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

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

the small 0.38-kg slider is known to move from position a to position b along the vertical-plane slot. determine (a) the work done on the body by its weight and (b) the work done on the body by the spring. the distance r = 0.70 m, the spring modulus k = 154 n/m, and the unstretched length of the spring is 0.53 m

explanation:

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