Body A, a 50 kg train signal, is attached to beam B of length 3.5 in and mass 20

Question

Body A, a 50 kg train signal, is attached to beam B of length 3.5 in and mass 200 kg. B is. in turn, connected to a fixed vertical post by a hinge and a cable C. The geometry of the system is shown at right. What is the tension in C Be sure to include a detailed free body diagram to accompany your analysis. What is the magnitude of the force provided by (he hinge What is the moment of inertia of A + B as a rigid system, if rotating about the hinge For B, use calculus principles; treat with a single integral, without breaking the beam into two parts. If C were to break, what would be the angular acceleration of B at the moment of breakage

Explanation / Answer

a) Let TC be the tension in C.

Length of the beam to the right of hinge, LB,r = (22 + 1.52)1/2 = 2.5 m

Length of the beam to the left of hinge, LB,l = 3.5 - 2.5 = 1 m

Angle made by the beam with the vertical post is given by,

sin = 1.5 / 2.5 = 0.6

Moment of the forces acting on the (A + B) system about the hinge is zero.

=> WB(LB/2 - LB,l)sin + WA(LB,r - 0.5)sin = 2TC

=> TC = 0.5{[200 * 9.81 * (3.5/2 - 1) * 0.6] + [50 * 9.81 * (2.5 - 0.5) * 0.6]}

=> TC = 735.75 N

b) Let FH and Fv be the horizontal and vertical forces provided by the hinge respectively.

Horizontal force balance:

FH = TC

=> FH = 735.75 N

Vertical force balance:

FV = WA + WB

=> FV = (MA + MB)g = (50 + 200) * 9.81 = 2452.5 N

So, magnitude of force provided by hinge,

F = (FH2 + FV2)1/2 = (735.752 + 2452.52)1/2 = 2560.5 N

c) Moment of inertia of A about the hinge, IA = 50 * 2.52 = 312.5 kg-m2

Moment of inertia of B about the hinge,

IB = (dm r2)

Here, dm = (MB/LB)dx

So, IB = -12.5 (MB/LB)x2dx

=> IB = (MB/LB) * [x3/3]-12.5

=> IB = (200 / 3.5) * [2.53 - (-1)3] / 3 = 316.67 kg-m2

Total moment of inertia, I = IA + IB = 312.5 + 316.67 = 629.2 kg-m2

d) If C breaks, net torque on the system about the hinge,

Tnet = WB(LB/2 - LB,l)sin + WA(LB,r - 0.5)sin

=> Tnet = {[200 * 9.81 * (3.5/2 - 1) * 0.6] + [50 * 9.81 * (2.5 - 0.5) * 0.6]} = 1471.5 N-m

Hence, angular acceleration at the moment of breakage,

= Tnet / I = 1471.5 / 629.2 = 2.34 rad/s2

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