Two bodies with the same magnitude of acceleration – Application of Newton’s law of motion problems and solutions

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 kinetic friction between m₂ and incline is 0.1.

(a) Determine their acceleration

(b) Determine the tension force

Known :

Mass 1 (m₁) = 2 kg

Mass 2 (m₂) = 4 kg

Coefficient of the kinetic friction between m₁ and inclined plane (μ_k1) = 0.2

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

Acceleration due to gravity (g) = 9.8 m/s²

a) The magnitude and direction of the acceleration

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

w_1x = w₁ sin 30^o = (19.6 N)(0.5) = 9.8 Newton

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

N₁ = The normal force on 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

———

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

w_2x = w₂ sin 60^o = (39.2 N)(0.87) = 34.1 Newton

w_2y = w₂ cos 60^o = (39.2 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

———

The magnitude of the acceleration :

∑F_x = m a_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 – w_1x – F_k1 = (m₁ + m₂) a_x

w_2x – F_k2 – w_1x – F_k1 = (m₁ + m₂ ) a_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

Magnitude of the acceleration = 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 = 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 and m₂ (c) magnitude of the tension force which connecting pulley and roof.

Solution

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

∑F_y = m a_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 – w₂ = (m₁ + m₂) a_y

w₁ – w₂ = (m₁ + m₂) a_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 and m₂

Apply Newton’s second law on m₂ :

∑F_y = m a_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 :

∑F_y = m a_y —— a_y = 0

∑F_y = 0

Upward force are positive, downward forces are negative :

T₃ – T₁ – T₂ = 0

T₃ = T₁ + T₂

T₁ and 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.

Solution

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 N)(0.87) = 127.89 Newton

w_2x = w₂ sin 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

∑F_x = m a_x —— a_x = 0

∑F_x = 0

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

F – F_k2 – w_2x – w₁ – T₂ + T₁ = 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 :

∑F_y = m a_y —— a_y = 0

∑F_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

Magnitude of the tension force = 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 ?

Solution :

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

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

w_1x = w₁ sin 60^o = (156.8 N)(0.87) = 136.4 Newton

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

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

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

N₂₃ = The 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 and N₃₂ are action-reaction pair)

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

(F_s23 and F_s32 are action-reaction pair)

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

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

Newton = 166.6 Newton

F_k2 = The 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 :

∑F_x = m a_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.

∑F_x = m a_x

w_1x – T₁ + T₂ – F_k2 – F_s32 + F_s23 = (m₁ + m₂ + m₃) a_x

w_1x – F_k2 = (m₁ + m₂ + m₃ ) a_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

∑F_x = m a_x

w_1x – F_k2 = (m₁ + m₂) a_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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1. Mass and weight
2. Normal force
3. Newton’s second law of motion
4. Friction force
5. Motion on the horizontal surface without friction force
6. The motion of two bodies with the same acceleration on the rough horizontal surface with the friction force
7. Motion on the inclined plane without friction force
8. Motion on the rough inclined plane with the friction force
9. Motion in an elevator
10. The motion of bodies connected by cord and pulley
11. Two bodies with the same magnitude of accelerations
12. Rounding a flat curve – dynamics of circular motion
13. Rounding a banked curve – dynamics of circular motion
14. Uniform motion in a horizontal circle
15. Centripetal force in uniform circular motion