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 theta = 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 l = 7.50 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 950 kg. (a) Determine the tension in the cable. Be careful in determining the angle that the cable makes with the bridge. It is not 90. N (b) Determine the horizontal force component acting on the bridge at the hinge. magnitude N direction to the right (c) Determine the vertical force component acting on the bridge at the hinge. magnitude magnitude direction upwards

Explanation / Answer

help: tan? = (12 - 5sin20) / 5cos20 ? = 65.46° a) sum moments about the hinge point to zero. Tsin65.46(5cos20) + Tcos65.46(5sin20) - 9.81(2300(7cos20/2) +1050(6cos20)) = 0 Tsin65.46(5cos20) + Tcos65.46(5sin20) = 9.81(2300((7cos20)/2) +1050(6cos20)) 4.984T = 132284 T = 26.5 kN b) sum forces in the horizontal direction to zero. Let Fx be the hinge reaction force in the x direction. Let "from the bridge toward the wall" be the positive direction. Tcos? - Fx = 0 Fx = Tcos? Fx = 26, 500cos65.46 Fx = 10, 900 N toward the bridge c) sum forces in the vertical direction to zero. Let Fy be the hinge reaction force in the y direction. Let up be positive direction. Tsin? + Fy - g(m1 + m2) = 0 Fy = g(m1 + m2) - Tcos? Fy = 9.81(2400 + 1050) - 26,500sin65.46 Fx = 8800 N upward

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