Ekuilibri i trupave në një plan të pjerrët - zbatimi i problemeve dhe zgjidhjeve të ligjit të parë të Njutonit

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 1I njohur:

Masë (m) = 2 kg

Përshpejtimi për shkak të gravitetit (g) = 10 m/s2

Block’s peshë (w) = mg = (2)(10) = 20 Njuton

Mëkati 37o = 0.6

Kosto 37o = 0.8

Coefficient of the fërkim kinetikk) = 0.2

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

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

the normal force (N) = wy = 16 Njuton

Shtepi : The external force (F)

Zgjidhje :

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

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

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 Njuton

The external force F greater than 10.4 Newton.

Shih edhe  Zgjerimi i zonës - problemet dhe zgjidhjet

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 3I njohur:

The coefficient of the static friction (µs) = 0.4

Këndi (θ) = 45o

Përshpejtimi për shkak të gravitetit (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 Njuton

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

Shtepi : The magnitude of the force F

zgjidhje:

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 Njuton

the y-component of the weight :

wy = 10√2 Njuton

Forca normale :

N = wy = 10√2 Njuton

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 + 42

F ≥ 14√2 Newton

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  1. Particles in one-dimensional equilibrium
  2. Particles in two-dimensional equilibrium
  3. Equilibrium of bodies connected by cord and pulley
  4. Equilibrium of bodies on the inclined plane

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