Sir Lost-a-Lot dons his armor and sets out from the castle on his trusty steed i
ID: 1638117 • 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 degree 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 7.00 m long and has a mass of 2200 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 1000 kg. (a) Determine the tension in the cable. (b) Determine the horizontal force component acting on the bridge at the hinge. magnitude direction to the right to the left (c) Determine the vertical force component acting on the bridge at the hinge. magnitude direction upwards downwardsExplanation / Answer
If h is the distance from the hinge to the cable connection on the wall and
d is the distance from the hinge to the cable connection on the bridge and
L is the length of the cable, then
L² = d² + h² - 2dhcos
where L is the cable length
and = angle between wall and bridge = 90º + 20º = 110º. So
L² = (5.00m)² + (12.0m)² - 2*5.00m*12.0m*cos110º = 210 m²
L = 14.5 m
Then the law of sines gives
L / sin = h / sin
where is the angle between the cable and the bridge.
14.5m / sin110 = 12.0 / sin
sin = 0.778
= 51.1º
a) For a coordinate system oriented with the bridge, summing the moments about the hinge gives
M = 0 = T * 5.00m * sin51.1º - [2200kg*3.5m + 1000kg*6.00m] * 9.8m/s² * cos20.0º
tension T = 32422.5 N
(b) Fx = T*cos(+20º) = 32422.5 * cos71.1º = 10502.2 N
(c) Fy = 32422.5 * sin71.1º - (2200+1000)kg * 9.8m/s² = - 685.5 N