Please show all work and answer all questions! thanks Sir Lost-a-Lot dons his an

Question

Please show all work and answer all questions! thanks

Sir Lost-a-Lot dons his 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 theta = 20.0 degree above the horizontal. The knight and his horse stop when their combined center of mass is sigma = 1.00 m from the end of the bridge. The uniform bridge is l = 7.50 m long and has a mass of 2.200 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. (a) Determine the tension in the cable. Be careful in determining the angle that the cable makes with the bridge. It is net 90 degree. N (b) Determine the horizontal force component acting or the bridge at the hinge. magnitude N direction (c) Determine the vertical force component acting on the bridge at the hinge, magnitude magnitude N direction

Explanation / Answer

given , angle above the horizontal = theta = 20 degree

d = 1m from the end of the bridge

length of the bridge, l = 7.5 m

mass of bridge, m = 2200 kg

distance of lift cable form the hinge, s = 5 m

h = 12 m

mass of rider, M = 950 kg

now from torque balance about the hinge, assuming the tension in the rope is T

T*s = mg*lcos(theta)/2 + Mg(l-d)cos(theta)

T*5 = g[2200*7.5cos(20)/2 + 950(7.5-1)cos(20) ] = 1329751.9799

a. T = 26,595.0395 N

b. horizontal component of force acting on the bridge = H

H = Tsin(20) = 9096.039 N

c. vertical force component be V

then V = Tcos(20) - g(2200 + 950)

V = 5910.33 7 N

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.