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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

Block \(A\) has a mass of \(10 \mathrm{kg}\), and blocks \(B\) and \(C\) have masses of \(5 \mathrm{kg}\) each. Knowing that the blocks are initially at rest and that \(B\) moves through \(3 \mathrm{m}\) in \(2 \mathrm{s}\), determine (a) the magnitude of the force \(\mathrm{P},(b)\) the tension in the cord \(A D .\) Neglect the masses of the pulleys and axle friction.

A mother and her child are skiing together, and the mother is holding the end of a rope tied to the child's waist. They are moving at a speed of \(7.2 \mathrm{km} / \mathrm{h}\) on a gently sloping portion of the ski slope when the mother observes that they are approching a steep descent. She pulls on the rope with an average force of \(7 \mathrm{N}\). Knowing the coefficient of friction between the child and the ground is 0.1 and the angle of the rope does not change, determine \((a)\) the time required for the child's speed to be cut in half, \((b)\) the distance traveled in this time.

A \(5000-\) lb truck is being used to lift a 1000 -lb boulder \(B\) that is on a 200 -lb pallet \(A\). Knowing the acceleration of the truck is \(1 \mathrm{ft} / \mathrm{s}^{2}\), determine \((a)\) the horizontal force between the tires and the ground, (b) the force between the boulder and the pallet.

A small 250 -g collar \(C\) can slide on a semicircular rod which is made to rotate about the vertical \(A B\) at a constant rate of 7.5 rad/s. Determine the three values of \(\theta\) for which the collar will not slide on the rod, assuming no friction between the collar and the rod.

A light train made up of two cars is traveling at \(90 \mathrm{km} / \mathrm{h}\) when the brakes are applied to both cars. Knowing that car \(A\) has a mass of \(25 \mathrm{Mg}\) and car \(B\) a mass of \(20 \mathrm{Mg}\), and that the braking force is \(30 \mathrm{kN}\) on each car, determine \((a)\) the distance traveled by the train before it comes to a stop, \((b)\) the force in the coupling between the cars while the train is slowing down.
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