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Determine the compressive force on object $A$ due to squeezing forces $P$ when the vice grip pliers are in the position shown. (Assume that the forces acting on $A$ are vertical.)

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$A$ = [[0]] 

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{geogebra_applet('vugb9fzx', ggb_vars)}

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\\dfrac{\\var{q(x2+x3+x4)}}{\\var{q(dperp)}}

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x

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\\var{x2} \\cos \\theta - \\var{y2} \\sin \\theta

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 (\\var{d(bx)} P) \\left(dfrac{\\var{q(y1+y2)}}{\\var{q(x1)}}

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Start by drawing neat, correct, labeled free body diagrams for the lower handle and the lower jaw.  Recognize that piece $CE$ is a two force member in compression so the force on pin $E$ acts down and left along a line passing through $C$ and $E$.

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Geometry

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Determine angle $\\theta$ from the geometry of the pliers.  

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\\[\\theta = \\tan^{-1}\\left( \\dfrac{\\var{q(x3)}}{\\var{q(y1)}}\\right) = \\var{d(theta)}°\\]

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FBD II

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\\[\\begin{align}\\text{II: } \\Sigma M_B &= 0\\\\ CE_x \\,(\\var{q(y2)}) -CE_y\\,(\\var{q(x2)}) + P \\,( \\var{q(x2+x3+x4)})& = 0 \\\\ CE\\left( \\var{y2} \\sin \\theta - \\var{x2} \\cos \\theta \\right) &=  -P \\,( \\var{q(x2)}+\\var{q(x3)}+\\var{q(x4)})\\\\CE &=-P \\,\\left( \\dfrac{\\var{q(x2+x3+x4)}}{\\var{-q(dperp)}} \\right)\\\\ &= \\var{d(ce)} P \\\\ \\\\ \\text{II: }\\Sigma F_x &= 0\\\\ B_x &= CE_x\\\\ &= CE \\sin{\\theta}\\\\ &= ( \\var{d(ce)} P )\\,(\\sin \\var{d(theta)}°)\\\\&= \\var{d(bx)} P\\end{align}\\]

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FBD I

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\\[\\begin{align} \\text{I: }\\Sigma M_D &=0\\\\ A\\,(\\var{q(x1)}) &= B_x\\,(\\var{q(y1)} + \\var{q(y2)})\\\\ A &= (\\var{d(bx)} P) \\, \\left(\\dfrac{\\var{q(y1+y2)}}{\\var{q(x1)}}\\right)\\\\  &= \\var{d(A)} P\\end{align}\\]

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