Review Problem. Sir Lost-a-Lot dons his armor and sets out from the castle on hi

Question

Review Problem. 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 (Fig. P12.20). 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 7.00 m long and has a mass of 2400 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 1000 kg. Suddenly, the lift cable breaks! The hinge between the castle wall and the bridge is frictionless, and the bridge swings freely until it is vertical.

(a) Find the angular acceleration of the bridge once it starts to move.
rad/s2

(b) Find the angular speed of the bridge when it strikes the vertical castle wall below the hinge.
rad/s2

(c) Find the force exerted by the hinge on the bridge immediately after the cable breaks.
R = (  i +  j) kN

(d) Find the force exerted by the hinge on the bridge immediately before it strikes the castle wall.
R = (  i +  j) kN

Explanation / Answer

a) by newtons law of rotation,

i alpha = (1000*6+2400*3.5)* 9.8*cos 20 degree

alpha =9.8* (1000*6+2400*3.5)* cos 20 degree /[2400*7^2/3+1000*6^2]

= 1.763 rad/s^2

b) angular velocity just before striking w

0.5 mv^2 + 0.5 iw^2 = 0.5mgl

m (0.5wl)^2 + ml^2w^2/3 = mgl

0.25 w^2 + 0.333 w^2 = g/l

w = sqrt(9.8/[(0.333+0.250)*7])

= 1.55 rad/s

c) Rx = [2400*1.763*3.5 + 1000*1.763*6] sin 20 degree = 8683 N = 8.683 kN i

Ry = 2400*9.8+1000*9.8 - [2400*1.736*3.5 + 1000*1.736*6]*cos 20 degree = 9829 N = 9.829 KN j

hence R = (-8.683 i + 9.829 j )kN

d) F = mw^2r = 2400*1.55^2*3.5

= 20181 N =( 0i + 20.181 j ) kN

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