A uniform rectangular plate of length $l$ and height h has a small triangular wedge like support welded to its right bottom and has a small light rigid wheel fitted to its other bottom edge. There is no friction at the axle of the wheel. This plate is placed with its length horizontal as shown on a rough horizontal plank. If the plank moves with rightward acceleration of $a_{o}$ the plate just begins to slide on the plank. While sliding, the wheel rolls. Since the wheel has negligible inertia, net force required to accelerate its centre of mass and net torque required to impart them angular acceleration can be neglected.
EXPERIMENT-1: The plank and the plate are initially at rest. $a_{o}$ is large enough
EXPERIMENT-2: The plank and the plate in a new experiment are first given a constant velocity towards right. the plank is then given a constant retardation a

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

The maximum value of acceleration of the plank towards left in EXP-1 for which plate will not topple on the plank is $\frac{g l}{h}$

If $l$ is large enough, the maximum value of a for which plate will not slide in EXP-2 is $\frac{a_{o} g l}{g l - 2 a_{o} h}$

The maximum value of acceleration of the plank towards left in EXP-1 for which plate will not topple on the plank is $\frac{2 g l}{h}$

If $l$ is large enough, the maximum value of a for which plate will not slide in EXP-2 is $\frac{a_{o} g l}{g l - a_{o} h}$

answer is A, B.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

The F.B.D of plate w.r.t plank is shown
For toppling to take place, $(N_{2} \to 0)$
$τ_{About wheel} > 0 \Rightarrow m a \frac{h}{2} > m g \frac{l}{2} \Rightarrow a > \frac{g l}{h}$
In the above diagram
On the verge of sliding (but without topping)
$f = μ N_{2} = m a_{0}$ ………….(1)
And $N_{1} + N_{2} = m g$ ………..(2)
And $τ_{a b o u t c c e n t r e} = 0$
$\Rightarrow N_{1} \frac{l}{2} - N_{2} \frac{l}{2} - μ N_{2} \frac{h}{2} = 0$
$N_{1} l = N_{2} (l + μ h)$ ………….(3)
(2) and (3) $\Rightarrow N_{2} (1 + \frac{μ h}{l} + 1) = m g$
$\Rightarrow N_{2} = \frac{m g l}{2 l + μ h} \Rightarrow μ = \frac{(m a_{0})}{m g l} (2 l + μ h) = \frac{a_{0}}{g} [\frac{2 l + μ h}{l}] \Rightarrow μ (1 - \frac{a_{0} h}{g l}] = \frac{2 a_{0}}{g}$

$\Rightarrow μ = \frac{2 a_{0} l}{g l - a_{0} h}$ ………..(4)
Now when plank is given retardation
F.B.D will be
Question Image

Again

$N_{4} + N_{3} = m g .............. (5) f^{'} = μ N_{3} = m a .................... (6)$

And $τ_{a b o u t c c e n t r e} = 0$

$\Rightarrow N_{4} \frac{l}{2} - N_{3} \frac{l}{2} + μ N_{3} \frac{h}{2} = 0 \Rightarrow N_{4} l = N_{3} [l - μ h] \dots \dots . . (7)$

(5) and (7) $\Rightarrow N_{3} [l - \frac{μ h}{l} + 1] = m g$

$\Rightarrow N_{3} = \frac{m g l}{2 l - μ h} (4), (6) \Rightarrow μ N_{3} = m a \Rightarrow \frac{μ m g l}{2 l - μ h} = m a \Rightarrow a = \frac{a_{0} g l}{g l - 2 a_{0} h}$