# Thread: Basic Mechanics - dynamics of a particle

1. ## Basic Mechanics - dynamics of a particle

A particle of mass 0.5kg is suspended from a vertical string. By means of another string, a second particle is suspended from the first of mass 0.4kg. A force of 10N is applied to the upper string and the particle moves upwards. Find the tension in the lower string and the acceleration of the system.

Note: Assume resistances other than those provided, and g, which is to be approximated by 9.8N, are null.

Answer: Tension 4.44N, acceleration 1.31 m/s^2

2. Write the net force $\displaystyle F_1$ on the first particle and $\displaystyle F_2$ on the second particle in terms of the unknown tension. Then, note that their accelerations are the same, so note that $\displaystyle F_i=ma$ for both particles, and then solve for the tension.

3. Originally Posted by Quacky
A particle of mass 0.5kg is suspended from a vertical string. By means of another string, a second particle is suspended from the first of mass 0.4kg. A force of 10N is applied to the upper string and the particle moves upwards. Find the tension in the lower string and the acceleration of the system.

Note: Assume resistances other than those provided, and g, which is to be approximated by 9.8N, are null.

Answer: Tension 4.44N, acceleration 1.31 m/s^2
forces acting on the first mass ... $\displaystyle T_1$ upward , $\displaystyle T_2$ and $\displaystyle m_1g$ downward

net force ... $\displaystyle T_1 - T_2 - m_1g = m_1a$

forces acting on the lower mass ... $\displaystyle T_2$ upward and $\displaystyle m_2g$ downward

net force ... $\displaystyle T_2 - m_2g = m_2a$

$\displaystyle T_1 - (m_1 + m_2)g = (m_1 + m_2)a$
$\displaystyle \frac{T_1 - (m_1+m_2)g}{m_1+m_2} = a$
$\displaystyle \frac{10 - (0.9)(9.8)}{0.9} = a$
calculate $\displaystyle a$ , then determine $\displaystyle T_2$ , the tension in the lower string from one of the first two equations.