A cubical block of mass M and edge a slides down a rough inclined plane of inclination θ with a uniform velocity. Choose the correct statement(s):
A
Torque of normal force on the block about centre is Mgasinθ2
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B
Torque of normal force on the block about the centre is Mgacosθ2
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C
Normal reaction force shifts by a distance atanθ2
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D
Normal reaction force shifts by a distance atanθ4
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Solution
The correct option is C Normal reaction force shifts by a distance atanθ2 To avoid toppling, let the normal reaction (N) shift by a distance x towards the right edge as shown in figure.
Applying rotational equilibrium (for just toppling) about the centre of the cube (A), torques in clockwise and anticlockwise direction should be equal. τMg=0,τf=fa2,τN=Nx, about point A. τclockwise=τanticlockwise ⇒f×a2=Nx...(i)
From translational equilibrium of cube, since it is moving with uniform velocity, acceleration a=0. f=Mgsinθ...(ii) N=Mgcosθ...(iii)
From Eq.(i),(ii),&(iii), we get, (Mgsinθ)×a2=(Mgcosθ)×x ∴x=atanθ2
Now, torque of normal reaction about its centre (A) is, τN=Nx=Mgcosθ×atanθ2 ∴τN=Mgasinθ2
Hence, option (a) and (c) are correct.