Motion on the rough inclined plane with the friction force – application of Newton’s law of motion problems and solutions

1. Object’s masa = 2 kg, paātrinājums gravitācijas dēļ = 9.8m/s², coefficient of the static friction = 0.2, coefficient of the kinetic friction = 0.1. Is the object at rest or accelerating? If the object is accelerated, find (a) the net force (b) magnitude and direction of the box’s paātrinājums!

Šķīdums

Zināms:

Masa (m) = 2 kg

Brīvās krišanas paātrinājums (g) = 9.8 m/s²

Coefficient of the static friction (μ_s) = 0.2

Coefficient of the kinetic friction (μ_k) = 0.1

Weight (w) = m g = (2)(9.8) = 19.6 Newton

The horizontal component of the svars (w_x) = w sin 30^o = (19.6)(0.5) = 9.8 ņūtoni

The vertical component of th weight (w_y) = w cos 30^o = (19.6)(0.5√3) = 9.8√3 Newton

Normālais spēks (N) = w_y = 9.8√3 Ņūtons

Force of the static friction (f_s) = (0.2)(9.8√3) = 1.96√3 Newton = 3.39 Newton

Force of the kinetic friction (f_k) = (0.1)(9.8√3) = 0.98√3 Newton = 1.69 Newton

šķīdums:

Object is at rest if w_x < f_s, object is moving down if w_x > f_s.

w_x = 9.8 Newton and f_s = 3.39 ņūtoni.

(a) the net force

∑F = w_x - f_k = 9.8 – 1.69 = 8.11 ņūtoni

(b) magnitude and direction of the acceleration

∑F = ma

8.11 = (2) a

a = 4.05

Paātrinājuma lielums = 4.05 m/s² and direction of the acceleration = downward.

2. Object’s mass = 4 kg, acceleration due to gravity = 9,8 m/s². Coefficient of the kinetic friction = 0.2 and coefficient of the static friction = 0.4. Magnitude of the force F = 40 Newton. The object is at rest or slides down ? If the object slides down, find (a) the net force (b) magnitude and direction of the acceleration!

Šķīdums

Zināms:

Masa (m) = 4 kg

Brīvās krišanas paātrinājums (g) = 9.8 m/s²

The coefficient of the static friction (μ_s) = 0.4

The coefficient of the kinetic friction (μ_k) = 0.2

Svars (w) = mg = (4)(9.8) = 39.2 ņūtoni

The horizontal component of the weight (w_x) = w sin 30^o = (39.2)(0.5) = 19.6 ņūtoni

The vertical component of the weight (w_y) = w cos 30^o = (392)(0..5√3) = 19.6√3 Newton

The normal force (N) = w_y = 19.6√3 Newton = 33.95 Newton

the static friction force (f_s) = μ_s N= (0,4)(33.95) = 13.58 Newton

The kinetic friction force (f_k) = μ_k N= (0.2)(33.95) = 6.79 Newton

F = 40 ņūtoni

šķīdums:

The object slides down if F < w_x +f_s. The object slides up if F > w_x +f_s.

F = 40 Newton, w_x = 19.6 Newton and f_s = 13.58 ņūtoni.

F is greater than w_x +f_s so the object slides up.

(a) The net force

∑F = F – w_x - f_k = 40 – 19.6 – 6.79 = 13.61 ņūtoni

(b) The magnitude and direction of the acceleration

∑F = ma

6.4 = (4) a

a = 1.6

The magnitude of the acceleration is 1.6 m/s² and direction of the acceleration is upward.

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1. Masa un svars
2. Normāls spēks
3. Ņūtona otrais kustības likums
4. Berzes spēks
5. Kustība uz horizontālas virsmas bez berzes spēka
6. Divu ķermeņu kustība ar vienādu paātrinājumu uz nelīdzenas horizontālas virsmas ar berzes spēku
7. Kustība slīpā plaknē bez berzes spēka
8. Kustība uz aptuvenas slīpas plaknes ar berzes spēku
9. Kustība liftā
10. Ķermeņu kustību savieno auklas un skriemeļi
11. Divi ķermeņi ar vienādu paātrinājuma lielumu
12. Plakanas līknes noapaļošana – apļveida kustības dinamika
13. Slīpas līknes apļošana – apļveida kustības dinamika
14. Vienmērīga kustība horizontālā aplī
15. Centripetāls spēks vienmērīgā apļveida kustībā