Y 8.-15 pointssenPSE9 12 P019 Sir Lost-a-Lot dons his armor and sets out from th

Question

Y 8.-15 pointssenPSE9 12 P019 Sir Lost-a-Lot dons his armor and sets out from the castle on his trusty steed (see figure below). stone usually, drawbridge is lowed to a horizontal position so that the end of the bridge rests on the 0- ledge. the far enough and stopped it at Lost-a-Lot's squire didn't lower the drawbridge m2/ not 20.0° above the horizontal. The knight and his horse stop when their combined center of mass is d kg. 1.00 m from the end of the bridge. The uniform bridge ist 8.50 m long and has a mass of 1 800 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. mass combined with his armor and steed is 900 kg. (a) Determine the tension in the cable. Page 7 9 net web/Student Assignment Responses/last?dep-16085961

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 = (1800kg * ½*8.5m + 900kg * 7.5m)*9.8m/s²*cos20º - T*sin85.46º*5m
where T is the cable tension. Solving, find
T = 26605.363 N
(b) ß = 180º - 70º - 85.46º = 24.54º angle between cable and wall
horizontal force Fx = T*sin24.54 = 11 049.963 N
away from wall

(c) vertical force Fy = (1800 + 900)kg * 9.8m/s² - T * cos24.54 = 2258.188 N
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