilibrium condition is satisfied, that is, that the net torque is zero. We will
compute the torques about the point,PinFigure8.13. Doing so,FAwill provide a clockwise torque and the weight
of the bridge and truck will provide counterclockwise torques. Again, the angle betweenrandFis 90◦.
Equation 2:∑τ= 0 →−FAL+mtgxt+mbgL 2 = 0
Equation 2 is used to solve forFA; useg=10m/s^2.