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

1. Two masses m_{1 }= 2 kg and m_{2} = 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 m_{1} and incline is 0.2 and the coefficient of the kinetic friction between m_{2} and incline is 0.1.

(a) Determine their acceleration

(b) Determine the tension force

__Known :__

Mass 1 (m_{1}) = 2 kg

Mass 2 (m_{2}) = 4 kg

Coefficient of the kinetic friction between m_{1} and inclined plane (μ_{k1}) = 0.2

Coefficient of the kinetic friction between m_{2} and inclined plane (μ_{k2}) = 0.1

Acceleration due to gravity (g) = 9.8 m/s^{2}

a) The magnitude and direction of the acceleration

w_{1} = weight 1 = m_{1 }g = (2 kg)(9.8 m/s^{2}) = 19.6 Newton

w_{1x} = w_{1} sin 30^{o} = (19.6 N)(0.5) = 9.8 Newton

w_{1y} = w_{1} cos 30^{o }= (19.6 N)(0.87) = 17 Newton

N_{1 }= The normal force on m_{1} = w_{1y} = 17 Newton

F_{k1} = The force of the kinetic friction on m_{1} = μ_{k1 }N_{1} = (0.2)(17 N) = 3.4 Newton

———

w_{2} = weight 2 = m_{2 }g = (4 kg)(9.8 m/s^{2}) = 39.2 Newton

w_{2x} = w_{2 }sin 60^{o} = (39.2 N)(0.87) = 34.1 Newton

w_{2y} = w_{2} cos 60^{o }= (39.2 N)(0.5) = 19.6 Newton

N_{2} = The normal force on m_{2} = w_{2y }= 19.6 Newton

F_{k2 }= The force of the kinetic friction on m_{2} = μ_{k2} N_{2} = (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} – T_{2} + T_{1} – w_{1x} – F_{k1 }= (m_{1} + m_{2}) a_{x}

w_{2x} – F_{k2} – w_{1x} – F_{k1} = (m_{1} + m_{2} ) 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/s^{2}

Magnitude of the acceleration = 3.16 m/s^{2} . Direction of the acceleration = direction of T_{1 }= direction of w_{2x}

b) Magnitude of the tension force

Apply Newton’s second law on the object 2 :

w_{2x }– F_{k2 }– T_{2} = m_{2 }a_{x}

34.1 N – 1.96 N – T_{2 }= (4 kg)(3.16 m/s^{2})

32.14 N – T_{2} = 12.64 N

T_{2 }= 32.14 N – 12.64 N = 19.5 Newton

The tension force = T = T_{1} = T_{2} = 19.5 Newton

2. m_{1 }= 4 kg, m_{2} = 2 kg. Determine (a) magnitude and direction of the acceleration (b) Magnitude of the tension force which connecting m_{1} and m_{2} (c) magnitude of the tension force which connecting pulley and roof.

Solution

w_{1} = m_{1 }g = (4 kg)(9.8 m/s^{2}) = 39.2 Newton

w_{2 }= m_{2} g = (2 kg)(9.8 m/s^{2}) = 19.6 Newton

a) Magnitude and direction of the acceleration

∑F_{y} = m a_{y}

*w*_{1 }*> w*_{2 }*so the direction of the object is same as the direction of the weight 1 (**w*_{1}*)**. Forces which has the same direction with acceleration is positive and forces which has opposite direction with acceleration is negative.*

w_{1 }– T_{1} + T_{2} – w_{2 }= (m_{1} + m_{2}) a_{y}

w_{1} – w_{2} = (m_{1} + m_{2}) 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/s^{2}

Magnitude of acceleration = 3.26 m/s^{2}. Direction of acceleration = direction of w_{1} .

b) Magnitude of tension force which connecting m_{1} and m_{2}

Apply Newton’s second law on m_{2} :

∑F_{y} = m a_{y}

w_{1} – T_{1} = m_{1 }a_{y}

39.2 N – T_{1 }= (4 kg)( 3.26 m/s^{2})

39.2 N – T_{1} = 13.04 N

T_{1 }= 39.2 N – 13.04 N

T_{1} = 26.16 Newton

Magnitude of the tension force which connection objects = T = T_{1} = T_{2} = 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 :

T_{3} – T_{1} – T_{2} = 0

T_{3 }= T_{1} + T_{2}

T_{1} and T_{2} have the same magnitude, T_{1} = T_{2} = T = 26.16 N :

T_{3} = 2T = 2(26.16 N) = 52.32 Newton

3. Block 1 (m_{1} = 10 kg) and block 2 (m_{2 }= 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

w_{1} = The weight of the block 1 = m_{1} g = (10 kg)(9.8 m/s^{2}) = 98 Newton

w_{2} = The weight of the block 2 = m_{2} g = (15 kg)(9.8 m/s^{2}) = 147 Newton

w_{2y} = w_{2} cos 30^{o} = (147 N)(0.87) = 127.89 Newton

w_{2x} = w_{2 }sin 30^{o} = (147 N)(0.5) = 73.5 Newton

N_{2} = 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} N_{2} = (0.42)(127.89 N) = 53.7 Newton

F_{s2} = The force of the static friction on the block 2 = μ_{s2} N_{2} = (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} – w_{1} – T_{2} + T_{1} = 0

F – F_{k2} – w_{2x }– w_{1} = 0

F = F_{k2 }+ w_{2x} + w_{1}

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

T_{1} – w_{1} = 0

T_{1} = w_{1} = 98 Newton

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

F – F_{k2} – w_{2x} – T_{2} = 0

T_{2 }= F – F_{k2 }– w_{2x}

T_{2 }= 225.2 N – 53.7 N – 73.5 N

T_{2} = 98 Newton

Magnitude of the tension force = T_{1} = T_{2} = T = 98 Newton

4. Block 1 (m_{1} = 16 kg) lies on a horizontal surface and the block 2 (m_{2} = 12 kg) lies on a smooth inclined plane, connected by a cord that passes over a small, frictionless pulley. Block 3 (m_{3} = 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?

w_{1} = The weight of the block 1 = m_{1} g = (16 kg)(9.8 m/s^{2}) = 156.8 Newton

w_{1x} = w_{1} sin 60^{o} = (156.8 N)(0.87) = 136.4 Newton

w_{1y} = w_{1} cos 60^{o} = (156.8 N)(0.5) = 78.4 Newton

N_{1} = The normal force exerted on the block 1 by the inclined plane = w_{1y} = 78.4 Newton

w_{3} = The weight of the block 3 = m_{3} g = (5 kg)(9.8 m/s^{2}) = 49 Newton

N_{23} = The normal force exerted on the block 3 bythe block 2 = w_{3 }= 49 Newton

N_{32} = The normal force exerted on the block 2 by the block 3 = N_{23} = w_{3} = 49 Newton

(N_{23 }and N_{32} are action-reaction pair)

F_{s23} = The force of the static friction exerted on the block 3 by the block 2 = μ_{s} N_{23 }= (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_{s}_{23} = 14.7 Newton

(F_{s23} and F_{s32} are action-reaction pair)

w_{2} = The weight of the block 2 = m_{2} g = (12 kg)(9.8 m/s^{2}) = 117.6 Newton

N_{2} = The normal force exerted on the object 2 by the horizontal surface = w_{2} + N_{32} = 117.6 Newton + 49

Newton = 166.6 Newton

F_{k2 }= The force of the kinetic friction on the block 2 = μ_{k} N_{2} = (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 }=m_{3 }a_{x}

—–> F_{s23} = μ_{s }N_{23 }= μ_{s }w_{3 }= μ_{s }m_{3 }g

μ_{s} m_{3 }g = m_{3 }a_{x}

μ_{s} g = a_{x}

a_{x} = (0.3)(9.8 m/s^{2}) = 2.94 m/s^{2}

The maximum acceleration of the block 3 so that the block 3 and the block 2 still slide together is 2.94 m/s^{2}.

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 T_{2 }= the direction of w_{1x}.

∑F_{x} = m a_{x}

w_{1x} – T_{1 }+ T_{2 }– F_{k2} – F_{s32} + F_{s23} = (m_{1} + m_{2} + m_{3}) a_{x}

w_{1x} – F_{k2} = (m_{1} + m_{2 }+ m_{3} ) 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/s^{2}

a_{x} is positive, means direction of the block displacement or the direction of the acceleration is same as direction of T_{2} or direction of w_{1x}.

The magnitude of the acceleration is 2.11 m/s^{2} , lower than 2.94 m/s^{2 }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 }= (m_{1} + m_{2}) a_{x}

—–> F_{k2} = μ_{k} N_{2 }= μ_{k} w_{2} = μ_{k} m_{2 }g = (0.4)(12 kg)(9.8 m/s^{2}) = 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/s^{2}

**Ebook PDF two bodies with the same magnitude of acceleration sample problems with solutions 161.27 KB**

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