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Magnitude and Direction of Acceleration

Consider the ...

Question

Consider the system shown in figure. If A is moving with a velocity v towards right on a horizontal surface then the velocity of C is (Assume that block C can move in horizontal direction only while block B can move only in vertical direction):

\sqrt{3} \frac{V}{2}

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\frac{V}{2}

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V

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None of these

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Solution

The correct option is C $V$
From the wedge constraint (or motion constraint), we have
$\frac{V_{b}}{V} = tan 30^{0}$
Or, $V_{b} = \frac{V}{\sqrt{3}} (i)$
For block C, we have
$\frac{V_{b}}{V_{c}} = tan 30^{0}$
Or, $V_{b} = \frac{V_{c}}{\sqrt{3}} (i i)$
Form (i) and (ii),
$V_{c} = V$

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Consider the system shown in figure. If A is moving with a velocity v towards right on a horizontal surface then the velocity of C is (Assume that block C can move in horizontal direction only while block B can move only in vertical direction):

The correct option is C VFrom the wedge constraint (or motion constraint), we haveVbV=tan300Or, Vb=V√3(i) For block C, we haveVbVc=tan300 Or, Vb=Vc√3(ii) Form (i) and (ii),Vc=V

The correct option is C $V$
From the wedge constraint (or motion constraint), we have
$\frac{V_{b}}{V} = tan 30^{0}$
Or, $V_{b} = \frac{V}{\sqrt{3}} (i)$
For block C, we have
$\frac{V_{b}}{V_{c}} = tan 30^{0}$
Or, $V_{b} = \frac{V_{c}}{\sqrt{3}} (i i)$
Form (i) and (ii),
$V_{c} = V$