Sir Lost-a-Lot dons his armor and sets out from the castle on his trusty steed i
ID: 1409378 • Letter: S
Question
Sir Lost-a-Lot dons his 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 stopped it 20.0degree 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 9.50 m long and has a mass of 2300 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. (a) Determine the tension in the cable. (b)Determine the horizontal force component acting on the bridge at the hinge, magnitude to the right to the left (c)Determine the vertical force component acting on the bridge at the hinge, magnitude. upwards downwards.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(2300(9.5cos20/2) +800(6cos20)) = 0
Tsin65.46(5cos20) + Tcos65.46(5sin20) = 9.81(2300((9.5cos20)/2) +800(6cos20))
4.984T = 144959
T = 29.08 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 = 29080cos65.46
Fx = 12079N 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(2300 + 800) - 29080sin65.46
Fx = 3958 N upward