Sir Lost-a-Lot dons his armor and sets out from the castle on his trusty steed (
ID: 1458957 • 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 ? = 6.50 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 950 kg.
Explanation / Answer
bridge mass = M = 2100 kg
knight and horse mass = m = 950 kg
cable is attached horizontally =x = 5 * cos20 = 4.70 m
and vertucally up distance = y = 5 sin20 = 1.71 m
angle which cable make with the vertical wall
theta = tan^-1(4.70 / ( 12 - 1.71) ] = 24.5 degree
Call the force components at the hinge Hx (to the right) and Hy (upward)
part a )
Take torques about the hinge end of the bridge:
Hx(0) + Hy(0) - Mg(3.25)cos20 - Tsin24.5 * 1.71 m + T cos24.5 *4.70 - mg*5.5 * cos20 = 0
-2100*9.8*3.25 * cos20 - Tsin24.5 * 1.71 + T cos24.5 * 4.70 - 950 * 9.8 * 5.5 * cos20 = 0
-62851.34 - 0.709T + 4.28 T - 48116.96 = 0
3.571T = 110968.3
T = 31074.85 N
part b )
sum of force in x direction = 0
Hx - Tsin24.5 = 0
Hx = 12886.53 N (right)
part c )
sum of all force in y direction = 0
Hy - Mg + T cos24.5 - mg = 0
Hy = (M+m)g - Tcos24.5
Hy = 1613.087 N (upward)