Te taurite o ngā tinana i runga i te papa whakarara – te whakamahinga o ngā raruraru me ngā otinga o te ture tuatahi a 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 1Mōhiotia:

Mass (m) = 2 kg

Whakaterenga na te kaha o te kaha (g) = 10 m/s2

Block’s taimaha (w) = mg = (2)(10) = 20 Ngā Newton

Hara 37o = 0.6

Kohinga 37o = 0.8

Tauwehenga o te te waku nekenekek) = 0.2

The y-component of the weight (wy) =w whaimana 37o = (20)(0.8) = 16 Ngā 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 Niutona

hiahia : The external force (F)

otinga :

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

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

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.

A hi'o atoa  Te wehewehenga mā te kōhao kotahi – ngā raruraru me ngā otinga

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 3Mōhiotia:

The coefficient of the static friction (µs) = 0.4

Koki (θ) = 45o

Te whakaterenga nā te kaha ā-papa (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 Niutona

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

hiahia : The magnitude of the force F

Rongoā:

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 Nūtene

the y-component of the weight :

wy = 10√2 Nūtene

Te kaha noa :

N = wy = 10√2 Nūtene

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. Ngā matūriki i roto i te taurite kotahi-ahu
  2. Ngā matūriki i roto i te taurite rua-ahu
  3. Te taurite o ngā tinana e honoa ana e ngā taura me ngā pūwero
  4. Equilibrium of bodies on the inclined plane

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