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Chapter 7: Problem 62

A block is lying on the horizontal frictionless surface. One end of a uniform rope is fixed to the block which is pulled in the horizontal direction by applying a force \(F\) at the other end. If the mass of the rope is half the mass of the block, the tension in the middle of the rope will be a. \(F\) b. \(2 F / 3\) c. \(3 F / 5\) d. \(5 F / 6\)

Short Answer

Expert verified
The tension in the middle of the rope is \( \frac{2F}{3} \).

Step by step solution

01

Analyze the System

We have a system consisting of a block and a rope with no friction. The force \( F \) is applied horizontally at one end of the rope.
02

Understand Mass Distribution

Let the mass of the block be \( M \) and the mass of the rope is given to be half of that of the block, \( m = \frac{M}{2} \).
03

Calculate the Total Mass

The total mass \( m_{total} \) of the system (block plus rope) is \( M + \frac{M}{2} = \frac{3M}{2} \).
04

Determine the Acceleration of the System

Using Newton's second law, the acceleration \( a \) of the entire system is given by \( F = m_{total} \cdot a = \frac{3M}{2} \cdot a \), so the acceleration \( a \) is \( a = \frac{2F}{3M} \).
05

Calculate the Force on the Rope's Half-Mass

The force required to accelerate the mass \( m = \frac{M}{2} \) of half the rope, using \( F = m \cdot a \), is \( F = \frac{M}{2} \cdot \frac{2F}{3M} = \frac{F}{3} \).
06

Determine Tension in the Center of the Rope

The tension at the center of the rope handles half the rope mass's force, which is the remaining part of \( F \). Therefore, the tension \( T \) in the middle is \( F - \frac{F}{3} = \frac{2F}{3} \).

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Key Concepts

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

Newton's Second Law in Action
In the case of the block and rope system, Newton's Second Law plays a crucial role. This law states that the force acting on an object is equal to the mass of the object multiplied by its acceleration. Mathematically, this is represented as \( F = m \times a \).
In our situation, the system consists of a block and a rope. A force \( F \) is applied horizontally, causing the system to accelerate. To find this acceleration, we need to consider the total mass of the entire system, which includes both the block and the rope. Since the mass of the rope is half of the mass of the block, we sum them up to find the total mass as \( M + \frac{M}{2} = \frac{3M}{2} \).
This total mass is then used in Newton's formula to calculate the acceleration: \[ a = \frac{F}{\frac{3M}{2}} = \frac{2F}{3M} \]. This gives the whole system an acceleration based on the combined effect of the block and rope.
Understanding Mass Distribution
Mass distribution refers to how mass is spread out in a system. In our specific exercise, the block and rope form a simple system where mass distribution is crucial for calculating tension.
The block has a mass \( M \) and the rope has mass \( \frac{M}{2} \). Understanding the value and division of mass is important because it determines how different parts of this system will respond to applied forces.
When you apply a force \( F \) at one end of such a system, the entire mass must be evenly accounted for in how it accelerates. Since the mass of the rope is half that of the block, the distribution tells us how much force is needed to move each part of the system, especially when considering tension in the rope.
Frictionless Surface Considerations
A frictionless surface presents a special scenario in mechanics, as it means there are no opposing forces slowing down the movement of objects. In this exercise, the block and rope lie on a frictionless surface, simplifying calculations.
Without friction, any force applied is solely responsible for the movement of the system. This means the force \( F \) moves everything directly without energy losses due to friction. Calculations thus become more straightforward because the force used equals the force applied, minus any internal forces within the system, like tension in parts of the rope.
For our system, the absence of friction allows us to focus on how force transmits across the rope. The force's effect is seen clearly in the tension, which is maintained equally through the rope's length until the block at the other end. Understanding the lack of friction helps simplify our approach to problems involving tension within connected bodies.

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Most popular questions from this chapter

A man is raising himself and the crate on which he stands with an acceleration of \(5 \mathrm{~m} / \mathrm{s}^{2}\) by a massless rope-andpulley arrangement. Mass of the man is \(100 \mathrm{~kg}\) and that of the crate is \(50 \mathrm{~kg}\). If \(g=10 \mathrm{~m} / \mathrm{s}^{2}\), then the tension in the rope is a. \(2250 \mathrm{~N}\) b. \(1125 \mathrm{~N}\) c. \(750 \mathrm{~N}\) d. \(375 \mathrm{~N}\)

A block of metal weighing \(2 \mathrm{~kg}\) is resting on a frictionless plane. It is struck by a jet releasing water at a rate of \(1 \mathrm{~kg} / \mathrm{s}\) and at a speed of \(5 \mathrm{~m} / \mathrm{s}\). The initial acceleration of the block is a. \(\frac{5}{3} \mathrm{~m} / \mathrm{s}^{2}\) b. \(\frac{25}{4} \mathrm{~m} / \mathrm{s}^{2}\) c. \(\frac{25}{6} \mathrm{~m} / \mathrm{s}^{2}\) d. \(\frac{5}{2} \mathrm{~m} / \mathrm{s}^{2}\)

A wooden block of mass \(M\) resting on a rough horizontal floor is pulled with a force \(F\) at an angle \(\phi\) with the horizontal. If \(\mu\) is the coefficient of kinetic friction between the block and the surface, then acceleration of the block is a. \(\frac{F}{M} \sin \phi\) b. \(\frac{F}{M}(\cos \phi+\mu \sin \phi)-\mu g\) c. \(\frac{\mu F}{M} \cos \phi\) d. \(\frac{F}{M}(\cos \phi-\mu \sin \phi)-\mu g\)

Two persons are holding a rope of negligible weight tightly at its ends so that it is horizontal. A \(15 \mathrm{~kg}\) weight in attached to rope at the mid point which now no more remains horizontal. The minimum tension required to completely straighten the rope is a. \(150 \mathrm{~N}\) b. \(75 \mathrm{~N}\) c. \(50 \mathrm{~N}\) d. infinitely large

A block of mass \(M\) is pulled along a horizontal frictionless surface by a rope of mass \(m\). Force \(P\) is applied at one end of rope. The force which the rope exerts on the block \(\mathrm{s}\) a. \(\frac{P}{(M-m)}\) b. \(\frac{P}{M(m+M)}\) c. \(\frac{P M}{(m+M)}\) d. \(\frac{P M}{(M-m)}\)
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