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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.

Step-by-step

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

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

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

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

At the instant x_{A}=4 m

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

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

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