Sir Lost-a-Lot dons hi armor and sets out from the castle on his trusty steed in

Question

Sir Lost-a-Lot dons hi 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 is stopped it 20.0° 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 9.00 m long and has a mass of 1800 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. Hint: First determine all the angles and lengths of the triangle made by the wall, the cable, and the drawbridge. (a) Determine the tension in the cable. 3092.05 N X Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. (b) Determine the horizontal force component acting on the bridge at the hinge. magnitude 1494.3 N X 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 (c) Determine the vertical force component acting on the bridge at the hinge. magnitude 106.98 N X Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. direction upwards downwards

Explanation / Answer

Ans:-

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 - [1800kg*4.00m + 800kg*7.00m] * 9.8m/s² * cos20.0º
tension T = 30292.6 N

(b) Fx = T*cos(+20º) = 30292.6N * cos71.1º = 9812.3 N
upward
(c) Fy = 30292.6N * sin71.1º - (1800+800)kg * 9.8m/s² =- 22614.12 N

Downward direction

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