After an unfortunate accident at a local warehouse you have been contracted to d

Question

After an unfortunate accident at a local warehouse you have been contracted to determine the cause. A jib crane collapsed and injured a worker. An image of this type of crane is shown in the figure below. The horizontal steel beam had a mass of 80.00 kg per meter of length and the tension in the cable was T = 11650 N. The crane was rated for a maximum load of 454.5 kg. If d = 6.160 m, s = 0.450 m, x = 1.300 m and h = 1.980 m, what was the magnitude of WL (the load on the crane) before the collapse? What was the magnitude of force at the attachment point P? The acceleration due to gravity is g = 9.810 m/s2.

Explanation / Answer

here,

mass = 80 kg / m
T = 11650 N
Max load = 454.4
d = length of beam = 6.160 m
s = 0.450
x = 1.300 m
h = 1.980 m

Assuming a beam in connected with string at height H and making nagle A with Beam.Some Weight is hanging at distnce X from Right End of beam.

Therefore angle made by string on contact point on beam :
A = arcTan(h/(d-s))
A = arcTan(1.980/(6.160-0.450))
A = 19.12 degrees

Part A:
Taking moment about hinged point :
0 = T(d - s)sinA - W(d - x) - F(d/2)
Where,
T = Tension in string
F = Weight in beam

0 = 11650(6.160 - 0.450)sin19.12 - W(6.160 - 1.300) - 80(6.160/2)
W = 4432.63 N

The load on crane is 4432.63 N

Part B:
From newton Second law we have Fnet = mass * Acceleration

Fy + TsinA - W = M*a
Fy = M*a - TsinA + W
Fy = ( 80 * 6.160 * 9.8 ) - 11650*Sin19.62 + 4432.63
Fy = 4829.44 - 11650*Sin19.12 + 4432.63
Fy = 5446.14 N

Similarly Fx = M*ax
Fx = T*CosA
Fx = 11650 * Cos19.12
Fx = 11007.32 N

Fnet = Sqrt(Fx^2 + Fy^2)
Fnet = Sqrt(11007.32^2 + 5446.14^2)
Fnet = 12280.94 N

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