The man pulls the boy up to the tree limb C by walking backward at a constant speed of 1.5 m/s . Determine the speed at which the boy is being lifted at the instant xA = 4 m . Neglect the size of the limb. When xA = 0 , xB = 8 m , so that A and B are coincident, i.e., the rope is 16 m long.

Question

The man pulls the boy up to the tree limb C by walking backward at a constant speed of 1.5 m/s . Determine the speed at which the boy is being lifted at the instant {x}_{A} = 4 m . Neglect the size of the limb. When {x}_{A} = 0 , {x}_{B} = 8 m , so that A and B are coincident, i.e., the rope is 16 m long.

Accepted Answer

Position-Coordinate Equation: Using the Pythagorean theorem to determine [latex]l_{A C}[/latex] we have [latex]l_{A C}[/latex]=[latex]\sqrt{x_{A}^{2}+8^{2}}[/latex]. Thus,

l=[latex]l_{A C}[/latex]+[latex]y_{B}[/latex]
16=[latex]\sqrt{x_{A}^{2}+8^{2}}[/latex]+[latex]y_{B}[/latex]
[latex]y_{B}[/latex]=16-[latex]\sqrt{x_{A}^{2}+64}[/latex]

Time Derivative: realizing that [latex]v _ { A }[/latex] = [latex]\frac { d x _ { A } } { d t }[/latex] and [latex]v_{B}[/latex]=[latex]\frac{d y_{B}}{d t}[/latex], we have

[latex]v_{B}[/latex]=[latex]\frac{d y_{B}}{d t}[/latex]=-[latex]\frac{x_{A}}{\sqrt{x_{A}^{2}+64}}[/latex] [latex]\frac{d x_{A}}{d t}[/latex]
[latex]v_{B}[/latex]=-[latex]\frac{x_{A}}{\sqrt{x_{A}^{2}+64}}[/latex] [latex]v_{A}[/latex]

At the instant [latex]x_{A}[/latex]=4 m

[latex]v_{B}[/latex]=-[latex]\frac{4}{\sqrt{4^{2}+64}}[/latex](1.5)=-0.671 m/s=0.671 m/s [latex]\uparrow[/latex]

Note: The negative sign indicates that velocity [latex]v_{B}[/latex] is in the opposite direction to that of positive [latex]y_{B}[/latex]

Question : The man pulls the boy up to the tree limb C by walking backw...

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