/*! 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 92 Block \(B,\) with mass \(5.00 \m... [FREE SOLUTION] | 91Ó°ÊÓ

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Block \(B,\) with mass \(5.00 \mathrm{~kg},\) rests on block \(A,\) with mass \(8.00 \mathrm{~kg}\), which in turn is on a horizontal tabletop (Fig. P5.92). There is no friction between block \(A\) and the tabletop, but the coefficient of static friction between blocks \(A\) and \(B\) is \(0.750 .\) A light string attached to block \(A\) passes over a friction less, massless pulley, and block \(C\) is suspended from the other end of the string. What is the largest mass that block \(C\) can have so that blocks \(A\) and \(B\) still slide together when the system is released from rest?

Short Answer

Expert verified
The greatest mass that block C can have so that blocks A and B still slide together is approximately 11.75 kg.

Step by step solution

01

Identify the forces on block B

Gravity force on block B is: \(F_{g_B} = m_B \cdot g = 5 \cdot 9.8 = 49N\). The frictional force, that is the force that block B applies to block A, is: \(f = \mu \cdot F_{g_B} = 0.75 \cdot 49 = 36.75N\). This is the maximum force block B can apply to block A to prevent relative motion, depending on block A's and C's forces.
02

Identify the forces on block A

Gravity force on block A is: \(F_{g_A} = m_A \cdot g = 8 \cdot 9.8 = 78.4N\). Now let's denote by F_T the tension force from the string connected to block C, and it's equal to \(F_{T} = m_C \cdot g\), where \(m_C\) is the mass of block C. For block A, there are 3 forces: the friction from block B (f), its gravity (F_{g_A}) and tension force (F_{T}). So \(F_{T}= f + F_{g_A} = 36.75 + 78.4 = 115.15N\).
03

Calculate the required mass of block C

Now, since \(F_{T} = m_C \cdot g\), the required mass of block C can be determined as: \(m_C = \frac{F_{T}}{g} = \frac{115.15}{9.8} = 11.75kg\). This is the maximum mass that block C can have so that blocks A and B still slide together when the system is released from rest.

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Key Concepts

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

Understanding the Static Friction Coefficient
When dealing with problems involving surfaces in contact, it's essential to comprehend the concept of the static friction coefficient. This value, represented by the Greek letter \(\text{\mu}_s\), plays a pivotal role in determining whether an object will move or remain at rest when a force is applied to it.

The static friction coefficient is a measure between two specific surfaces and represents the maximum force of static friction that can be produced before the object starts sliding. It is a dimensionless number derived from the ratio of the maximum static frictional force to the normal force exerted by an object. In simpler terms, it's the 'stickiness' between two surfaces at rest.

For instance, in our exercise, block B will resist sliding on block A thanks to static friction, which can be calculated using \(\text{\mu}_s\) and the gravitational force on block B. Whenever block A tries to slide, the static friction force is the force that needs to be overcome for block B to move with it. The use of the static friction coefficient makes it easy to predict when objects will stick together or slide apart, and is crucial for appropriately setting up the equations that solve physics problems.
Tension Force Calculation in Physics
Another fundamental aspect of physics problems involving strings and pulleys is tension force calculation. Tension can be thought of as the 'pulling' force transmitted through a string, cable, or chain when it is pulled tight by forces acting from opposite ends.

In our textbook problem, the tension in the string arises due to the weight of block C and the frictional force between blocks A and B. By knowing the mass of block C and the acceleration due to gravity (\(g\)), we can calculate the tension force (\(F_T\)) using the equation \(F_T = m_C \times g\).

It's essential to understand that the tension force throughout an ideal massless and frictionless string remains constant. That is to say, the tension on block A's side of the pulley is equal to the tension on block C's side. Calculating the correct tension force is critical for understanding the equilibrium of forces and motion of the system in problems of this nature.
System of Masses Physics
Delving into systems of masses in physics, specifically with our exercise, one needs to consider how different masses interact with each other through forces and motions. In such systems, the connection between the masses (via a string in our case) dictates that they don't operate independently but rather as a cohesive unit.

Here, block A's movement is influenced by block B's static friction and the gravitational pull on block C. Understanding how these masses and forces interact is critical for predicting how the system will behave when it is released from rest. Block C's mass directly affects block A's tension force and, consequently, the frictional force that block B can exert. If block C is too heavy, the static friction threshold between blocks A and B would be exceeded, causing block B to slide off.

This interconnectedness is the beauty of physics: how one mass can influence another, which in turn affects the entire system. Grasping these relationships allows students to analyze and predict the outcome of complex physical scenarios, making sense of the interaction between multiple objects and forces.

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Most popular questions from this chapter

An \(8.00 \mathrm{~kg}\) box sits on a ramp that is inclined at \(33.0^{\circ}\) above the horizontal. The coefficient of kinetic friction between the box and the surface of the ramp is \(\mu_{\mathrm{k}}=0.300 .\) A constant horizontal force \(F=26.0 \mathrm{~N}\) is applied to the box (Fig. \(\mathbf{P} 5.73),\) and the box moves down the ramp. If the box is initially at rest, what is its speed \(2.00 \mathrm{~s}\) after the force is applied?

Force on a Skater's Wrist. A \(52 \mathrm{~kg}\) ice skater spins about a vertical axis through her body with her arms horizontally outstretched; she makes 2.0 turns each second. The distance from one hand to the other is \(1.50 \mathrm{~m} .\) Biometric measurements indicate that each hand typically makes up about \(1.25 \%\) of body weight. (a) Draw a free-body diagram of one of the skater's hands. (b) What horizontal force must her wrist exert on her hand? (c) Express the force in part (b) as a multiple of the weight of her hand.

Jack sits in the chair of a Ferris wheel that is rotating at a constant \(0.100 \mathrm{rev} / \mathrm{s}\). As Jack passes through the highest point of his circular path, the upward force that the chair exerts on him is equal to one- fourth of his weight. What is the radius of the circle in which Jack travels? Treat him as a point mass.

Two ropes are connected to a steel cable that supports a hanging weight (Fig. P5.59). (a) Draw a freebody diagram showing all of the forces acting at the knot that connects the two ropes to the steel cable. Based on your diagram, which of the two ropes will have the greater tension? (b) If the maximum tension either rope can sustain without breaking is \(5000 \mathrm{~N},\) determine the maximum value of the hanging weight that these ropes can safely support. Ignore the weight of the ropes and of the steel cable.

Two blocks, with masses \(4.00 \mathrm{~kg}\) and \(8.00 \mathrm{~kg},\) are connected by a string and slide down a \(30.0^{\circ}\) inclined plane (Fig. \(\mathbf{P 5 . 9 6}\) ). The coefficient of kinetic friction between the \(4.00 \mathrm{~kg}\) block and the plane is \(0.25 ;\) that between the \(8.00 \mathrm{~kg}\) block and the plane is \(0.35 .\) Calculate (a) the acceleration of each block and (b) the tension in the string. (c) What happens if the positions of the blocks are reversed, so that the \(4.00 \mathrm{~kg}\) block is uphill from the \(8.00 \mathrm{~kg}\) block?

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