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

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 = 6.50 m long and has a mass of 1 600 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 1 100 kg.

(a) Determine the tension in the cable.
N

(b) Determine the horizontal force component acting on the bridge at the hinge.

(c) Determine the vertical force component acting on the bridge at the hinge.
magnitude

Explanation / Answer

The angle between the bridge and wall (vertical) = 90-20 = 700
The law of cosines gives us
C² = 5² + 12² - 2X5X12Xcos70º = 128
C = 11.3 m

This is the length of cable

Now we can use the law of sine’s to get the angle between the bridge and cable:
sin/ 12m = sin70º / 11.3m
sin = 0.997
= sin-1 0.997 = 85.46º

(a) Now sum the moments about the hinge:
M = 0 = (1600kg X1/2X6.5m + 1100kg X 1/2X12m)X9.8m/s² X cos20º - T X sin85.46º X 5m
where T is the cable tension. Solving, find
T = 13237.3N-m / (5m X sin85.46) = 21801N

This is the required tension

(b) ß = 180º - 70º - 85.46º = 24.54º

This is the angle between cable and wall
horizontal force Fx = T X sin24.54 = 10300 N

away from wall

(c) vertical force Fy = (1600 + 1100)kg X 9.8m/s² - T X cos24.54 = 29087 N
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