# Motion of two bodies with the same accelerations on the rough horizontal surface with the friction force – problems and solutions

1. Mass of the box 1 is 2 kg, the mass of the box 2 is 4 kg, acceleration of gravity is 10 m/s^{2}, the magnitude of the force F is 40 Newton. The coefficient of the kinetic friction between the box 1 with the floor is 0.2 and the coefficient of the kinetic friction between the box 2 and floor is 0.3. Find (a) The magnitude and direction of the box’s acceleration (b) Magnitude of the force exerted by the box 1 on the box 2 (F_{12}) and the magnitude of the force exerted by the box 2 on the box 1 (F_{21}).

Solution

__Known :__

Mass of the box 1 (m_{1}) = 2 kg

Mass of the box 2 (m_{2}) = 4 kg

Acceleration of gravity (g) = 10 m/s^{2},

The force F = 40 Newton,

Coefficient of the kinetic friction between the box 1 with floor (μ_{k1}) = 0.2

Coefficient of the kinetic friction between the box 2 with floor (μ_{k2}) = 0.3

The weight of the box 1 (w_{1}) = m_{1} g = (2)(10) = 20 Newton

The weight of the box 2 (w_{2}) = m_{2} g = (4)(10) = 40 Newton

The normal force exerted on the box 1 (N_{1}) = w_{1} = 20 Newton

The normal force exerted on the box 2 (N_{2}) = w_{2} = 40 Newton

The force of the kinetic friction exerted on the box 1 (f_{k1}) = (μ_{k1})(N_{1}) = (0.2)(20) = 4 Newton

The force of the kinetic friction exerted on the box 2 (f_{k2}) = (μ_{k1})(N_{2}) = (0.3)(40) = 12 Newton

__Solution :__

(a) Magnitude and direction of the box’s acceleration

ΣF = m a

F – f_{k1} – f_{k2 }= (m_{1} + m_{2}) a

40 – 4 – 12 = (2 + 4) a

24 = 6 a

a = 24 / 6

a = 4 m/s^{2}

Direction of the acceleration = direction of the net force = rightward.

(b) Magnitude of the force exerted by the box 1 on the box 2 (F_{12}) and the magnitude of the force exerted by the box 2 on the box 1 (F_{21}).

Calculate the magnitude of F_{12} :

ΣF = m a

F_{12} – f_{k2 }= (m_{2}) a

F_{12} – 12 = (4)(4)

F_{12} – 12 = 16

F_{12} = 16 + 12

F_{12} = 28 Newton

F_{12} and F_{21 }are action and reaction forces that act on the different objects. F_{12} and F_{21} has the same magnitude and opposite direction.

F_{12} = 28 Newton = F_{21} = 28 Newton.

2. Mass of the box 1 is 2 kg, mass of the box 2 is 4 kg, acceleration of gravity is 10 m/s^{2}, the force F is 40 N. The coefficient of the kinetic friction between the box 1 with the floor is 0.2 and the coefficient of the kinetic friction between box 2 and floor is 0.3. Determine (a) Magnitude and direction of the acceleration (b) The tension in the cord connecting the boxes. Ignore cord’s mass.

__Known :__

Mass of the box 1 (m_{1}) = 2 kg

Mass of the box 2 (m_{2}) = 4 kg

Acceleration of gravity (g) = 10 m/s^{2},

The force F = 40 Newton,

Coefficient of the kinetic friction between the box 1 with floor is 0.2 (μ_{k1}) = 0.2

Coefficient of the kinetic friction between the box 2 with floor is 0.2 (μ_{k2}) = 0.3

The weight of the box 1 (w_{1}) = m_{1} g = (2)(10) = 20 Newton

The weight of the box 2 (w_{2}) = m_{2} g = (4)(10) = 40 Newton

The normal force exerted on the box 1 (N_{1}) = w_{1} = 20 Newton

The normal force exerted on the box 2 (N_{2}) = w_{2} = 40 Newton

The force of the kinetic friction exerted on the box 1 (f_{k1}) = (μ_{k1})(N_{1}) = (0.2)(20) = 4 Newton

The force of the kinetic friction exerted on the box 2 (f_{k2}) = (μ_{k1})(N_{2}) = (0.3)(40) = 12 Newton

__Solution :__

(a) magnitude and direction of the acceleration

ΣF = m a

F – f_{k1} – f_{k2 }= (m_{1} + m_{2}) a

40 – 4 – 12 = (2 + 4) a

24 = 6 a

a = 24 / 6

a = 4 m/s^{2}

Magnitude of the acceleration is 4 m/s^{2}, direction of the acceleration = direction of the net force = rightward.

(b) Tension in the cord

Forces acts on the box 1 in the horizontal direction are the tension 1 (T_{1}) rightward and force of the kinetic friction 1 (f_{k1}) leftward. Apply Newton’s second law :

ΣF = m a

T_{1} – f_{k1 }= m_{1} a

T_{1} – 4 = (2)(4)

T_{1} – 4 = 8

T_{1 }= 8 + 4 = 12 Newton

The forces acts on the box 2 in the horizontal direction are the tension 2 (T_{2}) leftward and force of the kinetic friction 2 (f_{k2}) rightward. Apply Newton’s second law :

ΣF = m a

F – T_{2} – f_{k2 }= m_{2} a

40 – T_{2} – 12 = (4)(4)

28 – T_{2} = 16

T_{2 }= 28 – 16 = 12 Newton

The tension in the cord connecting the boxes = T_{1 }= T_{2} = T = 12 Newton.

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

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