The man pulls the boy up to the tree limb C by walking backward. If he starts from rest when xA = 0 and moves backward with a constant acceleration aA = 0.2 m/s^2 , determine the speed of the boy at the instant yB = 4 m . Neglect the size of the limb. When xA = 0 , yB = 8 m ,so that A and B are

Question

The man pulls the boy up to the tree limb C by walking backward. If he starts from rest when {x}_{A} = 0 and moves backward with a constant acceleration {a}_{A} = 0.2 m/{s}^{2} , determine the speed of the boy at the instant {y}_{B} = 4 m . Neglect the size of the limb. When {x}_{A} = 0 , {y}_{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]

I=[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: Taking the time derivative [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]y_{B}[/latex]=4 m, 4=16-[latex]\sqrt{x_{A}^{2}+64}[/latex], [latex]x_{A}[/latex]=8.944 m . The
velocity of the man at that instant can be obtained.

[latex]v_{A}^{2}[/latex]=[latex]\left(v_{0}\right)_{A}^{2}+2\left(a_{c}\right)_{A}\left[s_{A}-\left(s_{0}\right)_{A}\right][/latex]
[latex]v_{A}^{2}[/latex]=0+2(0.2)(8.944-0)
[latex]v_{A}[/latex]=1.891 m/s

[latex]v_{B}[/latex]=-[latex]\frac{8.944}{\sqrt{8.944^{2}+64}}[/latex](1.891)=-1.41 m/s=1.41 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...

This Question has been Answered.

Already have an account?

Login