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

Two bodies with the same magnitude of acceleration – Application of Newton's law of motion problems and solutions 1

Known :

Mass 1 (m1) = 2 kg

Mass 2 (m2) = 4 kg

Coefficient of the kinetic friction between m1 and inclined planek1) = 0.2

Coefficient of the kinetic friction between m2 and inclined plane (μk2) = 0.1

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

a) The magnitude and direction of the acceleration

Two bodies with the same magnitude of acceleration – Application of Newton's law of motion problems and solutions 2

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

w1x = w1 sin 30o = (19.6 N)(0.5) = 9.8 Newton

w1y = w1 cos 30o = (19.6 N)(0.87) = 17 Newton

N1 = The normal force on m1 = w1y = 17 Newton

Fk1 = The force of the kinetic friction on m1 = μk1 N1 = (0.2)(17 N) = 3.4 Newton

———

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

w2x = w2 sin 60o = (39.2 N)(0.87) = 34.1 Newton

w2y = w2 cos 60o = (39.2 N)(0.5) = 19.6 Newton

N2 = The normal force on m2 = w2y = 19.6 Newton

Fk2 = The force of the kinetic friction on m2 = μk2 N2 = (0.1)(19.6 N) = 1.96 Newton

———

The magnitude of the acceleration :

Fx = m ax

w2x > w1x so direction of the acceleration is the same as direction of w2x.

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

w2x – Fk2 – T2 + T1 – w1x – Fk1 = (m1 + m2) ax

w2x – Fk2 – w1x – Fk1 = (m1 + m2 ) ax

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

18.94 N = (6 kg) ax

ax = 18.94 N : 6 kg

ax = 3.16 m/s2

Magnitude of the acceleration = 3.16 m/s2 . Direction of the acceleration = direction of T1 = direction of w2x

b) Magnitude of the tension force

Apply Newton’s second law on the object 2 :

w2x – Fk2 – T2 = m2 ax

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

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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.

Two bodies with the same magnitude of acceleration – Application of Newton's law of motion problems and solutions 3

Solution

Two bodies with the same magnitude of acceleration – Application of Newton's law of motion problems and solutions 4

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

Fy = m ay

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) ay

w1 – w2 = (m1 + m2) ay

39.2 N – 19.6 N = (4 kg + 2 kg) ay

19.6 N = (6 kg) ay

ay = 19.6 N : 6 kg

ay = 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 :

Fy = m ay

w1 – T1 = m1 ay

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.

Two bodies with the same magnitude of acceleration – Application of Newton's law of motion problems and solutions 5Pulley is at rest :

Fy = m ay —— ay = 0

Fy = 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

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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.

Two bodies with the same magnitude of acceleration – Application of Newton's law of motion problems and solutions 6

Solution

Two bodies with the same magnitude of acceleration – Application of Newton's law of motion problems and solutions 7

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

w2y = w2 cos 30o = (147 N)(0.87) = 127.89 Newton

w2x = w2 sin 30o = (147 N)(0.5) = 73.5 Newton

N2 = The normal force on the block 2 = w2y = 127.89 Newton

Fk2 = The force of the kinetic friction on the block 2 = μk2 N2 = (0.42)(127.89 N) = 53.7 Newton

Fs2 = The force of the static friction on the block 2 = μs2 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

Fx = m ax —— ax = 0

Fx = 0

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

F – Fk2 – w2x – w1 – T2 + T1 = 0

F – Fk2 – w2x – w1 = 0

F = Fk2 + w2x + 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 :

Fy = m ay —— ay = 0

Fy = 0

T1 – w1 = 0

T1 = w1 = 98 Newton

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

F – Fk2 – w2x – T2 = 0

T2 = F – Fk2 – w2x

T2 = 225.2 N – 53.7 N – 73.5 N

T2 = 98 Newton

Magnitude of the tension force = T1 = T2 = T = 98 Newton

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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 ?

Two bodies with the same magnitude of acceleration – Application of Newton's law of motion problems and solutions 8

Solution :

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

Two bodies with the same magnitude of acceleration – Application of Newton's law of motion problems and solutions 9

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

w1x = w1 sin 60o = (156.8 N)(0.87) = 136.4 Newton

w1y = w1 cos 60o = (156.8 N)(0.5) = 78.4 Newton

N1 = The normal force exerted on the block 1 by the inclined plane = w1y = 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)

Fs23 = The force of the static friction exerted on the block 3 by the block 2 = μs N23 = (0.3)(49 N) = 14.7 Newton

Fs32 = The force of the static friction exerted on th block 2 by the block 3 = Fs23 = 14.7 Newton

(Fs23 and Fs32 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

Fk2 = The force of the kinetic friction on the block 2 = μk N2 = (0.4)(166.6 N) = 66.64 Newton

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

Fx = m ax

Fs23 =m3 ax

—–> Fs23 = μs N23 = μs w3 = μs m3 g

μs m3 g = m3 ax

μs g = ax

ax = (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 w1x.

Fx = m ax

w1x – T1 + T2 – Fk2 – Fs32 + Fs23 = (m1 + m2 + m3) ax

w1x – Fk2 = (m1 + m2 + m3 ) ax

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

69.76 N = (33 kg) ax

ax = 2.11 m/s2

ax is positive, means direction of the block displacement or the direction of the acceleration is same as direction of T2 or direction of w1x.

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

Fx = m ax

w1x – Fk2 = (m1 + m2) ax

—–> Fk2 = μk N2 = μk w2 = μk m2 g = (0.4)(12 kg)(9.8 m/s2) = 47.04 Newton

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

89.36 N = (28 kg) ax

ax = 89.36 N : 28 kg = 3.19 m/s2

  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