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Question: If the coefficient of friction between A and B is \(\mu \). The maximum horizontal acceleration of t...

If the coefficient of friction between A and B is μ\mu . The maximum horizontal acceleration of the wedge A for which B will remain at rest w.r.t the wedge is

A. μg\mu g
B. g(1+μ1μ)g(\dfrac{{1 + \mu }}{{1 - \mu }})
C. gμ\dfrac{g}{\mu }
D. g(1μ1+μ)g(\dfrac{{1 - \mu }}{{1 + \mu }})

Explanation

Solution

Try to solve by taking A as a frame of reference and use pseudo force concept it will be easy that way you can also solve by taking ground as a frame of reference but then you will have to apply constraint equations which will make it little complex.

Complete step by step answer:
We will solve this question by taking wedge A as a frame of reference.
So free body diagram according as we observe B from A will be

balancing force perpendicular to incline we have,
N=mgcosθ+masinθN = mg\cos \theta + ma\sin \theta so
Friction f=μN=μ(mgcosθ+masinθ)f = \mu N = \mu (mg\cos \theta + ma\sin \theta )
And balancing force along the incline we have,
f+mgsinθ=macosθf + mg\sin \theta = ma\cos \theta
μ(mgcosθ+masinθ)+mgsinθ=macosθ\Rightarrow \mu (mg\cos \theta + ma\sin \theta ) + mg\sin \theta = ma\cos \theta
a=gsinθ+μgcosθacosθμasinθ\Rightarrow a = \dfrac{{g\sin \theta + \mu g\cos \theta }}{{a\cos \theta - \mu a\sin \theta }} putting θ=45\theta = {45^ \circ }we have
a=g(1+μ1μ)\therefore a = g(\dfrac{{1 + \mu }}{{1 - \mu }})

So, the correct answer is “Option B”.

Note:
In these types of questions, choosing a frame of reference is critical because choosing a better frame of reference makes a lot of things easy here. For example, once we take wedge A as a frame of reference we won’t have to worry about constraint equations and it makes it easier to solve such questions.