Lost-a-Lot\'s squire didn\'t lower the draw-brdige far enough and stopped it at

Question

Lost-a-Lot's squire didn't lower the draw-brdige far enough and stopped it at theta=20.0 degrees above the horizontal. The knight and his horse stop when their combined center of mass is d=1.00m from the end of the bridge. The uniform bridge is L=8.00m long and has mass 2,000kg. The lift cable is attached to the bridge 5.00m 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 1,000kg. Determine

a) the tension in the cable

b) the horizontal force component acting on the bridge at the hinge

c) the vertical force component acting on the bridge at the hinge.

Answers: a) 27.7kN. b) 11.5 kN c) 4.19 kN

I need help on the step-by-step to obtaining these answers.

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.

M = 0

Tsin65.46(5cos20) + Tcos65.46(5sin20) - 9.81[(2000(8cos20/2) +1000(7cos20))] = 0
Tsin65.46(5cos20) + Tcos65.46(5sin20) = 9.81(2000(4cos20)) +1000(7cos20))
4.984T = 138275.77

T = 27,743.93 N
T = 27.74 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 = 27, 744cos65.46
Fx = 11,523 N = 11.5 kN 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(2000 + 1000) - 27,744sin65.46
Fy = 4192 N = 4.192 kN upward

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