Visual SolutionTricky

Question

In an Atwood machine, two masses

m_1 = 5 \text{ kg}

and

m_2 = 3 \text{ kg}

are connected by a light inextensible string passing over a smooth, massless pulley. The system is released from rest. Find the tension in the string. Take

g = 10 \text{ m/s}^2

(A)

37.5 \text{ N}

(B)

40 \text{ N}

(C)

30 \text{ N}

(D)

35 \text{ N}

Solution Path

Atwood machine:

a = \frac{(m_1-m_2)g}{m_1+m_2} = 2.5 \text{ m/s}^2

T = \frac{2m_1 m_2 g}{m_1+m_2} = 37.5 \text{ N}

01Question Setup

1/4

Atwood machine with

m_1 = 5 \text{ kg}

and

m_2 = 3 \text{ kg}

connected by a light string over a smooth pulley. Find the tension in the string.

T = \;?

02Free Body DiagramsKEY INSIGHT

2/4

Draw FBD for each mass. For

m_1

(heavier, moves down):

m_1 g - T = m_1 a

. For

m_2

(lighter, moves up):

T - m_2 g = m_2 a

. The constraint: both have the same acceleration magnitude.

50 - T = 5a

and

T - 30 = 3a

03Solve Simultaneously

3/4

Add both equations:

(50 - 30) = (5 + 3)a

, so

a = 20/8 = 2.5 \text{ m/s}^2

. Substitute back:

T = 30 + 3(2.5) = 37.5 \text{ N}

a = 2.5 \text{ m/s}^2, \quad T = 37.5 \text{ N}

04Verify with Direct Formula

4/4

For Atwood machine:

T = \frac{2m_1 m_2 g}{m_1 + m_2} = \frac{2 \times 5 \times 3 \times 10}{8} = \frac{300}{8} = 37.5 \text{ N}

\checkmark

. Common trap: assuming

T = m_1 g

T = m_2 g

\boxed{37.5 \text{ N}}

Want more practice?

Try more PYQs from this chapter →