Tension and Force Problem?
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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 8.00 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 1200 kg
a) Determine the tension in the cable.
Answer in kN
(b) Determine the horizontal force component acting on the bridge at the hinge.
magnitude
Answer in kN
direction (pick one from below)
to the right
to the left
(c) Determine the vertical force component acting on the bridge at the hinge.
magnitude
Answer in kN
direction (pick one from below)
upwards
downwards
Judi
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Tagged with: Communication • Damsels • Horizontal Force
Filed under: Drawing Dragons
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Lanell
Be careful with all the angles.
The first condition of equilibrium is that the vector sum of all the forces equals zero.
Fx = 0
Fx = Rx – T cos 64.2 o = 0
Rx = T cos 64.2 o = 0.435 T
Fy = 0
Fy = Ry + T sin 64.2 o – W – w = 0
W = (2 000 kg) (9.8 m/s2) = 19 600 N
w = (1 000 kg) (9.8 m/s2) = 9 800 N
Ry + 0.900 T sin 64.2 o = 29 400 N
As we should expect by now, at this stage, we have three unknowns — Rx, Ry, and T — but only two equations. We will get the third equation from applying the second condition of equilibrium, that the sum of the torques must equal zero. We will calculate the torques about the hinge of the drawbridge.
Rx: = 0
Ry: = 0
T: = ccw = (5 m) T sin 44.2o = (5 m) (T) (0.697) =
= 3.49 m T
W: = cw = (4 m) (19 600 N) sin 90o = 78 400 N-m
w: = cw = (7 m) (9 800 N) sin 90o = 68 600 N-m
ccw = cw
3.49 m T = 147 000 N-m
T = 42 120 N
Now that the tension is known, we can go back and determine the “reaction force” Rx and Ry,
Rx = 0.435 T
Rx =18 322 N
Ry = 29 400 N – 0.900 T sin 64.2 o
Ry = 29 400 N – 34 129 N
Ry = – 4 729 N