The beam in Fig. 5.41 weighs 200 lb and is supported by a pin support at A and the wire BC. The wire behaves like a linear spring with spring constant k = 60 lb/ft and is unstretched when the beam is in the position shown.Determine the reactions at A and the tension in the wire when the beam is in

Question

The beam in Fig. 5.41 weighs 200 lb and is supported by a pin support at A and the wire BC. The wire behaves like a linear spring with spring constant k = 60 lb/ft and is unstretched when the beam is in the position shown.Determine the reactions at A and the tension in the wire when the beam is inequilibrium.

Strategy
When the beam is in equilibrium, the sum of the moments about A due to the beam's weight and the force exerted by the wire equals zero. We will obtain a graph of the sum of the moments as a function of the angle of rotation of the beam relative to the horizontal to determine the position of the beam when it is in equilibrium. Once we know the position, we can determine the tension in the wire and the reactions at A.

Accepted Answer

Let α be the angle from the horizontal to the centerline of the beam (Fig. a). The distances b and h are

b = 8(1 — cos α),

h = 2 + 8 sin α,

and the length of the stretched wire is

[latex]L = \sqrt{b^{2} + h^{2} } [/latex]

The tension in the wire is

T = k(L - 2).

We draw the free-body diagram of the beam in Fig. b. In terms of the components of the force exerted by the wire.

[latex]T_{x} = \frac{b}{L} T , T_{y} = \frac{h}{L} T , [/latex]

the sum of the moments about A is

[latex]\sum{M_{A} } = ( 8\sin α )T_{x} + ( 8\cos α)T_{y} - ( 8\cos α)W [/latex].

If we choose a value of α, we can sequentially evaluate these quantities. Computing [latex]\sum{M_{A} } [/latex] as a function of α, we obtain the graph shown in Fig. 5.42. From the graph we estimate that [latex]\sum{M_{A} } = 0 [/latex] when α= 12°. By examining computed results near 12°,

α	[latex] \sum{M_{A} }(ft-lb) [/latex]
11.87°	-1.2600
11.88°	-1.2600
11.89°	0.0750
11.90°	0.7424
11.91°	1.4099

we estimate that the beam is in equilibrium when α = 11.89°. The corresponding value of the tension in the wire is T = 99. 1 lb.

To determine the reactions at A, we use the equilibrium equations

[latex]\sum{F_{x} } = A_{x} + T_{x} = 0[/latex],

[latex]\sum{F_{x} } = A_{x} + T_{x} - W = 0[/latex],

obtaining [latex] A_{x} = -4.7 lb and A_{y} = 101.0 lb[/latex].

Question 5.13: The beam in Fig. 5.41 weighs 200 lb and is supported by a pi...

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