Sir Lost-a-Lot dons his armor and sets out from the castle on his trusty steed (
ID: 1504538 • Letter: S
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 = 7.50 m long and has a mass of 2 100 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
(b) Determine the horizontal force component acting on the bridge at the hinge.
magnitude and direction
(c) Determine the vertical force component acting on the bridge at the hinge.
magnitude and direction
magnitude and direction
(c) Determine the vertical force component acting on the bridge at the hinge.
magnitude and direction
Explanation / Answer
Let T = tension in the lift cable(s)
If we assume the wall is vertical and draw a horizontal line from the cable-to-bridge connection point to the wall, we can determine the angle of the cable with the horizontal.
tan = (12 - 5sin20) / 5cos20
= 65.46°
a) sum moments about the hinge point to zero.
Tsin65.46(5cos20) + Tcos65.46(5sin20) - 9.81(2100(7.5cos20/2) +1100(6cos20)) = 0
Tsin65.46(5cos20) + Tcos65.46(5sin20) = 9.81(2100((7.5cos20)/2) +1100(6cos20))
4.984T = 78797
T = 15810N = 15.8kN
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 = 15810cos65.46
Fx = 6566 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(2100 + 1100) - 15810sin65.46
Fx = 17010 N upward