Gläichgewiicht vu Kierper op enger schréieger Fläch – Uwendung vun de Problemer a Léisunge vum éischte Gesetz vum 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 1Bekannt:

Mass (m) = 2 kg

Beschleunegung duerch Schwéierkraaft (g) = 10 m/s2

Block an Gewiicht (w) = mg = (2)(10) = 20 Newton

Ouni 37o = 0.6

Kosch 37o = 0.8

Coefficient of the kinetesch Reiwungk) = 0.2

The y-component of the weight (wy) = w fir 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

Gewënschte : The external force (F)

Léisung :

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.

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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 3Bekannt:

The coefficient of the static friction (µs) = 0.4

Wénkel (θ) = 45o

Schwéierkraaftbeschleunigung (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

Gewënschte : The magnitude of the force F

Léisung:

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

Déi normal Kraaft :

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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  1. Partikelen am eendimensionalen Gläichgewiicht
  2. Partikelen am zweedimensionalen Gläichgewiicht
  3. Gläichgewiicht vu Kierper, déi duerch Schnouer a Riemscheiwen verbonne sinn
  4. Equilibrium of bodies on the inclined plane

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