Sir Lost-a-Lot dons his armor and sets out from the castle on his trusty steed i
ID: 1410230 • 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 20.0° 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 8.50 m long and has a mass of 2000 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 900 kg. Hint: First determine all the angles and lengths of the triangle made by the wall, the cable, and the drawbridge.
(a) Determine the tension in the cable.
(b) Determine the horizontal force component acting on the bridge at the hinge. magnitude direction to the left to the right
(c) Determine the vertical force component acting on the bridge at the hinge. magnitude direction downwards upwards
Explanation / Answer
A is where the cable attaches to the bridge, O the hinge of the bridge, B where the cable attaches to the castle wall. We have, in the triangle AOB, OB = 12 m, OA = 5 m.
Now, OB is vertical and OA is 20 deg below horizontal, that makes angle AOB 110 deg, hence, angles OAB + OBA (call them A and B) is supplement to 110 deg:
A + B = 70 or A = 70 - B
Also, applying Sine Rule of Triangles:
sinA / 12 = sinB / 5
12 sinB = 5 sin(70 - B) = 5 sin 70 cosB - 5 cos70 sinB
tanB =5 sin70/13.7=0.342
B=18.92 deg
angle between the bridge and horizontal is 20 deg, then angle between bridge and vertical (direction of weight) must be 70 deg. So here we go
T x 5 sin 51.1 = (2000 kg x 4.25 sin 70 + 900 kg x 7.5 sin 70 )*9.8
where T is cable tension
Solving T = 36083.6 N
Only horizontal force causing a reaction at the hinge is the horizontal component of the tension (the weights are vertical)
H = T sin B = 11690.4 N (compression)
Vertical component of tension = T cos A = 22663.4 N
That is not the only vertical force on the bridge, there are the weights to consider. So
V + T cos B - 2000 g - 900 g = 0
V = -5718.2 N
consider positive upwards so V is in downward direction