(a) Determine the tension in the cable. (c) Determine the vertical force compone

Question

(a) Determine the tension in the cable.

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

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 isd= 1.00 m from the end of the bridge. The uniform bridge is l= 6.50m long and has a mass of 1 900kg. 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 wallh= 12.0 m above the bridge. Lost-a-Lot's mass combined with his armor and steed is 950kg. (a) Determine the tension in the cable. (c) Determine the vertical force component acting on the bridge at the hinge. magnitude

Explanation / Answer

Insert drawing does not work, so work up a FBD for the loaded bridge.

Place vertical and horizontal reactions Ry and Rx at the hinge.

Locate Mg, the weight of the bridge, at its midpoint, directed down.

Locate mg, the weight of knight and horse 1 m from the free end of the bridge, directed down.

Locate T, the tension in the cable, 5m from the hinge, toward the wall at an angle to be determined with the law of cosines and law of sines.

Clockwise moments are positive, counter-clockwise moments are negative.

Rx - Tx = 0

Ry + Ty - Mg - mg = 0

Normal components of the bridge's weight and knight's weight are Mgcos and mgcos.

To determine the angle T makes with the bridge, first use law of cosines to calculate length of cable.

c2 = a2 + b2 - 2abcos(90 - ), where a = 5 m and b = 12 m.

c2 = 25 + 144 - 2(60)cos(70) = 169 - 41 = 128

c = 11.31 m

Determine the angle T makes with the bridge using law of sines.

70/11.31 = /12

= 74.3°

Components of T parallel to and perpendicular to the bridge are Tcos74.3 and Tsin74.3. Use the perpendicular component when summing moments about the hinge.

Mh = (Mgcos20)(3.25) - (Tsin74.3)(5) + (mgcos20)(5.5) = 0

[(1900)(9.8)cos20(3.25) + (950)(9.8)cos20(5.5)]/5sin74.3 = T

T = 20213 N (a)

The components of T parallel to and perpendicular to the bridge have components in the vertical direction. Sum them to form Ty in the vertical force equation and solve for the vertical reaction at the hinge.

Ty = T(sin74.3)cos20 - (Tcos74.3)sin20 = 20213(sin74.3cos20 - cos74.3sin20) = 20213(.812) = 16413 N

Ry + Ty - Mg - mg = 0

Ry = Mg + mg - Ty = 1900(9.8) + 950(9.8) - 16413 = 11517 N

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