Engineering Mechanics - PKRB: Work and Energy - Discussion
Discussion Forum : PKRB: Work and Energy - General Questions (Q.No. 7)
7.
The small bridge consists of an 1,800-lb uniform deck EF (thin plate), two overhead beams AB (slender rods), each having a weight of 200 lb, and a 2,400-lb counterweight BC, which can be considered as a thin plate having the dimensions shown. The weight of the tie rods AE can be neglected. If the operator lets go of the rope when the bridge is at an at-rest position, 
= 45°, determine the speed at which the end of the deck E hits the roadway step at H, = 0°. The bridge is pin-connected at A, D, E, and F.
Discussion:
1 comments Page 1 of 1.
Bobby said:
5 years ago
T1+V1=T2+V2
since the system started from rest, T1=0
V1 = VEF + 2*VAB + VBC
= 15 * sin(45) * 1800 + 2 * 10 * sin(45) * 200-12.5 * sin(45) * 2400
= 707.106 lb.ft.
Since the system reaches datum at the horizontal position, V_2=0
V1=T2
T2=(1/2)(Itotal)(omega^2)
I total = 2*I AB+I BC+I EF.
T2 = 0.5(2898.55 + 11801.242 + 16770.186)(ω^2)
0.5(2898.55+11801.242+16770.186)(ω^2) = 707.106 lb.ft.
ω = 0.211.
vE = r * ω = 30 * 0.211 = 6.359 ft/s.
since the system started from rest, T1=0
V1 = VEF + 2*VAB + VBC
= 15 * sin(45) * 1800 + 2 * 10 * sin(45) * 200-12.5 * sin(45) * 2400
= 707.106 lb.ft.
Since the system reaches datum at the horizontal position, V_2=0
V1=T2
T2=(1/2)(Itotal)(omega^2)
I total = 2*I AB+I BC+I EF.
T2 = 0.5(2898.55 + 11801.242 + 16770.186)(ω^2)
0.5(2898.55+11801.242+16770.186)(ω^2) = 707.106 lb.ft.
ω = 0.211.
vE = r * ω = 30 * 0.211 = 6.359 ft/s.
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