Math  /  Algebra

Question42. The device shown below is the Atwood's machine considered in Example 6.5. Assuming that the masses of the string and the frictionless pulley are negligible, (a) find an equation for the acceleration of the two blocks; (b) find an equation for the tension in the string; and (c) find both the acceleration and tension when block 1 has mass 2.00 kg and block 2 has mass 4.00 kg .

Studdy Solution

STEP 1

1. The Atwood's machine consists of two blocks connected by a string over a pulley.
2. The pulley is frictionless, and the string is massless.
3. The system is only influenced by gravity.
4. The masses of the blocks are m1 m_1 and m2 m_2 .

STEP 2

1. Analyze the forces acting on each block.
2. Derive the equation for acceleration.
3. Derive the equation for tension in the string.
4. Calculate the specific values for acceleration and tension with given masses.

STEP 3

Consider the forces acting on each block. For block 1 with mass m1 m_1 , the force is m1g m_1 g downward. For block 2 with mass m2 m_2 , the force is m2g m_2 g downward.

STEP 4

Using Newton's second law, the net force on the system is (m2m1)g (m_2 - m_1)g , and the total mass of the system is m1+m2 m_1 + m_2 .
The acceleration a a of the system is given by:
a=(m2m1)gm1+m2a = \frac{(m_2 - m_1)g}{m_1 + m_2}

STEP 5

To find the tension T T in the string, consider the forces on one block. For block 1:
Tm1g=m1aT - m_1 g = m_1 a
Substitute the expression for a a :
T=m1g+m1((m2m1)gm1+m2)T = m_1 g + m_1 \left(\frac{(m_2 - m_1)g}{m_1 + m_2}\right)
Simplify:
T=2m1m2gm1+m2T = \frac{2m_1 m_2 g}{m_1 + m_2}

STEP 6

Substitute m1=2.00kg m_1 = 2.00 \, \text{kg} and m2=4.00kg m_2 = 4.00 \, \text{kg} into the equations for a a and T T .
Calculate acceleration a a :
a=(4.002.00)×9.812.00+4.00=2.00×9.816.00=3.27m/s2a = \frac{(4.00 - 2.00) \times 9.81}{2.00 + 4.00} = \frac{2.00 \times 9.81}{6.00} = 3.27 \, \text{m/s}^2
Calculate tension T T :
T=2×2.00×4.00×9.812.00+4.00=16.00×9.816.00=26.16NT = \frac{2 \times 2.00 \times 4.00 \times 9.81}{2.00 + 4.00} = \frac{16.00 \times 9.81}{6.00} = 26.16 \, \text{N}
The acceleration is 3.27m/s2 3.27 \, \text{m/s}^2 and the tension is 26.16N 26.16 \, \text{N} .

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