A rope with mass mr is attached to a block with mass mb as in Figure P 5.56 . The block rests on

step 1 All right, let's solve problem 56 in chapter 5. We have a block. We have this time a rope with m

A rope with mass mr is attached to a block with mass mb as...

A rope with mass $m_{r}$ is attached to a block with mass $m_{b}$ as in Figure $\mathrm{P} 5.56 .$ The block rests on a frictionless, horizontal surface. The rope does not stretch. The free end of the rope is pulled to the right with a horizontal force $\overrightarrow{\mathbf{F}}$ . (a) Draw force diagrams for the rope and the block, noting that the tension in the rope is not uniform. (b) Find the acceleration of the system in terms of $m_{b}, m_{r}$ and $F$ . (c) Find the magnitude of the force the rope exerts on the block. (d) What happens to the force on the block as the rope's mass approaches zero? What can you state about the tension in a light cord joining a pair of moving objects?

A rope with mass $m_{r}$ is attached to a block with mass $m_{b}$ as in Figure $\mathrm{P} 5.56 .$ The
block rests on a frictionless, horizontal surface. The rope does not stretch. The free end of the rope is pulled to the right with a horizontal force $\overrightarrow{\mathbf{F}}$ . (a) Draw force diagrams for the rope and the block, noting that the tension in the rope is not uniform. (b) Find the acceleration of the system in terms of $m_{b}, m_{r}$ and $F$ . (c) Find the magnitude of the force the rope exerts on the block. (d) What happens to the force on the block as the rope's mass approaches zero? What can you state about the tension in a light cord joining a pair of moving objects?

Key Concepts

Free-body Diagrams

Free-body diagrams are schematic representations used to visualize all the forces acting on individual components of a system. They help in isolating an object to account for its external forces, including applied forces, friction, tension, and normal forces, which is essential for formulating the equations of motion for each part of the system.

Newton's Second Law

Newton's Second Law states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration (F = ma). This fundamental principle is used to relate the forces acting on the system to the resulting acceleration, including when multiple masses or forces are involved in a connected system.

Variable Tension Distribution

In systems with a rope or chain having mass, the tension is not uniform throughout because each segment of the rope must accelerate its own mass in addition to transmitting the applied force. This concept is crucial in analyzing systems where the rope’s mass cannot be neglected, leading to a distribution of tension that varies along its length.

Inextensible Rope Assumption

An inextensible rope or cord is assumed to have a fixed length under tension, which implies that all parts of the rope move with the same acceleration. This simplifying assumption is often used in solving problems involving connected objects, as it links the motion of the objects through the constraint on the relative displacements of the points on the rope.

Limit Behavior in Idealized Systems

When analyzing systems with a rope that has mass, it is often instructive to consider the limiting case where the rope's mass approaches zero. In this limit, the rope becomes 'light', and the tension throughout the rope becomes uniform. This concept highlights how idealized assumptions simplify analysis by eliminating the effects of distributed mass in the connecting medium.

Transcript

00:01 All right, let's solve problem 56 in chapter 5.

00:04 We have a block.

00:05 We have this time a rope with mass, and we're pulling on it with force f.

00:12 And so let's sort of do an interesting shortcut.

00:16 Let's just say, here's our system.

00:19 We just consider the combined mass, right? the only force that's acting on them that we care about that's moving them in the horizontal direction is f.

00:27 And that's the combined mass of blocks plus the rope.

00:34 Times the acceleration of that system, right? and so that's just f over mb plus mr, right? and the acceleration of this total system, well, that's also the acceleration of each of the individual parts, right? if this whole thing's moving forward at rate a, this thing, this rope must be moving in a, this box must be moving at a.

01:04 So if we look at just the box, all we have is just the tension at that point.

01:12 I guess we'll call it t0 to indicate the tension at the beginning of the rope.

01:17 And so we'll get t0 is just the box mass times the acceleration, right? and if we consider the rope, this t0 is a reactionary pair, right? if the rope is pulling on the box, the box is pulling on the rope.

01:37 So this is t0 and it's getting pulled by f.

01:41 And so what we get us f minus t0 equals the mass of the rope times a.

01:51 And if you add these two equations together, you'll actually just retrieve that first equation we wrote when we consider the combined body, right? okay.

02:05 So without even having to draw the separate diagrams, we already pulled out a, but i showed you how to get another way in case you didn't trust the shenanigans i pulled to do it...

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