The mechanism shown in the figure consists of two pin-connected, homogeneous bars of weight W and length L each. The roller at B moves in a horizontal slot, located at the distance 1.5L below the pin at O. Determine the force P that will hold the system in equilibrium for θ1 = 30º

Question

The mechanism shown in the figure consists of two pin-connected, homogeneous bars of weight [latex]W[/latex] and length [latex]L[/latex] each. The roller at [latex]B[/latex] moves in a horizontal slot, located at the distance [latex]1.5L[/latex] below the pin at [latex]O[/latex]. Determine the force [latex]P[/latex] that will hold the system in equilibrium for [latex]θ_{1}=30^\circ[/latex]

Accepted Answer

The system has one [latex]DOF[/latex] because only one position coordinate, for example, [latex]θ_{1}[/latex] or [latex]θ_{2}[/latex], is required to specify its configuration. The figure shown is also an active-force diagram, because it shows only those forces than can perform virtual work on the system. The pin reactions at [latex]O[/latex] and [latex]A[/latex], and the roller reaction at [latex]B[/latex], are omitted because their virtual work is zero.

It is convenient to use the same figure for kinematic analysis. We introduce the [latex]xy[/latex]-coordinate system shown, the origin of which is located at the fixed point [latex]O[/latex]. The coordinates [latex]y_{1}[/latex] and [latex]y_{2}[/latex] locate the centers of gravity of the bars, and [latex]x_{B}[/latex] is the horizontal coordinate of end [latex]B[/latex].

If the system is given a virtual displacement consistent with the constraints, the principle of virtual work takes the form

[latex]δU=W\:δy_{1}+W\:δy_{2}+P\:δx_{B}=0[/latex] (a)

All the signs in Eq. (a) are positive because the direction of each force is the same as the positive coordinate direction of its point of application.

The next step is to express [latex]δy_{1}[/latex], [latex]δy_{2}[/latex], and [latex]δx_{B}[/latex] as functions of the virtual change in the kinematically independent coordinate. Let us choose [latex]θ_{1}[/latex] as the independent coordinate. Because the roller at [latex]B[/latex] is constrained to move in the horizontal slot, the equation of constraint is [latex]L\cos θ_{1}+L\cos θ_{2}=1.5L[/latex], or

[latex]\cos θ_{1}+\cos θ_{2}=1.5[/latex] (b)

which yields [latex]θ_{2}=50.66^\circ[/latex] when [latex]θ_{1}=30^\circ[/latex]. The evaluation of [latex]δy_{1}[/latex], [latex]δy_{2}[/latex], and [latex]δx_{B}[/latex] in terms of the virtual rotation [latex]δθ_{1}[/latex] now proceeds as follows.

Evaluation of [latex]δy_{1}[/latex]

From the figure, we see that [latex]y_{1}=(L/2)\cos θ_{1}[/latex]. Forming the differential of both sides and evaluating at [latex]θ_{1}=30^\circ[/latex], we find that

[latex]δy_{1}=\frac{dy_{1}}{dθ_{1}}δθ_{1}=−\frac{L}{2}\sin θ_{1}\:δθ_{1}=−0.2500L\:δθ_{1}[/latex] (c)

Evaluation of [latex]δy_{2}[/latex]

From the figure, [latex]y_{2}=L\cos θ_{1}+(L/2)\cos θ_{2}[/latex]. Substituting for [latex]θ_{2}[/latex] from Eq. (b), and simplifying, gives [latex]y_{2}=(L/2)\cos θ_{1}+0.75L[/latex]. Taking the differential of each side, and substituting [latex]θ_{1}=30^\circ[/latex], yields

[latex]δy_{2}=\frac{dy_{2}}{dθ_{1}}δθ_{1}=−\frac{L}{2}\sin θ_{1}\:δθ_{1}=−0.2500L\:δθ_{1}[/latex] (d)

Evaluation of [latex]δx_{B}[/latex]
From the figure, [latex]x_{B}=L\sin θ_{1}+L\sin θ_{2}[/latex]. In principle, [latex]θ_{2}[/latex] could be eliminated by using Eq. (b), but this would result in a rather cumbersome expression. It is much easier to form the differentials first and then carry out the substitution. From the chain rule for differentiation, we obtain

[latex]δx_{B}=\frac{∂x_{B}}{∂θ_{1}}δθ_{1}+\frac{∂x_{B}}{∂θ_{2}}δθ_{2}[/latex]

which gives

[latex]δx_{B}=L\cos θ_{1}\:δθ_{1}+L\cos θ_{2}\:δθ_{2}[/latex] (e)

From Eq. (b) we obtain the following equation of constraint in terms of [latex]δθ_{1}[/latex] and [latex]δθ_{2}[/latex]:

[latex]\frac{∂}{∂θ_{1}}(\cos θ_{1}+\cos θ_{2})\:δθ_{1}+\frac{∂}{∂θ_{2}}(\cos θ_{1}+\cos θ_{2})\:δθ_{2}=0[/latex]

or

[latex]−\sin θ_{1}\:δθ_{1}−\sin θ_{2}\:δθ_{2}=0[/latex]

which gives

[latex]δθ_{2}=−(\sin θ_{1}/\sin θ_{2})\:δθ_{1}[/latex] (f)

Substituting Eq. (f) into Eq. (e), we find that

[latex]δx_{B}=L\cos θ_{1}\:δθ_{1}+L\cos θ_{2}\left(-\frac{\sin θ_{1}}{\sin θ_{2}}\right)\:δθ_{1}[/latex]

which, when evaluated at [latex]θ_{1}=30^\circ[/latex] and [latex]θ_{2}=50.66^\circ[/latex], yields

[latex]δx_{B}=0.4562L\:δθ_{1}[/latex] (g)

Virtual Work Equation

Substituting Eqs. (c), (d), and (g) into Eq. (a) gives

[latex]δU=W(−0.2500L\:δθ_{1})+W(−0.2500L\:δθ_{1})+P(0.4562L\:δθ_{1})=0[/latex]

which simplifies to

[latex](−0.5W+0.4562P)\:δθ_{1}=0[/latex] (h)

This equation can be satisfied for a nonzero [latex]δθ_{1}[/latex] only if the term in parentheses vanishes, which yields

[latex]P=1.096\:W[/latex]

Because [latex]P[/latex] is positive, its direction is as shown in the figure.

Question 10.2: The mechanism shown in the figure consists of two pin-connec...

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