Daidaiton gawawwaki a kan wani babban titin da aka karkata - amfani da matsalolin da mafita na dokar farko ta Newton

1. A 2-kg block lies on a rough inclined plane at an angle 37o to the horizontal. Determine the magnitude of the external force exerted on the block, so the block is not slides down the plane. (sin 37o = 0.6, cos 37o = 0.8, g = 10 m.s-2, µk = 0.2)

Equilibrium of bodies on inclined plane – application of Newton's first law problems and solutions 1An sani:

Mass (m) = 2 kg

Hanzarta saboda nauyi (g) = 10 m/s2

Block’s nauyi (w) = mg = (2)(10) = 20 Newtons

Zunubi 37o = 0.6

Kos 37o = 0.8

Coefficient na gogayya ta motsik) = 0.2

The y-component of the weight (wy) = w kowa 37o = (20)(0.8) = 16 Newton

The x-component of the weight (wx) = w sin θ = (20)(sin 37) = (20)(0.6) = 12 Newton

the normal force (N) = wy = 16 Newton

Nema : The external force (F)

Magani :

Equilibrium of bodies on inclined plane – application of Newton's first law problems and solutions 2wx = 12 Newton

The force of the kinetic friction (fk) = µk N = (0.1)(16) = 1.6 Newton

The magnitude of the external force F exerted on the block :

F + fk - wx = 0

F = wx - fk

F = 12 – 1.6

F = 10.4 Newton

The external force F greater than 10.4 Newton.

Ka kuma duba  Faɗaɗa yanki - matsaloli da mafita

2. Mass of a block = 2 kg, coefficient of static friction µs = 0.4 and θ = 45o. Determine the magnitude of the force F so the block start to slides up.

Equilibrium of bodies on inclined plane – application of Newton's first law problems and solutions 3An sani:

The coefficient of the static friction (µs) = 0.4

Kusurwoyi (θ) = 45o

Saurin gudu saboda nauyi (g) = 10 m/s2

Block’s mass (m) = 2 kilogram

Block’s weight (w) = m g = (2 kg)(10 m/s2) = 20 kg m/s2 = 20 Newton

The x-component of the weight (wx) = w sin θ = (20)(sin 45) = (20)(0.5√2) = 10√2 Newton

The y-component of the weight (wy) = w cos θ = (20)(cos 45) = (20)(0.5√2) = 10√2 Newton

Nema : The magnitude of the force F

Magani:

Equilibrium of bodies on inclined plane – application of Newton's first law problems and solutions 4Block starts to slide up, if Fwx + fs.

The x-component of the weight :

wx = 10√2 Newton

the y-component of the weight :

wy = 10√2 Newton

Ƙarfin al'ada :

N = wy = 10√2 Newton

The force of the static friction :

fs = µs N = (0,4)(10√2) = 4√2

The magnitude of the force F so that the block starts to slide up :

Fwx + fs

F ≥ 10√2 + 4.2

F ≥ 14√2 Newton

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  3. Daidaiton jikin da aka haɗa ta hanyar igiyoyi da ƙwallo
  4. Equilibrium of bodies on the inclined plane

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