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Chapter 12: Problem 31

A \(10-\) b block \(B\) rests as shown on a 20 -lb bracket \(A\). The coefficients of friction are \(\mu_{s}=0.30\) and \(\mu_{k}=0.25\) between block \(B\) and bracket \(A,\) and there is no friction in the pulley or between the bracket and the horizontal surface. (a) Determine the maximum weight of block \(C\) if block \(B\) is not to slide on bracket \(A .\) (b) If the weight of block \(C\) is 10 percent larger than the answer found in \(a,\) determine the accelerations of \(A, B,\) and \(C .\)

Short Answer

Expert verified
Maximum weight of block C is 3 lb. If 10% larger, all blocks accelerate at 0.322 ft/s虏.

Step by step solution

01

Analyze forces on Block B

Block B with a weight of 10 lb rests on bracket A. The maximum frictional force that prevents B from sliding is given by the static friction formula, which is the product of the static friction coefficient and the normal force (weight of Block B). This is calculated as follows: \( F_{friction} = \mu_s \, m_B \cdot g = 0.30 \times 10 \text{ lb} = 3 \text{ lb} \).
02

Construct the force equation for equilibrium

For block B to not slide, the maximum frictional force must equal the force due to the tension in the rope caused by block C. Hence, \( T_{max} = F_{friction} = 3 \text{ lb} \). Block C must exert this maximum tension without exceeding it to maintain equilibrium.
03

Calculate the Maximum Weight of Block C

Using the equilibrium condition, the tension in the rope is due to the weight of block C, hence \( W_C = T_{max} = 3 \text{ lb} \). Thus, the maximum weight of block C is 3 lb.
04

Calculate the weight of Block C if it is 10% larger

If block C is 10% larger, its weight becomes \( W_{C_{new}} = 1.1 \times 3 \text{ lb} = 3.3 \text{ lb} \). This new weight causes an acceleration of B and A.
05

Determine acceleration of blocks using Newton's Second Law

When block C is heavier than 3 lb, sliding occurs. Applying Newton's law, \( F = m \cdot a \), where the net force acting on the system A, B, and C is the excess weight of block C over the maximum frictional force. The system's total weight is sum of both blocks A and B (30 lb). The net accelerating force is \( 3.3 \text{ lb} - 3 \text{ lb} = 0.3 \text{ lb} \). Thus: \( \frac{0.3}{30} \times 32.2 \text{ ft/s}^2 \approx 0.322 \text{ ft/s}^2 \).
06

Compute accelerations of individual blocks

Block A alone (20 lb) has the same acceleration as the entire system, so its acceleration is \( a = 0.322 \text{ ft/s}^2 \). Block B is subject to the same force, thus \( a = 0.322 \text{ ft/s}^2 \). Block C being a separate segment of the system and also accounting for the same force, also has \( a = 0.322 \text{ ft/s}^2\).

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Key Concepts

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

Friction
Friction is the resistance that one surface or object encounters when moving over another. In our exercise, the coefficient of static friction, represented as \( \mu_s \), is crucial. It determines the frictional force \( F_{friction} \) needed to prevent movement of Block B on Bracket A. This force can be calculated using the equation: \( F_{friction} = \mu_s \cdot m_B \cdot g \). Here, \( m_B \) is the mass of Block B and \( g \) is the acceleration due to gravity (approximated as 32.2 ft/s虏 for calculations in pounds).

To maintain static friction, the force applied by Block C through the tension in the pulley must not exceed 3 lb. In scenarios where the static friction limit is exceeded, kinetic friction takes over, but its coefficient (\( \mu_k \)) is typically lower, indicating less resistance once motion starts.
Equilibrium
Equilibrium occurs when all forces acting on a system are balanced, resulting in no net movement. In statics problems鈥攕uch as our exercise where Block B rests on Bracket A鈥攅quilibrium means that the static friction force equals the pulling force due to Block C. We say the system is in translational equilibrium.

Translational equilibrium implies that for Block B to remain stationary, the frictional force of 3 lb must balance with the force exerted by Block C via the tension in the cable. This concept is essential when calculating whether an additional weight or force will disrupt the motionlessness, leading to acceleration.
Newton's Second Law
Newton's Second Law provides a key formula in solving dynamics problems: \( F = m \cdot a \). In this formula, \( F \) is the net force acting on an object, \( m \) is its mass, and \( a \) is its acceleration. This principle defines how forces influence motion.

When the weight of Block C exceeds the maximum static friction, the system is no longer in equilibrium. Using Newton's Second Law, we determine the system's acceleration. The excess weight of 0.3 lb (compared to the maximum frictional force) causes the entire system to accelerate. The total mass considered includes both Block A and Half of B, applying the net force to this sum results in the calculated acceleration.
Acceleration
Acceleration is the rate of change of velocity over time, often a result of unbalanced forces. After Block C's weight surpasses the equilibrium limit, Block A, Block B, and Block C experience uniform acceleration of approximately 0.322 ft/s虏.

Since all three blocks are interconnected in this exercise, with Block B on A and C causing the force, they share the same acceleration. Therefore, even though Block C instigates the change, the entire system moves collectively, demonstrating a fundamental characteristic of mechanical systems where components are interrelated in function and behavior.

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Most popular questions from this chapter

Communication satellites are placed in a geosynchronous orbit, i.e., in a circular orbit such that they complete one full revolution about the earth in one sidereal day \((23.934 \mathrm{h})\), and thus appear stationary with respect to the ground. Determine \((a)\) the altitude of these satellites above the surface of the earth, (b) the velocity with which they describe their orbit. Give the answers in both SI and U.S. customary units.

To transport a series of bundles of shingles \(A\) to a roof, a contractor uses a motor-driven lift consisting of a horizontal platform \(B C\) which rides on rails attached to the sides of a ladder. The lift starts from rest and initially moves with a constant acceleration \(a_{1}\) as shown. The lift then decelerates at a constant rate \(a_{2}\) and comes to rest at \(D,\) near the top of the ladder. Knowing that the coefficient of static friction between a bundle of shingles and the horizontal platform is 0.30 , determine the largest allowable acceleration \(\mathbf{a}_{1}\) and the largest allowable deceleration \(\mathbf{a}_{2}\) if the bundle is not to slide on the platform.

A semicircular slot of 10 -in. radius is cut in a flat plate that rotates about the vertical \(A D\) at a constant rate of 14 rad/s. A small, \(0.8-\) lb block \(E\) is designed to slide in the slot as the plate rotates. Knowing that the coefficients of friction are \(\mu_{s}=0.35\) and \(\mu_{k}=0.25,\) determine whether the block will slide in the slot if it is released in the position corresponding to \((a) \theta=80^{\circ},(b) \theta=40^{\circ} .\) Also determine the magnitude and the direction of the friction force exerted on the block immediately after it is released.

Show that the angular momentum per unit mass \(h\) of a satellite describing an elliptic orbit of semimajor axis \(a\) and eccentricity \(\varepsilon\) about a planet of mass \(M\) can be expressed as $$h=\sqrt{G M a\left(1-\varepsilon^{2}\right)}$$

A \(0.5-\mathrm{kg}\) block \(B\) slides without friction inside a slot cut in arm \(O A\) that rotates in a vertical plane. The rod has a constant angular acceleration \(\dot{\theta}=10 \mathrm{rad} / \mathrm{s}^{2} .\) Knowing that when \(\theta=45^{\circ}\) and \(r=0.8 \mathrm{m}\) the velocity of the block is zero, determine at this instant, (a) the force exerted on the block by the arm, (b) the relative acceleration of the block with respect to the arm.
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