A-constant-force-F-20-N-acts-on-a-block-of-mass-2-kg-which-is-connected-to-two-blocks-of-masses-m-1-1-kg-and-m-2-2-kg-Calculate-the-accelerations-produced-in-all-the-three-blocks-Assume-pull

Question Number 21713 by Tinkutara last updated on 01/Oct/17

A constant force F = 20 N acts on a

block of mass 2 kg which is connected to

two blocks of masses m_{1} = 1 kg and

m_{2} = 2 kg . Calculate the accelerations

produced in all the three blocks . Assume

pulleys are frictionless and weightless .

Commented by Tinkutara last updated on 01/Oct/17

Commented by mrW1 last updated on 01/Oct/17

F - T = Ma

2 T - m_{1} g = m_{1} a_{1}

T - m_{2} g = m_{2} a_{2}

a_{1} = \frac{1}{2} (a - a_{2})

T = F - M

2 (F - Ma) = m_{1} (g + a_{1})

\Rightarrow a_{1} = \frac{2 (F - Ma)}{m_{1}} - g

F - Ma = m_{2} (g + a_{2})

\Rightarrow a_{2} = \frac{F - Ma}{m_{2}} - g

\frac{4 (F - Ma)}{m_{1}} - 2 g = a - \frac{F - Ma}{m_{2}} + g

\frac{4 F}{m_{1}} + \frac{F}{m_{2}} - 3 g = (1 + \frac{4 M}{m_{1}} + \frac{M}{m_{2}}) a

\Rightarrow a = \frac{(m_{1} + {4 m}_{2}) F - {3 m}_{1} m_{2} g}{(m_{1} + {4 m}_{2}) M + m_{1} m_{2}}

= \frac{(1 + 4 \times 2) \times 20 - 3 \times 1 \times 2 \times 9.8}{(1 + 4 \times 2) \times 2 + 1 \times 2}

= \frac{180 - 6 \times 9.8}{20}

= 6.06 m / s^{2} (\leftarrow)

\Rightarrow a_{1} = \frac{2 (20 - 2 \times 6.06)}{1} - 9.8 = 5.96 m / s^{2} (↑)

\Rightarrow a_{2} = \frac{20 - 2 \times 6.06}{2} - 9.8 = - 5.86 m / s^{2} (↓)

Commented by Tinkutara last updated on 01/Oct/17

I {don}^{'} t think all blocks have same

accelerations . If you take g = 9.8 m / s^{2},

then I can check your answer from book .

Commented by mrW1 last updated on 02/Oct/17

they have the same value of acceleration

just because of these specific parameters .

Commented by Tinkutara last updated on 02/Oct/17

Can you explain why a_{1} = \frac{1}{2} (a - a_{2}) ?

Commented by mrW1 last updated on 02/Oct/17

since the rope is inextensible, the

motion of mass m_{1} is dependent from

the motion of the other two masses .

position of mass M = x (+ ve if \leftarrow)

position of mass m_{2} = x_{2} (+ ve if ↑)

position of mass m_{1} = x_{1} (+ ve if ↑)

we get x_{1} = \frac{x}{2} - \frac{x_{2}}{2}

a = \frac{d^{2} x}{{dt}^{2}}

a_{1} = \frac{d^{2} x_{1}}{{dt}^{2}}

a_{2} = \frac{d^{2} x_{2}}{{dt}^{2}}

\Rightarrow a_{1} = \frac{1}{2} (a - a_{2})

Commented by Tinkutara last updated on 02/Oct/17

Thank you very much Sir!