Twee lichamen met dezelfde versnelling – Toepassing van Newtons bewegingswetten: problemen en oplossingen

1. Two masses m₁ = 2 kg en m₂ = 5 kg are on inclined plane and are connected together by a string as shown in the figure. The coefficient of the kinetic friction between m₁ and incline is 0.2 and the coefficient of the kinetische wrijving between m₂ and incline is 0.1.

(a) Determine their versnelling

(b) Determine the tension force

Bekend:

Massa 1 (m₁) = 2 kg

Massa 2 (m₂) = 4 kg

Coefficient of the kinetic friction between m₁ en Hellend vlak (μ_k1) = 0.2

Coefficient of the kinetic friction between m₂ and inclined plane (μ_k2) = 0.1

Versnelling als gevolg van zwaartekracht (g) = 9.8 m/s²

a) The magnitude and direction of the acceleration

w₁ = gewicht 1 = m₁ g = (2 kg)(9.8 m/s²) = 19.6 Newton

w_1x = met₁ zonde 30^o = (19.6 N)(0.5) = 9.8 Newton

w_1y = met₁ cos 30^o = (19.6 N)(0.87) = 17 Newton

N₁ = Het normale kracht op m₁ = met_1y = 17 Newton

F_k1 = The force of the kinetic friction on m₁ =_k1 N₁ = (0.2)(17 N) = 3.4 Newton

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w₂ = weight 2 = m₂ g = (4 kg)(9.8 m/s²) = 39.2 Newton

w_2x = met₂ zonde 60^o = (39.2 N)(0.87) = 34.1 Newton

w_2y = met₂ cos 60^o = (39.2 N)(0.5) = 19.6 Newton

N₂ = The normal force on m₂ = met_2y = 19.6 Newton

F_k2 = The force of the kinetic friction on m₂ =_k2 N₂ = (0.1)(19.6 N) = 1.96 Newton

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De grootte van de versnelling:

ΣF_x = ma_x

w_2x > w_1x so direction of the acceleration is the same as direction of w_2x.

Forces which points along acceleration is positive and forces which has opposite direction with acceleration is negative.

w_2x - F_k2 - T₂ + T₁ - in_1x - F_k1 = (m₁ + m₂) om_x

w_2x - F_k2 - in_1x - F_k1 = (m₁ + m₂ ) om_x

34.1 N – 1.96 N – 9.8 N – 3.4 N = (2 kg + 4 kg) a_x

18.94 N = (6 kg) a_x

a_x = 18.94 N : 6 kg

a_x = 3.16 m/s²

Grootte van de versnelling = 3.16 m/s²² . Direction of the acceleration = direction of T₁ = direction of w_2x

b) Magnitude of the tension force

Apply Newton’s second law on the object 2 :

w_2x - F_k2 - T₂ = m₂ a_x

34.1 N – 1.96 N – T₂ = (4 kg) (3.16 m/s²)

32.14 N – T₂ = 12.64 N

T₂ = 32.14 N – 12.64 N = 19.5 Newton

The tension force = T = T₁ = T₂ = 19.5 Newton

2.m₁ = 4 kg, m₂ = 2 kg. Determine (a) magnitude and direction of the acceleration (b) Magnitude of the tension force which connecting m₁ en M₂ (c) magnitude of the tension force which connecting pulley and roof.

Het resultaat

w₁ = m₁ g = (4 kg)(9.8 m/s²) = 39.2 Newton

w₂ = m₂ g = (2 kg)(9.8 m/s²) = 19.6 Newton

a) Magnitude and direction of the acceleration

ΣF_y = ma_y

w₁ > w₂ so the direction of the object is same as the direction of the weight 1 (w₁). Forces which has the same direction with acceleration is positive and forces which has opposite direction with acceleration is negative.

w₁ - T₁ + T₂ - in₂ = (m₁ + m₂) om_y

w₁ - in₂ = (m₁ + m₂) om_y

39.2 N – 19.6 N = (4 kg + 2 kg) a_y

19.6 N = (6 kg) a_y

a_y = 19.6 N : 6 kg

a_y = 3.26 m/s²

Magnitude of acceleration = 3.26 m/s². Direction of acceleration = direction of w₁ .

b) Magnitude of tension force which connecting m₁ en M₂

Toepassen Newton's tweede wet op m₂ :

ΣF_y = ma_y

w₁ - T₁ = m₁ a_y

39.2 N – T₁ = (4 kg)( 3.26 m/s²)

39.2 N – T₁ = 13.04 N

T₁ = 39.2 N – 13.04 N

T₁ = 26.16 Newton

Magnitude of the tension force which connection objects = T = T₁ = T₂ = 26.16 Newton

c) Magnitude of the tension force which connecting pulley and roof.

Pulley is at rest :

ΣF_y = ma_y —— a_y = 0

ΣF_y = 0

Upward force are positive, downward forces are negative :

T₃ - T₁ - T₂ = 0

T₃ = T₁ + T₂

T₁ en T₂ have the same magnitude, T₁ = T₂ = T = 26.16 N :

T₃ = 2T = 2(26.16 N) = 52.32 Newton

3. Block 1 (m₁ = 10 kg) and block 2 (m₂ = 15 kg) connected by a cord over frictionless pulley. Coefficient of the static friction between the block 2 with incline = 0.6. The coefficient of the kinetic friction between the block 2 with incline = 0.42. Determine (a) The magnitude of the minimum force F exerted on the objects so the objects accelerated upward (b) Determine the magnitude of the tension force.

Het resultaat

w₁ = The weight of the block 1 = m₁ g = (10 kg)(9.8 m/s²) = 98 Newton

w₂ = The weight of the block 2 = m₂ g = (15 kg)(9.8 m/s²) = 147 Newton

w_2y = met₂ cos 30^o = (147 N)(0.87) = 127.89 Newton

w_2x = met₂ zonde 30^o = (147 N)(0.5) = 73.5 Newton

N₂ = The normal force on the block 2 = w_2y = 127.89 Newton

F_k2 = The force of the kinetic friction on the block 2 = μ_k2 N₂ = (0.42)(127.89 N) = 53.7 Newton

F_s2 = The force of the static friction on the block 2 = μ_s2 N₂ = (0.6)(127.89 N) = 76.7 Newton

a) The magnitude of the minimum force F exerted on the objects so the objects accelerated upward

ΣF_x = ma_x —— a_x = 0

ΣF_x = 0

Upward forces and rightward forces are positive, downward forces and leftward forces are negative.

F – F_k2 - in_2x - in₁ - T₂ + T₁ = 0

F – F_k2 - in_2x - in₁ = 0

V = V_k2 + met_2x + met₁

F = 53.7 N + 73.5 N + 98 N

F = 225.2 Newton

b) The magnitude of the tension force

Apply Newton’s law of the motion on the block 1 :

ΣF_y = ma_y —— a_y = 0

ΣF_y = 0

T₁ - in₁ = 0

T₁ = met₁ = 98 Newton

Apply Newton’s law of the motion on the block 2 :

F – F_k2 - in_2x - T₂ = 0

T₂ = F – F_k2 - in_2x

T₂ = 225.2 N – 53.7 N – 73.5 N

T₂ = 98 Newton

Grootte van de trekkracht = T₁ = T₂ = T = 98 Newton

4. Block 1 (m₁ = 16 kg) lies on a horizontal surface and the block 2 (m₂ = 12 kg) lies on a smooth inclined plane, connected by a cord that passes over a small, frictionless pulley. Block 3 (m₃ = 5 kg) lies on the block 2. The coefficient of the kinetic friction between the block 2 and the horizontal surface is 0,4. The coefficient of the static friction between the block 2 with the block 3 is 0,3.

(A) When the system is released from rest, the block 3 and the block 2 still slide together ?

(B) If there is no block 3, what is the acceleration of the block 1 and the block 2 ?

oplossing:

a) When the system is released from rest, the block 3 and the block 2 still slide together?

w₁ = Het weight of the block 1 = m₁ g = (16 kg)(9.8 m/s²) = 156.8 Newton

w_1x = met₁ zonde 60^o = (156.8 N)(0.87) = 136.4 Newton

w_1y = met₁ cos 60^o = (156.8 N)(0.5) = 78.4 Newton

N₁ = Het normal force exerted on the block 1 by the inclined plane = met_1y = 78.4 Newton

w₃ = Het weight of the block 3 = m₃ g = (5 kg)(9.8 m/s²) = 49 Newton

N₂₃ = Het normal force exerted on the block 3 bythe  block 2 = met₃ = 49 Newton

N₃₂ = The normal force exerted on the block 2 by the block 3 = N₂₃ = met₃ = 49 Newton

(N₂₃ en N₃₂ are action-reaction pair)

F_s23 = Het force of the static friction exerted on the block 3 by the block 2 =_s N₂₃ = (0.3)(49 N) = 14.7 Newton

F_s32 = Het force of the static friction exerted on th block 2 by the block 3 =F_s23 = 14.7 Newton

(F_s23 en F_s32 are action-reaction pair)

w₂ = Het weight of the block 2 = m₂ g = (12 kg)(9.8 m/s²) = 117.6 Newton

N₂ = Het normal force exerted on the object 2 by the horizontal surface = met₂ + N₃₂ = 117.6 Newton + 49

Newton = 166.6 Newton

F_k2 = Het force of the kinetic friction on the block 2 =_k N₂ = (0.4)(166.6 N) = 66.64 Newton

Apply Newton’s law of motion on the block 3 :

ΣF_x = ma_x

F_s23 =m₃ a_x

—–> F_s23 =_s N₂₃ =_s w₃ =_s m₃ g

μ_s m₃ g = m₃ a_x

μ_s g = een_x

a_x = (0.3)(9.8 m/s²) = 2.94 m/sec²

The maximum acceleration of the block 3 so that the block 3 and the block 2 still slide together is 2.94 m/s².

Now we calculate the magnitude of the system’s acceleration after released from rest.

The direction of the block displacement = the direction of the block’s acceleration = the direction of T₂ = the direction of w_1x.

ΣF_x = ma_x

w_1x - T₁ + T₂ - F_k2 - F_s32 + F_s23 = (m₁ + m₂ + m₃) om_x

w_1x - F_k2 = (m₁ + m₂ + m₃ ) om_x

136.4 N – 66.64 N = (16 kg + 12 kg + 5 kg) a_x

69.76 N = (33 kg) a_x

a_x = 2.11 m/s²

a_x is positive, means direction of the block displacement or the direction of the acceleration is same as direction of T₂ or direction of w_1x.

The magnitude of the acceleration is 2.11 m / s² , Theower than 2.94 m / s² so we can conclude that block 3 and block 2 still slide together after released from rest.

b) The magnitude of the acceleration of the block 1 and the block 2

ΣF_x = ma_x

w_1x - F_k2 = (m₁ + m₂) om_x

—–> F_k2 =_k N₂ =_k w₂ =_k m₂ g = (0.4)(12 kg)(9.8 m/s²) = 47.04 Newton

136.4 N – 47.04 N = (16 kg + 12 kg) a_x

89.36 N = (28 kg) a_x

a_x = 89.36 N : 28 kg = 3.19 m/s²

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1. Massa en gewicht
2. Normale kracht
3. Newton's tweede bewegingswet
4. Wrijvingskracht
5. Beweging op een horizontaal oppervlak zonder wrijvingskracht
6. De beweging van twee lichamen met dezelfde versnelling op een ruw horizontaal oppervlak met wrijvingskracht.
7. Beweging op een hellend vlak zonder wrijvingskracht
8. Beweging op het ruwe hellende vlak met wrijvingskracht
9. Beweging in een lift
10. De beweging van lichamen wordt mogelijk gemaakt door touwen en katrollen.
11. Twee lichamen met dezelfde versnellingsgrootte.
12. Het nemen van een vlakke bocht – dynamiek van cirkelbeweging
13. Het nemen van een hellende bocht – dynamiek van cirkelbeweging
14. Gelijkmatige beweging in een horizontale cirkel
15. Middelpuntzoekende kracht bij eenparige cirkelbeweging