/*! 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 28 Block \(A\) has a mass of \(10 \... [FREE SOLUTION] | 91Ó°ÊÓ

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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.

Short Answer

Expert verified
(a) Force P is 15 N. (b) Tension in cord AD is 114.3 N.

Step by step solution

01

Identify Given Data and Diagram Setup

We know the masses are: block A is 10 kg, block B is 5 kg, and block C is 5 kg. Block B moves 3 meters in 2 seconds. We need to determine the force P and the tension in cord AD. First, draw a diagram showing blocks A, B, and C connected over pulleys.
02

Calculate the Acceleration of Block B

Block B covers a distance of 3 meters in 2 seconds starting from rest. Using the formula for acceleration \(s = ut + \frac{1}{2}at^2\), where \(u = 0\), we solve for acceleration \(a\): \(3 = \frac{1}{2} \times a \times (2)^2\). We find \(a = \frac{3}{2}\), so \(a = 1.5 \, \mathrm{m/s}^2\).
03

Apply Newton's Second Law to Block B

For block B, the formula is \(F = ma\). For block B, this becomes \(P - T = m_B \cdot a\). Substitute known values: \(P - T = 5 \times 1.5\). This gives \(P - T = 7.5 \, \mathrm{N}\).
04

Apply Newton's Second Law to Block C

For block C, which only experiences tension \(T\) acting upwards, we have \(T = m_C \cdot a\). Substituting in for mass and acceleration gives \(T = 5 \times 1.5 = 7.5 \, \mathrm{N}\).
05

Solve Tension Equation for Block B

Since \(P - T = 7.5 \, \mathrm{N}\) and \(T = 7.5 \, \mathrm{N}\), substitute \(T\) into equation: \(P = 7.5 + 7.5 = 15 \, \mathrm{N}\).
06

Apply Newton's Second Law to Block A for Cord AD Tension

For block A, gravity and tension act on it. Using \(ma = T_{AD} - mg\), we solve for \(T_{AD}\). Substituting in \(m = 10\) and \(a = 1.5\), and \(g = 9.81\) gives:\[T_{AD} = 10 \, (1.5 + 9.81) = 114.3 \, \mathrm{N}.\]

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Key Concepts

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

Newton's Second Law
Newton's Second Law is a fundamental principle that governs the motion of objects and is crucial in understanding dynamics. It states that the acceleration of an object is dependent on two main factors: the net force acting on the object and the object's mass. In simple terms, the law can be expressed by the equation \( F = ma \), where \( F \) is the force applied to the object, \( m \) is the mass of the object, and \( a \) is the acceleration produced.In the context of our exercise, Newton's Second Law helps us find unknown forces and accelerations in a system of interconnected blocks and pulleys. By applying this law, we can determine the relationships between the forces acting on each block and compute the necessary quantities such as tension and applied force. This shows how a force causes acceleration and how mass modulates this effect.
Newton's Second Law forms the backbone of solving many physics problems involving motion.
Acceleration calculation
Calculating acceleration is essential in understanding how quickly an object's velocity changes over time. Acceleration describes not just the speed of change but also the direction. To compute acceleration, we use the formula: \[ s = ut + \frac{1}{2}at^2 \]where:
  • \( s \) is the distance traveled,
  • \( u \) is the initial velocity (zero if the object starts from rest),
  • \( a \) is the acceleration, and
  • \( t \) is the time taken.
In our problem, block B covered a distance of 3 meters in 2 seconds, starting from rest. By reorganizing the equation and solving for \( a \), we find: \[ 3 = \frac{1}{2}a(2)^2 \]Solving gives \( a = 1.5 \, \mathrm{m/s}^2 \).
This value of acceleration is then used in subsequent steps to find forces such as tension in the cords.
Tension in a cord
Tension in a cord is a force that is transmitted through a string, rope, or cable when it is pulled tight by forces acting from opposite ends. In many physics problems, understanding tension is key to solving dynamics questions involving pulleys or suspended weights.To analyze tension in the context of our exercise, consider how it acts as an intermediary force between connected blocks. In the exercise, block B is connected to block C via a cord. The tension, denoted as \( T \), can be specifically computed for block C by applying Newton's Second Law: \[ T = m_C \cdot a \]Substituting the mass of block C (5 kg) and the previously calculated acceleration (1.5 m/s²), we find \[ T = 5 \times 1.5 = 7.5 \, \mathrm{N} \].Understanding tension helps clarify how forces are distributed throughout a system and is crucial for solving for unknown forces or accelerations in complex systems with multiple pulleys or masses.

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Most popular questions from this chapter

A 1 -kg collar can slide on a horizontal rod that is free to rotate about a vertical shaft. The collar is initially held at \(A\) by a cord attached to the shaft. A spring of constant \(30 \mathrm{N} / \mathrm{m}\) is attached to the collar and to the shaft and is undeformed when the collar is at \(A .\) As the rod rotates at the rate \(\dot{\theta}=16 \mathrm{rad} / \mathrm{s}\), the cord is cut and the collar moves out along the rod. Neglecting friction and the mass of the rod, determine (a) the radial and transverse components of the acceleration of the collar at \(A,(b)\) the acceleration of the collar relative to the rod at \(A,(c)\) the transverse component of the velocity of the collar at \(B .\)

An airplane has a mass of \(25 \mathrm{Mg}\) and its engines develop a total thrust of \(40 \mathrm{kN}\) during take-off. If the drag \(D\) exerted on the plane has a magnitude \(D=2.25 \mathrm{v}^{2}\), where \(v\) is expressed in meters per second and \(D\) in newtons, and if the plane becomes airborne at a speed of \(240 \mathrm{km} / \mathrm{h}\), determine the length of runway required for the plane to take off.

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 \(10-\) 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 \(C .\)

A baggage conveyor is used to unload luggage from an airplane. The 10 -kg duffel bag \(A\) is sitting on top of the \(20-\) kg suitcase \(B .\) The conveyor is moving the bags down at a constant speed of \(0.5 \mathrm{m} / \mathrm{s}\) when the belt suddenly stops. Knowing that the cofficient of friction between the belt and \(B\) is 0.3 and that bag \(A\) does not slip on suitcase \(B\), determine the smallest allowable coefficient of static friction between the bags.

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