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A rigid rod is placed against the wall as shown in the figure. When velocity of its lower end is 10 m/s and its base makes an angle \(\alpha\) =60° with horizontal, then the vertical velocity of its end B will be
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|dw:1408458384702:dw|
|dw:1408474683192:dw| \(x^2+y^2=l^2\) differentiate \(2x\dfrac{dx}{dt}+2y\dfrac{dy}{dt}=0\) \(2x*v_a+2y*v_b=0\) use trig to get rid of x and y \(x=l*cos \alpha\) \(y=l*sin \alpha\) \(2(l*cos \alpha)*v_a+2(l*sin \alpha)*v_b=0\) \(v_b=-\dfrac{2(l*cos \alpha)*v_a}{2(l*sin \alpha)}=-\dfrac{v_a}{tan\alpha}=-v_a*cot\alpha\)
Thanx @aaronq !
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