Question

Hey
And its,a physics ques
From laws of motion-motion in connected bodies

There was a,ques of 2 bodies suspended from a pully m1>m2 with acceleration a
Find a
And tension

Ans was a=(m1-m2)g/(m1+m2)

And T=2m1m2g/(m1+m2)



For mass m1
T=m1g-m1a (i)

For m2
T =m2a+m2g

My doubt
Why acc is taken in opp dir.?
Since in m1 acc is downward thus eqn should be
For mass m1
T=m1g+m1a (i)

and for m2 acc is upward
T =m2g-m2a

Then f=ma is taken opp
In direction
When it is along the direction of mg in m1 case and along t in m2 case

Answer 1

Answer on Question #42019 – Physics - Mechanics | Kinematics | Dynamics

Hey

And its, a physics ques

From laws of motion-motion in connected bodies

There was a, ques of 2 bodies suspended from a pully $m1 > m2$ with acceleration a

Find a

And tension

Ans was $a = (m1 - m2)g / (m1 + m2)$

And $T = 2m1m2g / (m1 + m2)$

For mass m1

T=m1g-m1a (i)

For m2

T = m2a + m2g

My doubt

Why acc is taken in opp dir.?

Since in m1 acc is downward thus eqn should be

For mass m1

T=m1g+m1a (i)

and for m2 acc is upward

T = m2g - m2a

Then $f = ma$ is taken opp

In direction

When it is along the direction of $mg$ in $m1$ case and along $t$ in $m2$ case

Solution:

Forces acting on the two masses and pulley:

mg - the force of gravity,

T - string tension force,

a - acceleration of the system;

Accelerations is taken in opposite directions because acceleration of the heavy bodie $\mathfrak{m}_1$ is directed downward (bodie is moving down) and the other's bodie $\mathfrak{m}_1$ is directed upward (moving up).

Tension force to each end of the string:

$$\left| \overrightarrow {\mathrm {T} _ {1}} \right| = \left| \overrightarrow {\mathrm {T} _ {1} ^ {\prime}} \right|; \left| \overrightarrow {\mathrm {T} _ {2}} \right| = \left| \overrightarrow {\mathrm {T} _ {2} ^ {\prime}} \right|; \left| \overrightarrow {\mathrm {a} ^ {\prime}} \right| = | \vec {\mathrm {a}} |$$

If we consider that we have ideal pulley:

$$\overrightarrow {\mathrm {T} _ {1} ^ {\prime}} = \overrightarrow {\mathrm {T} _ {2} ^ {\prime}}$$

Newton's second law for the first mass:

$$\overrightarrow {\mathrm {T} _ {1}} + \overrightarrow {\mathrm {m} _ {1} \vec {\mathrm {g}}} = \mathrm {m} _ {1} \vec {\mathrm {a}}$$ $$y: m _ {1} g - T _ {1} = m _ {1} a$$ $$T _ {1} = m _ {1} g - m _ {1} a$$

Newton's second law for the second mass:

$$\overrightarrow {T _ {2}} + \overrightarrow {m _ {2} g} = m _ {1} \vec {a}$$ $$y: T _ {2} - m _ {2} g = m _ {2} a$$ $$T _ {2} = m _ {2} a - m _ {2} g = T _ {1} \text{ (from (1)) (2)}$$ $$(1) = (2):$$ $$m _ {2} a - m _ {2} g = m _ {1} g - m _ {1} a$$ $$a (m _ {1} + m _ {2}) = g (m _ {1} - m _ {2})$$ $$a = \frac {g \left(m _ {1} - m _ {2}\right)}{m _ {1} + m _ {2}}$$

We do not know the values of the masses but we know that $m_{1} > m_{2}$, so we can take the absolute difference $m_{1} - m_{2}$ for the positive value of the acceleration:

$$(3) \text{in}(1):$$ $$T _ {1} = T _ {2} = T = m _ {1} g - m _ {1} \frac {g (m _ {1} - m _ {2})}{m _ {1} + m _ {2}} = m _ {1} g \left(\frac {m _ {1} + m _ {2} - m _ {1} + m _ {2}}{m _ {1} + m _ {2}}\right) =$$ $$= \frac {2 m _ {1} m _ {2} g}{m _ {1} + m _ {2}}$$

Answer: $a = \frac{g(m_1 - m_2)}{m_1 + m_2}, T = \frac{2m_1m_2g}{m_1 + m_2}$

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