As shown in the figure with Problem 4.94 , a block of mass M1=0.250 kg is initially at rest on

step 1 In this problem we have three masses M1, M2 and M3. They said to us that the mass M1 is initiall

As shown in the figure with Problem 4.94 , a block of mass...

As shown in the figure with Problem 4.94 , a block of mass $M_{1}=0.250 \mathrm{~kg}$ is initially at rest on a slab of mass $M_{2}=0.420 \mathrm{~kg}$, and the slab is initially at rest on a level table. A string of negligible mass is connected to the slab, runs over a frictionless pulley on the edge of the table, and is attached to a hanging mass $M_{3}=1.80 \mathrm{~kg} .$ The block rests on the slab but is not tied to the string, so friction provides the only horizontal force on the block. The slab has a coefficient of kinetic friction $\mu_{\mathrm{k}}=0.340$ with both the table and the block. When released, $M_{3}$ pulls on the string, which accelerates the slab so quickly that the block starts to slide on the slab. Before the block slides off the top of the slab: a) Find the magnitude of the acceleration of the block. b) Find the magnitude of the acceleration of the slab.

As shown in the figure with Problem 4.94 , a block of mass $M_{1}=0.250 \mathrm{~kg}$ is initially at rest on a slab of mass $M_{2}=0.420 \mathrm{~kg}$, and the slab is initially at rest on a level table. A string of negligible mass is connected to the slab, runs over a frictionless pulley on the edge of the table, and is attached to a hanging mass $M_{3}=1.80 \mathrm{~kg} .$ The block rests on the slab but is not tied to the string, so friction provides the only horizontal force on the block. The slab has a coefficient of kinetic friction $\mu_{\mathrm{k}}=0.340$ with both the table and the block. When released, $M_{3}$ pulls on the string, which accelerates the slab so quickly that the block starts to slide on the slab. Before the block slides off the top of the slab:
a) Find the magnitude of the acceleration of the block.
b) Find the magnitude of the acceleration of the slab.

Key Concepts

Newton's Second Law

Newton's Second Law, which states that the net force acting on an object is equal to the mass of the object times its acceleration, is crucial in analyzing the motion of systems with multiple interacting bodies. It provides the fundamental relationship that allows us to write equations of motion for each object in the system based on the forces acting upon them.

Frictional Forces

Frictional forces, particularly kinetic friction in situations where objects slide relative to each other, play a key role in the dynamics of systems. The magnitude of kinetic friction is typically given by the product of the coefficient of kinetic friction and the normal force, and it acts to oppose the relative motion between surfaces in contact.

Free-Body Diagrams

Free-body diagrams are an essential tool for visualizing and resolving the forces acting on each component of a system. By representing all forces, including gravitational, frictional, tension, and normal forces, these diagrams enable a systematic approach to applying Newton's laws and solving for unknown quantities such as acceleration.

Relative Motion Analysis

Relative motion analysis involves understanding the movement of one object with respect to another. In scenarios where one surface is moving (or accelerating) while another object slides on it, distinguishing between the absolute acceleration and the relative acceleration is critical for correctly applying force and motion equations.

Connected Systems

In problems involving connected objects, such as those linked by a string over a pulley, it is important to recognize the constraints imposed by the connection. The acceleration and tension in the string are interdependent across the different bodies, which must be incorporated into the analysis using Newton's laws to fully describe the system's motion.

Transcript

00:01 In this problem we have three masses m1, m2 and m3.

00:04 They said to us that the mass m1 is initially addressed on a slab m2, so m2 is a slab.

00:11 This slab is also initially addressed.

00:15 We have a string which connect the mass m2 with the mass m3.

00:21 This string is of negligible mass and runs over a frictionless pulley, which is on the edge of the table.

00:34 So they said to us that there is some friction force, some kinetic friction force between, there's going to be some kinetic friction force between the mass m2 and the table and block.

00:48 Okay, and when the system is released, m3 pulse in the stream, which accelerate the system and also the slab so quickly that the block starts to slide on the slide.

01:00 So before the block slides off the top of the slab, we need to find the magnitude of the acceleration of this block m1 and we need to find the magnitude of the acceleration of the system.

01:12 Okay, so the first thing i'm going to do is a free body diagram for each of these masses.

01:20 So i'm going to start with m1.

01:23 So we're going to have the normal force, okay, which fills m1 because of its contact with m2.

01:31 The weight.

01:34 I'm going to assume that when the system starts to move in, to move, sorry, there is going to be, this block over here is going to move to the right, okay? to left, sorry.

01:48 So we are going to have some kinetic friction force in here, okay? and there is nothing more in here.

01:56 So the equation of motion for this mass over here, it's going to be the kinetic friction force, which is going to be equal to mass times its acceleration.

02:08 And also, normal one, which is going to be equal to the weight one.

02:14 This is going to be equation one, and this is going to be equation two.

02:18 So now i'm going to do, sorry about this.

02:22 Now i'm going to do the same for the mass three.

02:28 So this is going to be x, this is going to be y.

02:33 In here it's going to be tension force, which is going to, i'm going to denote it by tension 3 and the weight.

02:40 Okay.

02:42 So the equation of motion for this mass is going to be in x, there's nothing.

02:48 And in y, it's going to be motion.

02:52 So this is going to be like this.

02:54 And this is going to be equal to the mass three times the asceration.

03:01 I'm going to put it like this.

03:04 Okay.

03:06 So, and i'm going to put a one in here.

03:11 So now i'm going to do the same for the mass m2.

03:18 So m2.

03:21 So i'm going to have the normal force, okay, that feels m2 because of the table, which is going to be pointing to, which is going to be pointing up.

03:38 And there is also going to be a normal force, okay, because of its contact, because of the contact between m2 and m1, okay, which i'm going to call it n1 because it's going to be equal in models to these other and normal force in here.

03:55 But there is going to be, there is, there are going to be opposed in direction.

04:03 So i'm going to call it n1.

04:07 There is also going to be a kinetic friction force.

04:10 So if the system is moving in this direction, there is going to be a kinetic friction force pointing to the right in here.

04:21 Okay.

04:22 There is going to be the tension over here, which i'm going to denote it like this.

04:26 And there is also going to be a kinetic friction force because of the presence of the mass m1.

04:34 Okay.

04:34 So the equation of motion for this is going to be tension 2 minus friction kinetic force 1 minus friction kinetic force 2 equal to the mass 2 times this duration 2 y and this is going to be like this the normal 2 minus the normal 1 of course the weight of the mass 2 minus the weight of the mass 2 and this is going to be equal to 0 so i'm going to put it like this so this is going to be equation 3 and this is going to be equation 4 and this is going to be equation 5 so from equation 1 so i'm going to do letter a so from equation 1 we have that the kinetic friction force 1 is equal to the this.

05:40 So in this case we're going to have the acceleration is going to be minus the kinetic friction divided by its mass.

05:48 Okay.

05:49 So in this case, this cannot be negative because this is a modulus in here.

05:53 So these minus that to us, the assumption that we do at the beginning, okay, that the direction of the velocity is going to be pointing in the left direction...

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