- #1
Saitama
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Homework Statement
Rod AB is placed against a block which is moving towards right with a speed of 1m/s. If at an instant when the rod makes an angle ##60^{\circ}## with the horizontal and end A is sliding towards left with a speed of 1m/s. Then the speed of the point of contact B of the rod is?
(Ans: ##\sqrt{3}/2\,\,\text{m/s}##)
Homework Equations
The Attempt at a Solution
Let the horizontal distance from the dashed line of point A be ##x_A## and that of B be ##x_B##. Also, let the distance of B from ground be ##y##. Clearly, ##dx_A/dt=-1\,\,\text{m/s}## and ##dx_B/dt=1\,\,\text{m/s}##. From Pythagoras theorem,
$$(x_B-x_A)^2+y^2=l^2 \Rightarrow 2(x_B-x_A)\left(\frac{dx_B}{dt}-\frac{dx_A}{dt}\right)+2y\frac{dy}{dt}=0$$
where ##l## is the length of rod.
$$\Rightarrow -y\frac{dy}{dt}=2(x_B-x_A) \Rightarrow \frac{dy}{dt}=-\frac{2}{\sqrt{3}}\,\,\text{m/s}$$
where I have used ##\tan(60^{\circ})=y/(x_B-x_A)=\sqrt{3}##.
Hence, the net velocity is
$$\sqrt{1^2+\left(-\frac{2}{\sqrt{3}}\right)^2}=\sqrt{\frac{7}{3}}\,\,\text{m/s}$$
Where did I go wrong?
Any help is appreciated. Thanks!