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Chapter 4: Problem 74

Two blocks of masses \(m_{1}\) and \(m_{2}\) are suspended by a massless string over a frictionless pulley with negligible mass, as in an Atwood machine. The blocks are held motionless and then released. If \(m_{1}=3.50 \mathrm{~kg}\). what value does \(m_{2}\) have to have in order for the system to experience an acceleration \(a=0.400 g\) ? (Hint: There are two solutions to this problem.

Short Answer

Expert verified
For the Atwood machine with an acceleration of 0.400g and m1 = 3.50 kg, the mass m2 required for the system to experience this acceleration is 8.17 kg.

Step by step solution

01

Analyze the forces acting on both blocks

Since the blocks are suspended by a massless string over a frictionless pulley, the only forces acting on the blocks are the gravitational force (weight) and tension in the string. Let \(T\) be the tension in the string, and let's denote the weight of the masses as \(W_1 = m_1g\) and \(W_2 = m_2g\), where \(g\) is the acceleration due to gravity (\(9.81 m/s^2\)).
02

Apply Newton's second law for both blocks

From Newton's second law of motion, we can write the equations of motion for both blocks as: For block \(m_1\) (downward direction taken as positive): \(T - W_1 = m_1a\) For block \(m_2\) (upward direction taken as positive): \(W_2 - T = m_2a\)
03

Solve the equations to find \(m_2\)

Adding both equations, we get: \(W_2 - W_1 = (m_2 - m_1)a\) Now, substitute the values for \(W_1\) and \(W_2\) to get: \(m_2g - m_1g = (m_2 - m_1)a\) We're given \(m_1 = 3.50\) kg, and \(a = 0.400g\). Therefore, the equation becomes: \(m_2 - 3.50 = (m_2 - 3.50)(0.400)\).
04

Solve for \(m_2\)

To solve the equation above, we can use the following steps: 1. Divide both sides by \((m_2 - 3.50)\): \(\frac{m_2 - 3.50}{m_2 - 3.50} = 0.400\) 2. We get: \(1 = 0.400\) This result is invalid, which indicates that there is no physical solution for this case.
05

Consider the alternative case

Since there was no physical solution for the previous case, let's consider the alternative case where block \(m_1\) goes up and block \(m_2\) goes down. The equations of motion for both blocks will be: For block \(m_1\) (upward direction taken as positive): \(T - W_1 = -m_1a\) For block \(m_2\) (downward direction taken as positive): \(W_2 - T = m_2a\) Adding both equations, we get: \(W_2 - W_1 = (m_2 + m_1)a\) \(m_2g - m_1g = (m_2 + m_1)a\) Substitute the given values for \(m_1\) and \(a\): \(m_2 - 3.50 = (m_2 + 3.50)(0.400)\). Now, you have a linear equation to solve for \(m_2\).
06

Solve for \(m_2\)

To solve the equation in step 5, follow these steps: 1. Expand the equation: \(m_2 - 3.50 = 0.400m_2 + 0.400 \times 3.50\) 2. Rearrange the terms to have \(m_2\) on one side: \(m_2 - 0.400m_2 = 0.400 \times 3.50 + 3.50\) 3. Simplify and solve for \(m_2\): \(m_2(1 - 0.400) = 4.900\) \(m_2 = \frac{4.900}{0.600}\) \(m_2 = 8.17\) kg So, for the system to experience an acceleration of \(0.400g\), \(m_2\) must have a mass of \(8.17\) kg.

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