Sir Lost-a-Lot dons his armor and sets out from the castle on his trusty steed (
ID: 1502862 • Letter: S
Question
Sir Lost-a-Lot dons his armor and sets out from the castle on his trusty steed (see figure below). Usually, the drawbridge is lowed to a horizontal position so that the end of the bridge rests on the stone ledge. Unfortunately Lost-a-Lot's squire didn't lower the drawbridge far enough and stopped it at = 20.0° above the horizontal. The knight and his horse stop when their combined center of mass is d = 1.00 m from the end of the bridge. The uniform bridge is = 7.50 m long and has a mass of
2 100 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 h = 12.0 m above the bridge. Lost-a-Lot's mass combined with his armor and steed is 1 050 kg. (a) Determine the tension in the cable.Be careful in determining the angle that the cable makes with the bridge. It is not 90°. N
(b) Determine the horizontal force component acting on the bridge at the hinge. magnitude N direction
(c) Determine the vertical force component acting on the bridge at the hinge.
magnitude magnitude N direction
Explanation / Answer
We need the angle the cable makes with the bridge.
The angle between the bridge and the wall (vertical) is 70º.
The law of cosines gives us
C² = 5² + 12² - 2*5*12*cos70º = 128
C = 11.3 m length of cable
Now we can use the law of sines to get the angle between the bridge and cable:
sin / 12m = sin70º / 11.3m
sin = 0.997
= arcsin0.997 = 85.46º
(a) Now sum the moments about the hinge:
M = 0 = (2100kg * ½*7.5m + 1050kg * 6.5m)*9.8m/s²*cos20º - T*sin85.46º*5m
where T is the cable tension. Solving, find
T = 135372,11895N·m / (5m*sin85.46) = 27159.6421644 N 27 160 N
(b) ß = 180º - 70º - 85.46º = 24.54º angle between cable and wall
horizontal force Fx = T*sin24.54 = 11 280.171 N
away from wall
(c) vertical force Fy = (2100 + 1050)kg * 9.8m/s² - T * cos24.54 = 6163.32 N
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