/*! 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 27 Three forces are acting on a par... [FREE SOLUTION] | 91Ó°ÊÓ

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Three forces are acting on a particle of mass \(m\) initially in equilibrium. If the first 2 forces \(\left(R_{1}\right.\) and \(R_{2}\) ) are perpendicular to each other and suddenly the third force \(\left(R_{3}\right)\) is removed, then the acceleration of the particle is (a) \(\frac{R_{3}}{m}\) (b) \(\frac{R_{1}+R_{2}}{m}\) (c) \(\frac{R_{1}-R_{2}}{m}\) (d) \(\frac{R_{1}}{m}\)

Short Answer

Expert verified
The acceleration of the particle is \(\frac{R_1 + R_2}{m}\), option (b).

Step by step solution

01

Understanding the Initial Equilibrium Condition

Since the particle is initially in equilibrium, the net force acting on it is zero. This implies that the vector sum of the forces \( R_1, R_2, \) and \( R_3 \) is zero. In equation form: \( R_1 + R_2 + R_3 = 0 \).
02

Perpendicular Forces Analysis

Given that \( R_1 \) and \( R_2 \) are perpendicular to each other, we can deduce that \( R_3 \) should balance the vector sum of \( R_1 \) and \( R_2 \) to maintain equilibrium. Let \( R_1 \) be along the x-axis and \( R_2 \) along the y-axis.
03

Net Force upon Removal of \( R_3 \)

When the third force \( R_3 \) is removed, the net force on the particle becomes \( R_1 + R_2 \). Since \( R_1 \) and \( R_2 \) are originally balanced by \( R_3 \), removing \( R_3 \) results in a net force equal to \( R_1 + R_2 \).
04

Using Newton's Second Law

According to Newton's Second Law, the acceleration \( a \) of the particle is given by \( a = \frac{F_{net}}{m} \), where \( F_{net} \) is the net force acting on the particle. Since \( R_1 + R_2 \) is the net force after removing \( R_3 \), the acceleration is \( a = \frac{R_1 + R_2}{m} \).
05

Conclusion

Therefore, the acceleration of the particle after the third force is removed is \( \frac{R_1 + R_2}{m} \). This corresponds to option (b).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

equilibrium of forces
A fundamental idea in physics is the equilibrium of forces, which occurs when all the forces acting on an object sum up to zero. This means the object is at rest or moves with constant velocity. In this exercise, three forces—\(R_1\), \(R_2\), and \(R_3\)—act on a particle maintaining it in equilibrium. Here, equilibrium signifies that the net force is zero, represented by the equation \(R_1 + R_2 + R_3 = 0\). No acceleration occurs when forces are balanced. However, when one force is removed, like \(R_3\) in our scenario, the equilibrium is disturbed, causing changes in motion. Understanding equilibrium conditions is crucial for predicting how objects will move when one or more forces are altered.
vector addition
In this context, vector addition is a method to combine multiple force vectors into a single resultant vector. When forces act on an object, they must be added as vectors because they have both magnitude and direction.
  • Force \(R_1\) acts along the x-axis.
  • Force \(R_2\) acts along the y-axis.
These two forces can be added using the Pythagorean theorem when they are perpendicular. This is done by calculating the magnitude of the resulting vector. With \(R_3\) balancing the sum of \(R_1\) and \(R_2\), all three forces ensure the object remains in equilibrium when combined vectorially. Removing one vector, such as \(R_3\), disrupts this balance, leading to motion.
perpendicular forces analysis
Analyzing perpendicular forces involves examining how two forces at right angles affect each other and work together to maintain equilibrium.
In this exercise, \(R_1\) and \(R_2\) are perpendicular, meaning they affect different dimensions independently.
  • Force \(R_1\) affects horizontal movement.
  • Force \(R_2\) affects vertical movement.
The resultant of these two forces forms a right-angled triangle, with the hypotenuse being the vector sum that force \(R_3\) cancels to maintain equilibrium. Knowing that these forces act perpendicularly helps predict how their removal or change would alter the system's dynamics.
net force calculation
Net force calculation is a primary step when dealing with multiple forces acting upon an object. In this particular scenario, calculating net force involves removing one of the initially balanced forces. Here, upon the removal of \(R_3\), the net force becomes \(R_1 + R_2\). Newton's Second Law states that the net force drives an object's acceleration: \[a = \frac{F_{net}}{m}\]where \(F_{net}\) is the sum of remaining forces after \(R_3\) is removed, leading to accelerated motion. Since the net force from \(R_1\) and \(R_2\) wasn't counterbalanced, the particle accelerates, with its acceleration determined by the equation \(a = \frac{R_1 + R_2}{m}\). Mastering net force calculation is essential for understanding how forces influence motion.

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Most popular questions from this chapter

A car is going at a speed of \(6 \mathrm{~m} / \mathrm{s}\) when it encounters a 15 \(\mathrm{m}\) slope of angle \(30^{\circ} .\) The friction coefficient between the road and tyre is \(0.5\). The driver applies the brakes. The minimum speed of the car with which it can reach the bottom is \(\left(g=10 \mathrm{~m} / \mathrm{s}^{2}\right)\) (a) \(4 \mathrm{~m} / \mathrm{s}\) (b) \(3 \mathrm{~m} / \mathrm{s}\) (c) \(7.49 \mathrm{~m} / \mathrm{s}\) (d) \(8.45 \mathrm{~m} / \mathrm{s}\)

Two weights \(w_{1}\) and \(w_{2}\) are suspended from the ends of a light string passing over a smooth fixed pulley. If the pulley is pulled up with acceleration \(g\), the tension in the string will be (a) \(\frac{4 w_{1} w_{2}}{w_{1}+w_{2}}\) (b) \(\frac{2 w_{1} w_{2}}{w_{1}+w_{2}}\) (c) \(\frac{w_{1}-w_{2}}{w_{1}+w_{2}}\) (d) \(\frac{w_{1} w_{2}}{2\left(w_{1}-w_{2}\right)}\)

A bullet comes out of the barrel of gun of length \(2 \mathrm{~m}\) with a speed \(80 \mathrm{~m} / \mathrm{s}\). The average acceleration of the bullet is (a) \(1.6 \mathrm{~m} / \mathrm{s}^{2}\) (b) \(160 \mathrm{~m} / \mathrm{s}^{2}\) (c) \(1600 \mathrm{~m} / \mathrm{s}^{2}\) (d) \(16 \mathrm{~m} / \mathrm{s}^{2}\)

A block rests on a smooth inclined plane making an angle of \(30^{\circ}\) with the horizontal. The coefficient of static friction between the block and the plane is \(0.8\). If the frictional force on the block is \(10 \mathrm{~N}\), the mass of the block (in \(\mathrm{kg}\) ) is (Take \(g=10 \mathrm{~m} / \mathrm{s}^{2}\) ) (a) \(2.0\) (b) \(4.0\) (c) \(1.6\) (d) \(2.5\)

A machine gun fires a bullet of mass \(40 \mathrm{~g}\) with a velocity \(1200 \mathrm{~m} / \mathrm{s}\). The man holding it can exert a maximum force of \(144 \mathrm{~N}\) on the gun. How many bullets can he fire per second at the most? (a) One (b) Four (c) Two (d) Three

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