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Chapter 4: Problem 444

A mass of \(\mathrm{M} \mathrm{kg}\) is suspended by a weight-less string, the horizontal force that is required to displace it until the string makes an angle of \(60^{\circ}\) with the initial vertical direction is (A) \(\mathrm{Mg} / \sqrt{3}\) (B) \(\mathrm{Mg} \cdot \sqrt{2}\) (C) \(\mathrm{Mg} / \sqrt{2}\) (D) \(\mathrm{Mg} \cdot \sqrt{3}\)

Short Answer

Expert verified
The horizontal force required to displace the mass until the string makes an angle of 60 degrees with the initial vertical direction is \(F = \frac{Mg}{\sqrt{3}}\).

Step by step solution

01

Draw the forces diagram

First, draw a diagram with the mass M, the string, and the angles. The forces acting on the mass are tension (T) in the string (that has a vertical and a horizontal component), gravitational force (Mg), and the horizontal force F.
02

Apply equilibrium of forces

We will analyze the vertical and horizontal forces when the mass is in equilibrium. In the vertical direction, the vertical component of tension balances the gravitational force: \(T_v = Mg\) In the horizontal direction, the horizontal component of tension balances the horizontal force: \(T_h = F\) Since the string makes an angle of 60 degrees with the initial vertical direction, the angle between the string and the horizontal direction is 30 degrees (90 - 60 degrees). Thus, we can find the relation between the vertical and horizontal components of tension using trigonometry.
03

Relate tension components

We can find the relationship between the vertical and horizontal components of tension using the angle between the string and the horizontal direction: Using sine function, we have: \(T_h = T \cdot \sin(30^{\circ})\) Using cosine function, we have: \(T_v = T \cdot \cos(30^{\circ})\) Now, we know that \(T_v = Mg\), and \(T_h = F\). Therefore, \(Mg = T \cdot \cos(30^{\circ})\) and \(F = T \cdot \sin(30^{\circ})\)
04

Solve for F

Now, we need to find the value of F in terms of M and g. Divide the equations: \(F = T \cdot \sin(30^{\circ})\) and \(Mg = T \cdot \cos(30^{\circ})\) We obtain: \(F / Mg = \sin(30^{\circ}) / \cos(30^{\circ})\) We know that \(\sin(30^{\circ}) = 1/2\) and \(\cos(30^{\circ}) = \sqrt{3}/2\), so: \(F / Mg = (1/2) / (\sqrt{3}/2)\)
05

Simplify and find the answer

Finally, simplify the expression for F: \(F = Mg \cdot ((1/2) / (\sqrt{3}/2))\) Multiply both sides by 2: \(F = Mg / \sqrt{3}\) This corresponds to the choice (A) in the problem. Therefore, the horizontal force required to displace the mass until the string makes an angle of 60 degrees with the initial vertical direction is: \(F = \frac{Mg}{\sqrt{3}}\)

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Most popular questions from this chapter

Two bodies of masses \(m_{1}\) and \(m_{2}\) have equal kinetic energies. If \(P_{1}\) and \(P_{2}\) are their respective momentum, what is ratio of \(\mathrm{P}_{2}: \mathrm{P}_{1}\) ? (A) \(\mathrm{m}_{1}: \mathrm{m}_{2}\) (B) \(\sqrt{\mathrm{m}}_{2} / \sqrt{\mathrm{m}_{1}}\) (C) \(\sqrt{m_{1}}: \sqrt{m_{2}}\) (D) \(\mathrm{m}_{1}^{2}: \mathrm{m}_{2}^{2}\)

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