Két azonos nagyságú gyorsulású test – Newton mozgástörvényének alkalmazása problémák és megoldások

1. Two masses m₁ = 2 kg and 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 kinetikus súrlódás between m₂ and incline is 0.1.

(a) Determine their gyorsulás

(b) Determine the tension force

Ismert:

Tömeg 1 (m₁) = 2 kg

Tömeg 2 (m₂) = 4 kg

Coefficient of the kinetic friction between m₁ és a ferde sík (μ_k1) = 0.2

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

A gravitációs gyorsulás (g) = 9.8 m/s²

a) The magnitude and direction of the acceleration

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

w_1x = w₁ bűn 30^o = (19.6⁻⁹⁴ N)(0.5) = 9.8 Newton

w_1y = w₁ cos 30^o = (19.6⁻⁹⁴ N)(0.87) = 17 Newton

N₁ = A normál erő a m₁ = w_1y = 17 Newton

F_k1 = The force of the kinetic friction on m₁ = μ_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 = w₂ bűn 60^o = (39.2⁻⁹⁴ N)(0.87) = 34.1 Newton

w_2y = w₂ cos 60^o = (39.2⁻⁹⁴ N)(0.5) = 19.6 Newton

N₂ = The normal force on m₂ = w_2y = 19.6 Newton

F_k2 = The force of the kinetic friction on m₂ = μ_k2 N₂ = (0.1)(19.6 N) = 1.96 Newton

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A gyorsulás nagysága:

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 + T.₁ - w_1x - F_k1 = (m₁ +m₂), hogy_x

w_2x - F_k2 - w_1x - F_k1 = (m₁ +m₂ ), hogy_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

A gyorsulás nagysága = 3.16 m/s² . 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 = T.₂ = 19.5 Newton

2.m₁ = 4 kg, m²₂ = 2 kg. Determine (a) magnitude and direction of the acceleration (b) Magnitude of the tension force which connecting m₁ és M₂ (c) magnitude of the tension force which connecting pulley and roof.

Megoldás

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 + T.₂ - w₂ = (m₁ +m₂), hogy_y

w₁ - w₂ = (m₁ +m₂), hogy_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 és M₂

Jelentkezem Newton második törvénye a 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 = T.₂ = 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 = T.₁ + T.₂

T₁ és T₂ have the same magnitude, T₁ = T.₂ = 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.

Megoldás

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 = w₂ cos 30^o = (147⁻⁹⁴ N)(0.87) = 127.89 Newton

w_2x = w₂ bűn 30^o = (147⁻⁹⁴ 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 - w_2x - w₁ - T₂ + T.₁ = 0

F – F_k2 - w_2x - w₁ = 0

F = F_k2 +w_2x +w₁

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 - w₁ = 0

T₁ = w₁ = 98 Newton

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

F – F_k2 - w_2x - T₂ = 0

T₂ = F – F_k2 - w_2x

T₂ = 225.2 N – 53.7 N – 73.5 N

T₂ = 98 Newton

A húzóerő nagysága = T₁ = T.₂ = 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.

(1) 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 ?

megoldás:

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

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

w_1x = w₁ bűn 60^o = (156.8⁻⁹⁴ N)(0.87) = 136.4 Newton

w_1y = w₁ cos 60^o = (156.8⁻⁹⁴ N)(0.5) = 78.4 Newton

N₁ = A normal force exerted on the block 1 by the inclined plane = w_1y = 78.4 Newton

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

N₂₃ = A normal force exerted on the block 3 bythe  block 2 = w₃ = 49 Newton

N₃₂ = The normal force exerted on the block 2 by the block 3 = N₂₃ = w₃ = 49 Newton

(N₂₃ és a N₃₂ are action-reaction pair)

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

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

(F_s23 és a F_s32 are action-reaction pair)

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

N₂ = A normal force exerted on the object 2 by the horizontal surface = w₂ +N₃₂ = 117.6 Newton + 49

Newton = 166.6 Newton

F_k2 = A force of the kinetic friction on the block 2 = μ_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 = μ_s N₂₃ = μ_s w₃ = μ_s m₃ g

μ_s m₃ g = m₃ a_x

μ_s g = a_x

a_x = (0.3)(9.8 m/s²) = 2.94 m/s²

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 + T.₂ - F_k2 - F_s32 + F_s23 = (m₁ +m₂ +m₃), hogy_x

w_1x - F_k2 = (m₁ +m₂ +m₃ ), hogy_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 lower 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), hogy_x

—–> F_k2 = μ_k N₂ = μ_k w₂ = μ_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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