A block of mass m1=2.30 kg is placed in front of a block of mass m2=5.20 kg, as shown in the f

A block of mass m1=2.30 kg is placed in front of a block of...

Key Concepts

System Acceleration Constraint

When multiple objects are in contact and move together, they share a common acceleration unless relative motion occurs. Analyzing the system under the constraint that the blocks move together without slipping allows one to relate the forces between the blocks, including friction and contact forces, to the overall applied force, ensuring that internal forces balance appropriately to maintain the intended motion.

Static Friction

Static friction is the resistance force that prevents two contacting surfaces from sliding past one another until a certain threshold force is exceeded. Its maximum value is given by the product of the coefficient of static friction and the normal force. This concept is crucial when determining the minimum external force required to keep one block from moving relative to the other.

Contact Forces

Contact forces are the interactive forces that objects in physical contact exert on each other, such as normal forces and frictional forces. In the context of two blocks in contact, understanding the contact force helps in evaluating how force is transmitted from one block to the other, which is key in calculating the acceleration of the system when an external force is applied.

Newton's Laws of Motion

This concept explains how forces acting on an object lead to acceleration according to the relation F = ma, and how all forces acting on an object must be taken into account to determine its motion. It underpins the reasoning for calculating the net force on each block and their subsequent accelerations, ensuring that the dynamics of the system are properly analyzed.

Free-Body Diagrams

Free-body diagrams are graphical representations used to isolate an object and depict all the forces acting upon it. In problems involving multiple objects, drawing free-body diagrams for each block allows one to systematically account for individual forces like gravity, normal forces, friction, and applied forces, thereby simplifying the analysis of each object's motion.

Transcript

0:00 Hi there.

00:01 So for this problem, we have a block of mass m1 equal to 2 .3 kilograms and a mass m2 of a value of 5 .2 kilograms.

00:20 And the mass m1 is placed in front of the block of mass m2 as is shown in this figure.

00:29 Now, the coefficient of static friction between these masses is also given, and that is equal to 0 .65, and there is negligible friction between the large block and the tabletop.

00:46 So for part a of this problem, we are asked about what forces are opting on the mass m1.

01:00 Now to answer this question, what we can do is to draw a free body diagram for that mass.

01:11 So i'm going to draw that in here.

01:14 This is the mass m1.

01:17 We're going to draw the axis.

01:19 This is the y -axis.

01:21 This is the x -axis.

01:25 So the forces that are opting on these mass are, of course, the weight downward.

01:33 So that is the mass m1 times the acceleration due to gravity.

01:39 That's the weight.

01:41 And we also have the frictional force upward, so it prevents this to fall.

01:49 So this is the frictional force we'll call.

01:53 And we have a normal force to the left because it is always perpendicular to, the surface of comtet.

02:12 So these three are the forces that are acting on the block.

02:16 The normal force, the frictional, and the gravitational force.

02:21 So for part b, we are asked about what is the minimum external force f that can be applied to m2 so that m1 doesn't fault.

02:34 Now, to answer that question, we need first to use, newton's second law and first to the block one so for that we are going to have that the zoom of forces in the x component it's equal to well as you can see from the picture from this diagram of forces the only forces that are acting on the x component is the normal force and one and this of course is going to be equal to the mass m1 times the acceleration.

03:17 And for the sum of forces in the y component, we know that we have in here the frictional force upward minus the weight of this block.

03:31 And this is equal to zero because this block is not moving in that component for this case.

03:39 So we will have that the frictional force.

03:42 Is equal to the weight.

03:46 Now, there is a minimal force f when there is a minimum acceleration.

03:53 And we know that that is when the frictional force is equal to the product between the static frictional force and the normal force n1.

04:04 So what we can do in here is to solve for the normal force into this expression.

04:12 As we know this is also the frictional force.

04:16 So we will have equating that two equations, we have the mass m1 times the acceleration due to gravity is equal to the coefficient of static friction times the normal force n1.

04:27 So the normal force n1 is equal to the mass n1 times acceleration due to gravity over the coefficient of static friction.

04:37 And what we can do now is to substitute that expression in here so we can obtain the acceleration for this case.

04:51 So plotting that in dirt, we have the mass times acceleration due to gravity over the coefficient of study friction.

05:03 This is the mass mn1 and this is the mass mn1 times the acceleration.

05:10 So from here we obtain that the acceleration of this block.

05:15 Is equal to the acceleration due to gravity over the coefficient of static friction.

05:21 Now with this we now consider a block that is composed of the of the block one and the block two.

05:31 So in that sense we will have that the sum of forces in the x component are going to be just the sum the force that is being applied is equal to the sum of these masses times the acceleration...

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