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Chapter 7: Problem 13

Three forces are acting on a particle of mass \(m\) initially in equilibrium. If the first 2 forces \(\left(R_{1}\right.\) and \(\left.R_{2}\right)\) 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 \( \frac{R_3}{m} \), so the answer is option a.

Step by step solution

01

Analyzing Forces in Equilibrium

Initially, the particle is in equilibrium, which means that the net force acting on it is zero. Therefore, the sum of all forces equals zero: \( R_1 + R_2 + R_3 = 0 \).
02

Setup Perpendicular Forces

Since \( R_1 \) and \( R_2 \) are perpendicular to each other, they can be considered as the components of a resultant force in a right-angled triangle. Thus, the vector sum of \( R_1 \) and \( R_2 \) is the hypotenuse of this triangle, equivalent and opposite in direction to \( R_3 \).
03

Determine Resultant Force when R3 is Removed

When \( R_3 \) is removed, the resultant force on the particle is the sum of \( R_1 \) and \( R_2 \). Since they are initially balanced by \( R_3 \), their vector sum is equal in magnitude to \( R_3 \) but in its opposite direction: \( R_1 + R_2 = -R_3 \).
04

Calculate Acceleration Using Newton's Second Law

According to Newton's second law, acceleration \( a \) is given by \( a = \frac{F_{net}}{m} \). In this case, the net force is \( R_3 \) directed in the same direction as \( R_1 + R_2 \). Therefore, \( a = \frac{R_3}{m} \).
05

Conclusion and Option Selection

The removal of \( R_3 \) results in an imbalance causing acceleration. The net force equals \( R_3 \), giving \( a = \frac{R_3}{m} \). Thus, the correct answer is option a: \( \frac{R_3}{m} \).

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Key Concepts

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

Equilibrium of forces
Equilibrium occurs when all the forces acting on a particle sum up to zero. Simply put, the forces are balanced, so the object stays still or moves with a constant velocity. Imagine a tug-of-war game where both teams are pulling with equal strength. The rope stays in the same place due to balanced forces.
For an object in equilibrium, the equation can be written as:
  • Net force = 0
  • Sum of forces = 0
In this exercise, the forces involved are \(R_1\), \(R_2\), and \(R_3\). These forces acting together create a state of rest or uniform motion for the particle. However, when one of these forces (like \(R_3\)) is removed, the balance is disrupted, leading to the object accelerating in the direction of the resultant unopposed force.
Perpendicular forces
When two forces are perpendicular, they form a right-angled triangle. This allows us to apply the Pythagorean theorem to find the resultant force. Perpendicular forces are independent of each other because they act in different directions 鈥� one influences only the vertical motion, and the other the horizontal motion.
If you consider two forces \(R_1\) and \(R_2\) acting at a point:
  • \(R_1\) acts horizontally
  • \(R_2\) acts vertically
These can be resolved into one resultant force using vector addition. The resultant force forms the hypotenuse of the triangle, emphasizing how these forces interact spatially to influence the motion of the particle. If \(R_3\) balances them by being equal and opposite to this hypotenuse, all three forces keep the system in equilibrium.
Resultant force
The concept of a resultant force is crucial when dealing with multiple forces. It is the single force that effectively represents the combined impact of two or more forces acting on a body. The resultant can be visualized as the force needed to balance other forces in a system; in this case, what \(R_3\) did when \(R_1\) and \(R_2\) were present.
Forces \(R_1\) and \(R_2\) give rise to a resultant because they are perpendicular 鈥� they are added vectorially as components of a right-angled triangle. When \(R_3\) is removed, it leaves \(R_1\) and \(R_2\) as the unopposed resultant, leading the object to accelerate:
\[R_{net} = \sqrt{R_1^2 + R_2^2}\]
The link between force and acceleration comes from Newton's Second Law, where acceleration \(a\) is calculated by dividing the resultant force by mass \(m\):
\[a = \frac{R_{net}}{m}\]

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Most popular questions from this chapter

Starting from rest, a body slides down a \(45^{\circ}\) inclined plane in twice the time it takes to slide the same distance in the absence of friction. What is the coefficient of friction between the body and the inclined plane? a. \(\sqrt{3} / 2\) b. \(3 / 4\) c. \(1 / 2\) d. \(1 / 4\)

A block of metal weighing \(2 \mathrm{~kg}\) is resting on a frictionless plane. It is struck by a jet releasing water at a rate of 1 \(\mathrm{kg} / \mathrm{s}\) and a speed of \(5 \mathrm{~m} / \mathrm{s}\). The initial acceleration of the block will be a. \(2.5 \mathrm{~m} / \mathrm{s}^{2}\) b. \(5 \mathrm{~m} / \mathrm{s}^{2}\) c. \(10 \mathrm{~m} / \mathrm{s}^{2}\) d. \(20 \mathrm{~m} / \mathrm{s}^{2}\)

A professor holds an eraser against a vertical chalkboard by pushing horizontally on it. He pushes with a force that is much greater than it required to hold the eraser. The force of friction exerted by the board on the eraser increases if he a. pushes eraser with slightly greater force. b. pushes eraser with slightly less force. c. raises his elbow so that the force he exerts is slightly downward but has same magnitude. d. lowers his elbow so that the force he exerts is slightly upward but has the same magnitude.

A monkey of mass \(40 \mathrm{~kg}\) climbs on a massless rope of breaking strength \(600 \mathrm{~N}\). The rope will break if the monkey a. Climbs up with a uniform speed of \(5 \mathrm{~m} / \mathrm{s}\). b. Climbs up with an acceleration of \(6 \mathrm{~m} / \mathrm{s}^{2}\). c. Climbs down with an acceleration of \(4 \mathrm{~m} / \mathrm{s}^{2}\). d. Climbs down with a uniform speed of \(5 \mathrm{~m} / \mathrm{s}\).

Two blocks \(m\) and \(M\) tied together with an inextensible string are placed at rest on a rough horizontal surface with coefficient of friction \(\mu\). The block \(m\) is pulled with a variable force \(F\) at a varying angle \(\theta\) with the horizontal. The value of \(\theta\) at which the least value of \(F\) is required to move the blocks is given by a. \(\theta=\tan ^{-1} \mu\) b. \(\theta>\tan ^{-1} \mu\) c. \(\theta<\tan ^{-1} \mu\) d. Insufficient data
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