Sir Lost-a-Lot dons his armor and sets out from the castle on his trusty steed i
ID: 1410096 • Letter: S
Question
Sir Lost-a-Lot dons his armor and sets out from the castle on his trusty steed in his quest to improve communication between damsels and dragons. Unfortunately his squire lowered the draw bridge too far and finally stopped it 20.0 degrees below the horizontal. Lost-a-Lot and his horse stop when their combined center of mass is 1.00 m from the end of the bridge. The uniform bridge is 9.50 m long and has a mass of 2300 kg. The lift cable is attached to the bridge 5.00 m from the hinge at the castle end, and to a point on the castle wall 12.0 m above the bridge. Lost-a-Lot's mass combined with his armor and steed is 800 kg. Determine the tension in the cable. Your response differs from the correct answer by more than 10%. Double check your calculations. Determine the horizontal force component acting on the bridge at the hinge, magnitude Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully, direction to the right to the left Determine the vertical force component acting on the bridge at the hinge, magnitude Your response differs significantly from the correct answer. Rework your solution from the beginning and check step carefulExplanation / Answer
Fx = 0
Fx = Rx - T cos 64.2 o = 0
Rx = T cos 64.2 o = 0.435 T
Fy = 0
Fy = Ry + T sin 64.2 o - W - w = 0
W = (2300 kg) (9.8 m/s2) = 22540 N
w = (800 kg) (9.8 m/s2) = 7840 N
Ry + 0.900 T sin 64.2 o = 30380N
As we should expect by now, at this stage, we have three unknowns -- Rx, Ry, and T -- but only two equations. We will get the third equation from applying the second condition of equilibrium, that the sum of the torques must equal zero. We will calculate the torques about the hinge of the drawbridge.
Rx: tou = 0
Ry: tou = 0
T:tou = tou ccw = (5 m) T sin 44.2o = (5 m) (T) (0.697) =
= 3.49 m T
W: tou= tou cw = (4 m) (2254 N) sin 90o = 90160 N-m
w:tou = tou cw = (8.5 m) (7840 N) sin 90o = 66640 N-m
tou ccw = tou cw
3.49 m T = 156800 N-m
T = 44928.36N
Now that the tension is known, we can go back and determine the "reaction force" Rx and Ry,
Rx = 0.435 T
Rx =18 322 N
Ry = 30380 N - 0.900 T sin 64.2 o
Ry = 30380 N - 34 129 N
Ry = - 3749 N
What does this negative sign mean? When I guessed that Ry pointed up, I guessed wrong! The forces are such -- and they are located such -- that the hinge exerts a downward force so Ry points down and that is shown by the negative value we calculate for Ry.
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