/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 147 A light string passing over a sm... [FREE SOLUTION] | 91影视

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Chapter 3: Problem 147

A light string passing over a smooth light pulley connects two blocks of masses \(m_{1}\) and \(m_{2}\) (vertically). If the acceleration of the system is \(g / 8\), then the ratio of the masses is (A) \(8: 1\) (B) \(9: 7\) (C) \(4: 3\) (D) \(5: 3\)

Short Answer

Expert verified
The ratio of the masses is (B) \(9: 7\).

Step by step solution

01

Analyze the forces

To start solving this problem, we first need to analyze the forces acting on the two blocks. The forces acting on block \(m_1\) are the tension in the string \(T\) and the force due to gravity \(m_1g\). The forces acting on block \(m_2\) are the tension in the string \(T\) and the force due to gravity \(m_2g\).
02

Write the equations of motion

Now, we will write the equations of motion for the blocks using Newton's second law. Since the acceleration is \(g/8\), where \(g\) is the acceleration due to gravity, we will use that value in our equations. For block \(m_1\): \(m_1\frac{g}{8} = T - m_1g\) For block \(m_2\): \(m_2\frac{g}{8} = m_2g - T\)
03

Solve the system of equations

To find the ratio of the masses, we will now solve the system of equations. First, let's add the two equations of motion to eliminate the tension \(T\). \(m_1\frac{g}{8} + m_2\frac{g}{8} = -m_1g + m_2g\) Now, let's simplify the equation and solve for the ratio. \(\frac{m_1 + m_2}{8} = m_2 - m_1\) \(8(m_2 - m_1) = m_1 + m_2\) \(8m_2 - 8m_1 = m_1 + m_2\) \(9m_1 = 7m_2\) Now we can see that the ratio of the masses is: \(\frac{m_1}{m_2} = \frac{7}{9}\) So the correct answer is: (B) \(9: 7\)

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Most popular questions from this chapter

A car of mass \(m\) starts from rest and acquires a velocity along east, \(v=v \hat{i}(v>0)\) in two seconds. Assuming the car moves with uniform acceleration, the force exerted on the car is (A) \(\frac{m v}{2}\) eastward and is exerted by the car engine. (B) \(\frac{m v}{2}\) eastward and is due to the friction on the tyres exerted by the road. (C) more than \(\frac{m v}{2}\) eastward exerted due to the engine and overcomes the friction of the road. (D) \(\frac{m v}{\text { exerted by the engine. }}\)

Three blocks \(m_{1}, m_{2}\) and \(m_{3}\) of masses \(8 \mathrm{~kg}, 3 \mathrm{~kg}\) and \(1 \mathrm{~kg}\) are placed in contact on a smooth surface. Forces \(F_{1}=140 \mathrm{~N}\) and \(F_{2}=20 \mathrm{~N}\) are acting on blocks \(m_{1}\) and \(m_{3}\), respectively, as shown. The reaction between blocks \(m_{2}\) and \(m_{3}\) is (A) \(2.5 \mathrm{~N}\) (B) \(7.5 \mathrm{~N}\) (C) \(22.5 \mathrm{~N}\) (D) \(30 \mathrm{~N}\)

A pebble of mass \(0.05 \mathrm{~kg}\) is thrown vertically upwards. The direction and magnitude of the net force on the pebble is given below, choose the incorrect option. (A) During its upward motion, force is \(0.5 \mathrm{~N}\) in vertically upward. (B) During its downward motion, force is \(0.5 \mathrm{~N}\) in vertically downward. (C) At the highest point, where it is momentarily at rest, force is \(0.5 \mathrm{~N}\) in vertically downward. (D) If the pebble was thrown at an angle of say \(45^{\circ}\) with the horizontal direction, force is \(0.5 \mathrm{~N}\) in vertically downward (Ignoring air resistance).

For a particle rotating in a vertical circle with uniform speed, the maximum and minimum tension in the string are in the ratio \(5: 3\). If the radius of vertical circle is \(2 \mathrm{~m}\), the speed of revolving body is \(\left(g=10 \mathrm{~m} / \mathrm{s}^{2}\right)\) (A) \(\sqrt{5} \mathrm{~m} / \mathrm{s}\) (B) \(4 \sqrt{5} \mathrm{~m} / \mathrm{s}\) (C) \(5 \mathrm{~m} / \mathrm{s}\) (D) \(10 \mathrm{~m} / \mathrm{s}\)

With what minimum acceleration mass \(M\) must be moved on frictionless surface so that \(m\) remains stick to it as shown. The co-efficient of friction between \(M\) and \(m\) is \(\mu\). (A) \(\mu g\) (B) \(\frac{g}{\mu}\) (C) \(\frac{\mu m g}{M+m}\) (D) \(\frac{\mu m g}{M}\)
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