Sir Lost-a-Lot dons his armor and... Sir Lost-a-Lot dons his armor and sets out
ID: 1648318 • Letter: S
Question
Sir Lost-a-Lot dons his armor and...
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 theta = 20.0 degree 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 l = 7.50 m long and has a mass of 2 200 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 950 kg. (a) Determine the tension in the cable. IN (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 NExplanation / 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 = (2200kg * ½*7.5m + 950kg * 6.5m)*9.8m/s²*cos20º - T*sin85.46º*5m
where T is the cable tension. Solving, find
T = 132839.64N·m / (5m*sin85.46) = 26651N 26650 N
(b) ß = 180º - 70º - 85.46º = 24.54º angle between cable and wall
horizontal force Fx = T*sin24.54 = 11 358 N 11 400 N
away from wall
(c) vertical force Fy = (2200 + 950)kg * 9.8m/s² - T * cos24.54 = 33935N
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