Ligevægt af legemer på et skråplan – anvendelse af Newtons første lov, problemer og løsninger

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 1Kendt:

Masse (m) = 2 kg

Acceleration på grund af tyngdekraften (g) = 10 m/s2

Blok's vægt (w) = mg = (2)(10) = 20 Newton

Synd 37o = 0.6

For 37o = 0.8

Koefficienten for kinetisk friktion (µk) = 0.2

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

Wanted : The external force (F)

Løsning :

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 3Kendt:

The coefficient of the static friction (µs) = 0.4

Vinkel (θ) = 45o

Tyngdeacceleration (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

Wanted : The magnitude of the force F

opløsning:

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

The x-component of the weight :

wx = 10√2 Newton

the y-component of the weight :

wy = 10√2 Newton

Normalkraften :

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 :

F ≥ wx + fs

F ≥ 10√2 + 4√2

F ≥ 14√2 Newton

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  1. Partikler i endimensionel ligevægt
  2. Partikler i todimensionel ligevægt
  3. Ligevægt mellem legemer forbundet med snore og taljer
  4. Equilibrium of bodies on the inclined plane

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