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Question

A block of mass 2 kg is placed on an inclined surface, which is placed inside a truck. The coefficient of static friction (μ) between the block and the inclined plane is 0.25. If the truck accelerates at the rate of 2 m/s2, what should be the maximum angle of the inclined surface so that the block does not slide down the incline? Take g=10 m/s2 .

A
tan1(0.047)
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B
tan1(7.47)
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C
tan1(0.07)
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D
tan1(0.56)
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Solution

The correct option is A tan1(0.047)


From the FBD of block, let the truck be accelerated in direction as shown.
To prevent sliding of block down the incline, it must be in equilibrium from frame of reference of truck.
Hence Fpseudo=ma will act in opposite direction of a, static friction (Ff) will act up the incline at maximum value.
Ff=μN ......(1)
From the FBD, applying equilibrium condition;
Perpendicular to inclined plane:

N+masinθ=mgcosθ ........(2)
Ff=μ(mgcosθmasinθ) ....(3)
Along the inclined plane:
macosθ+mgsinθ=μ(mgcosθmasinθ) ....(4)
mcosθ(a+gtanθ)=μmcosθ(gatanθ)
tanθ(g+μa)=μga
tanθ=μgag+μa

Or, tanθ=(0.25×10)210+(0.25×2)
θ=tan1(0.047)

Hence obtained value of θ corresponds to θmax, for which block will not slip.

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