Question

a rectangular block of mass 6 kg is to be held against a rough vertical wall bt applying a force perpendicular to the wall what is the minimum for ce to be applied if the coefficient of Friction is 0.42​

Answer 1

The minimum force applied is 140 N.

Explanation:

F(g) = Force due to gravity

F(N) = Applied force ( Normal force)

F(r) = Frictional force

F(f) = μ F(N)

μ = 0.4

F(f) > F(g)   [ To cancel the downward force]

For minimum

F(f) = F(g) = 0.42 x F(n)

M(g) = 0.42 x F(N)

6 x 9.8 = 0.42  x F(N)

F(N) = 140 N

Hence the minimum force applied is 140 N.

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A block of mass 10 kg placed on rough horizontal surface having coefficient of friction 0.5,if a horizontal force of 100 N is acting on it then acceleration of the block will be(g=10 m/s2) ?

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