Question

A block of mass m is placed on a smooth wedge of inclination θ . The whole system is accelerated horizontally so that the block does not slip on the wedge. The force exerted by the wedge on the block has a magnitude: (A) mg tan θ (B) mg sin θ (C) mg cos θ (D) mg sec θ

Answer 1

Answer: mg/Cosθ

Explanation:

Taking horizontal acceleration of wedge as 'a'

FCosΘ = FsinΘ

F = mass(m) × acceleration(a) = ma

For horizontal resolution g = 0

Therefore,

Horizontal = Vertical

maCosΘ = mgSinΘ

aCosΘ = gSinΘ

a = gSinΘ/CosΘ

Recall from trigonometry :

SinΘ/Cosθ = tanΘ

Therefore,

a = gtanΘ

Normal force acing on the wedge:

mgCosΘ + maSinΘ - - - - (y)

Substitute a = gtanΘ into (y)

mgCosΘ + mgtanΘsinΘ

tanΘ = sinΘ/cosΘ

mgCosΘ + mgsinΘ/cosΘsinΘ

mgCosΘ + mgsin^2Θ/cosΘ

Factorizing

mg(Cosθ + sin^2Θ/cosΘ)

Taking the L. C. M

mg[(Cos^2θ + sin^2Θ) /Cosθ]

Recall: Cos^2θ + sin^2Θ = 1

mg[ 1 /Cosθ]

mg/Cosθ