# Bodies connected by the cord and pulley – application of Newton’s law of motion problems and solutions

1. Two boxes are connected by a cord running over a pulley. Ignore the mass of the cord and pulley and any friction in the pulley. Mass of the box 1 = 2 kg, mass of the box 2 = 3 kg, acceleration due to gravity = 10 m/s^{2}. Find (a) The acceleration of the system (b) The tension in the cord!

Solution

__Known :__

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

Mass of the box 2 (m_{2}) = 3 kg

Acceleration due to gravity (g) = 10 m/s^{2}

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

Weight of the box 2 (w_{2}) = m_{2} g = (3)(10) = 30 Newton

__Solution :__

__(a) magnitude and direction of the acceleration __

w_{2 }> w_{1} so the box 2 accelerates downward and the box 1 accelerates upward.

Forces that has the same direction with acceleration (w_{2} and T_{1}), its sign positive. Forces that has opposite direction with acceleration (T_{2} and w_{1}), its sign negative.

∑F = m a

w_{2} – T_{2} + T_{1} – w_{1} = (m_{1} + m_{2}) a ——-> T_{1} = T_{2} = T

w_{2} – T + T – w_{1} = (m_{1} + m_{2}) a

w_{2} – w_{1} = (m_{1} + m_{2}) a

30 – 20 = (2 + 3) a

10 = 5 a

a = 10 / 5

a = 2 m/s^{2}

Magnitude of the acceleration is 2 m/s^{2}.

__(b) The tension force__

__The box 2 :__

There are two forces acts on the box 2 : first, weight of the box 2 (w_{2}), points downward so it’s positive. Second, tension force exerted on the box 2 (T_{2}), points upward so it’s negative. Apply Newton’s second law of motion.

∑F = m a

w_{2} – T_{2} = m_{2} a

30 – T_{2} = (3)(2)

30 – T_{2} = 6

T_{2} = 30 – 6

T_{2} = 24 Newton

__Box 1 :__

There are two forces acts on the box 1. *First*, weight of the box 1 (w_{1}), points downward so it’s negative. *Second*, the tension force exerted on the box 1 (T_{1}) points upward so it’s positive. Apply Newton’s second law of motion :

∑F = m a

T_{1} – w_{1} = m_{1} a

T_{1} – 20 = (2)(2)

T_{1} – 20 = 4

T_{1} = 20 + 4

T_{1} = 24 Newton

Magnitude of the tension force = T_{1} = T_{2} = T = 24 Newton

2. An object on a rough horizontal surface. Mass of the object 1 = 2 kg, mass of the object 2 = 4 kg, acceleration due to gravity = 10 m/s^{2}, coefficient of the static friction = 0.4, coefficient of the kinetic friction = 0.3. The system is at rest or accelerated ? If the system is accelerated, find the magnitude and direction of the system’s acceleration!

Solution

__Known :__

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

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

Acceleration due to gravity (g) = 10 m/s^{2}

Coefficient of the static friction (μ_{s}) = 0.4

The coefficient of the kinetic friction (μ_{k}) = 0.3

Weight of the object 1 (w_{1}) = m_{1} g = (2)(10) = 20 Newton

Weight of the object 2 (w_{2}) = m_{2} g = (4)(10) = 40 Newton

Normal force exerted on the object 1 (N) = w_{1} = 20 Newton

Force of the static friction exerted on the object 1 (f_{s}) = μ_{s }N = (0.4)(20) = 8 Newton

Force of the kinetic friction exerted on the object 1 (f_{k}) = μ_{k }N = (0.3)(20) = 6 Newton

__Wanted:__ acceleration (a)

__Solution :__

w_{2} > f_{s} (40 Newton > 8 Newton) so the object 2 is accelerated vertically downward and the object 1 is accelerated horizontally rightward. The friction force that acts on the objects 1 is the force of the kinetic friction (f_{k}). Apply Newton’s second law of motion :

∑F = m a

w_{2} – the = (m_{1} + m_{2}) a

40 – 6 = (2 + 4) a

34 = 6 a

a = 34 / 6 = 17 / 3

a = 5.7 m/s^{2}

Magnitude of the acceleration = 5.7 m/s^{2}

**Ebook PDF motion of bodies connected by cord and pulley sample problems with solutions 79.76 KB**

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