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Q.

A block remains stationary relative to an accelerating sledge. The angle of friction between the block and sledge is $ϕ$ . Then the minimum acceleration A of the sledge to cause relative sliding between the block and sledge is

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a

g sin ϕ

b

g \tan ϕ

c

g \cos ϕ

d

g \cot ϕ

answer is A.

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Detailed Solution

Question Image

For relative motion between block and sledge

$mA \geq f$

$\geq μ N$

$\geq μ mg$

${So A}_{\min} = μ g$

$A_{min} = (\tan ϕ) g$

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