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Chapter 3: Problem 64

A 300 -g mass hangs at the end of a string. A second string hangs from the bottom of that mass and supports a 900 -g mass. \((a)\) Find the tension in each string when the masses are accelerating upward at \(0.700 \mathrm{~m} / \mathrm{s}^{2}\). Don't forget gravity. ( \(b\) ) Find the tension in each string when the acceleration is \(0.700 \mathrm{~m} / \mathrm{s}^{2}\) downward.

Short Answer

Expert verified
(a) Upward: Top String: 12.921 N, Bottom String: 9.981 N. (b) Downward: Top String: 11.421 N, Bottom String: 8.991 N.

Step by step solution

01

Identify the System

We have two masses: \(m_1 = 300\) g and \(m_2 = 900\) g, or in kilograms, \(m_1 = 0.3\) kg and \(m_2 = 0.9\) kg. They are connected by strings, and the system is accelerating upward in part (a) and downward in part (b). The acceleration due to gravity \(g\) is \(9.81\, \text{m/s}^2\).
02

Newton's Second Law for Upward Acceleration

When the acceleration is upward, the net force on each mass is the tension minus the gravitational force. The equations for \(m_1\) (top mass) and \(m_2\) (bottom mass) are:\[T_1 - m_1g = m_1a \quad \text{and} \quad T_2 - m_2g = m_2a\]where \(T_1\) is the tension in the upper string, and \(T_2\) is the tension in the lower string.
03

Calculate Tension in Each String (Upward Acceleration)

First, solve for \(T_2\):\[T_2 = m_2g + m_2a = 0.9\times 9.81 + 0.9\times 0.7 = 9.981\, \text{N}\]Then, solve for \(T_1\):\[T_1 = m_1g + m_1a + T_2 = 0.3\times 9.81 + 0.3\times 0.7 + 9.981 = 12.921\, \text{N}\]
04

Newton's Second Law for Downward Acceleration

When the acceleration is downward, the net force equation reverses direction:\[T_1 = m_1g - m_1a \quad \text{and} \quad T_2 = m_2g - m_2a\]
05

Calculate Tension in Each String (Downward Acceleration)

First, solve for \(T_2\):\[T_2 = m_2g - m_2a = 0.9\times 9.81 - 0.9\times 0.7 = 8.991\, \text{N}\]Then, solve for \(T_1\):\[T_1 = m_1g - m_1a + T_2 = 0.3\times 9.81 - 0.3\times 0.7 + 8.991 = 11.421\, \text{N}\]

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Key Concepts

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

Newton's Second Law
When facing physics problems involving motion and force, Newton's Second Law is a fundamental concept that helps us understand the relationship between these elements.
Simply put, Newton's Second Law states that the force acting on an object is equal to the mass of that object multiplied by its acceleration. In equation form, this is expressed as:
  • \(F = ma\)
This concept is crucial when dealing with moving systems like the one in our exercise, where masses are bound by strings.
Understanding this law allows us to determine how the tension within the strings will change based on the movement and acceleration of the masses.
For our specific problem, we apply this law separately to each mass in our two-mass system, taking into account both the forces of tension and gravity acting on them.
Tension in Strings
In physics, tension is a force exerted by a string, rope, or any other type of connector. It's an indirect force because it isn't applied directly by a solid object but transmitted through a medium.
When dealing with tension in strings, especially in multi-mass systems, it's important to consider how tension varies along the string depending on additional forces acting on the system, such as gravity.
In our exercise, the strings support two masses that are either accelerating upward or downward. The tension can be broken down into two parts:
  • Tension in the upper string, \(T_1\), affects both masses and needs to counter the combined forces of gravity and acceleration.
  • Tension in the lower string, \(T_2\), impacts only the lower mass, so it is responsible for countering both the mass's gravitational force and its acceleration.
For upward acceleration, forces due to gravity reduce net tension, whereas for downward acceleration, they're subtracted from net tension. This dual consideration ensures correct calculation of string tension.
Acceleration and Gravity
In our physics problem, understanding acceleration and gravity is essential. Both are vector forces that cause changes in movement.
Acceleration is the change in velocity over time and is expressed as \(a\). It has both magnitude and direction.
In our problem, knowing whether the system is accelerating upwards or downwards is key, as this directionality affects how tension in the strings is calculated.
Gravity, with a standard value of \(9.81 \text{ m/s}^2\), affects everything on Earth. It is a force that pulls objects toward the ground, adding to the forces in calculations.
When the mass systems accelerate upwards, the effective force increases as gravity and acceleration work in the same direction. Conversely, a downward acceleration results in the decrement of net force according to Newton's Second Law.

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Most popular questions from this chapter

A constant force acts on a \(5.0\) kg object and reduces its velocity from \(7.0 \mathrm{~m} / \mathrm{s}\) to \(3.0 \mathrm{~m} / \mathrm{s}\) in a time of \(3.0\) s. Determine the force. We must first find the acceleration of the object, which is constant because the force is constant. Taking the direction of motion as positive, from Chapter 2 $$ a=\frac{v_{f}-v_{i}}{t}=\frac{-4.0 \mathrm{~m} / \mathrm{s}}{3.0 \mathrm{~s}}=-1.33 \mathrm{~m} / \mathrm{s}^{2} $$ Use \(F=m a\) with \(m=5.0 \mathrm{~kg}\) : $$ F=(5.0 \mathrm{~kg})\left(-1.33 \mathrm{~m} / \mathrm{s}^{2}\right)=-6.7 \mathrm{~N} $$ The minus sign indicates that the force is a retarding force, directed opposite to the motion.

A force of \(100 \mathrm{lb}\) acting on a body weighing \(500 \mathrm{lb}\) causes the body to accelerate uniformly. What would happen to the acceleration if the force is increased to \(200 \mathrm{lb}\) ? [Hint: Units are not important here as long as you are consistent.] Assume no friction.

Consider an essentially spherical homogeneous celestial body of mass \(M\). The acceleration due to gravity in its vicinity beyond its surface at a distance \(R\) from its center is \(g_{R}\). Show that $$ g_{R}=\frac{G M}{R^{2}} $$ Notice that the acceleration drops off as \(1 / R^{2}\). Imagine an object of mass \(m\) at a distance \(R\) from the center of our celestial body. Its weight is \(F_{W}=m g_{R}\), but that's also the gravitation force on it due to the mass \(M\), that is, \(F_{W}=F_{G} .\) Hence, $$ m g_{R}=G \frac{m M}{R^{2}} $$

A 20 -kg wagon is pulled along the level ground by a rope inclined at \(30^{\circ}\) above the horizontal. A friction force of \(30 \mathrm{~N}\) opposes the motion. How large is the pulling force if the wagon is moving with (a) constant speed and \((b)\) an acceleration of \(0.40 \mathrm{~m} / \mathrm{s}^{2}\) ?

Suppose, as depicted in Fig. 3-14, that a 70 -kg box is pulled by a 400-N force at an angle of \(30^{\circ}\) to the horizontal. The coefficient of kinetic friction is \(0.50\). Find the acceleration of the box. Fig. \(3-14\) Because the box does not move up or down, we have \(\sum F_{y}=m a_{y}=\) 0\. From Fig. \(3-14\), this equation is $$ +\uparrow \sum_{y}=F_{N}+200 \mathrm{~N}-m g=0 $$ But \(m g=(70 \mathrm{~kg})\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)=686.7 \mathrm{~N}\), and it follows that \(F_{N}=\) \(486.7 \mathrm{~N}\) Next find the friction force acting on the box: $$ F_{\mathrm{f}}=\mu_{k} F_{N}=(0.50)(486.7 \mathrm{~N})=243.4 \mathrm{~N} $$ Now write \(\sum F_{x}=m a_{x}\) for the box. It is $$ (346-243.4) \mathrm{N}=(70 \mathrm{~kg})\left(a_{X}\right) $$ from which \(a_{x}=1.466 \mathrm{~m} / \mathrm{s}^{2}\) or \(1.5 \mathrm{~m} / \mathrm{s}^{2}\)
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