Two blocks are stacked on a frictionless table, and a...

Two blocks are stacked on a frictionless table, and a horizontal force $F$ is applied to the top block (block 1). Their masses are $m_{1}=2.50 \mathrm{~kg}$ and $m_{2}=3.75 \mathrm{~kg}$. The coefficients of static and kinetic friction between the blocks are 0.456 and 0.380 , respectively a) What is the maximum applied force $F$ for which $m_{1}$ will not slide off $m_{2} ?$ b) What are the accelerations of $m_{1}$ and $m_{2}$ when $F=24.5 \mathrm{~N}$ is applied to $m_{1}$ ?

Two blocks are stacked on a frictionless table, and a horizontal force $F$ is applied to the top block (block 1). Their masses are $m_{1}=2.50 \mathrm{~kg}$ and $m_{2}=3.75 \mathrm{~kg}$. The coefficients of static and kinetic friction between the blocks are 0.456 and
0.380 , respectively
a) What is the maximum applied force $F$ for which $m_{1}$ will not slide off $m_{2} ?$
b) What are the accelerations of $m_{1}$ and $m_{2}$ when $F=24.5 \mathrm{~N}$ is applied to $m_{1}$ ?

Key Concepts

Relative Motion Analysis

Relative motion analysis involves examining the motion of objects with respect to one another. In this context, it is used to determine whether the top block will move relative to the bottom block. By comparing the acceleration provided by the applied force with the maximum acceleration that static friction can provide, one can find the condition under which no sliding occurs between the blocks.

Kinetic Friction

Kinetic friction is the resistive force that acts between two sliding surfaces. Once the applied force exceeds the maximum static friction, the block will begin to slide, and kinetic friction, which typically has a lower coefficient than static friction, takes effect. Its value is likewise determined by the kinetic friction coefficient and the normal force.

Free-Body Diagrams

Free-body diagrams are visual representations of all the forces acting on each object. They are essential tools in mechanics for setting up and solving equations based on Newton's Second Law. In this scenario, drawing free-body diagrams for both blocks helps in understanding the contributions of the applied force and frictional forces, which is critical for determining the accelerations and the threshold for slipping.

Static Friction

Static friction is the force that prevents relative motion between two surfaces in contact up to a certain limit. In this problem, static friction between the two blocks determines the maximum force that can be applied to the top block before it begins to slide over the bottom block. Its maximum value is calculated as the product of the static friction coefficient and the normal force between the blocks.

Newton's Second Law

Newton's Second Law of Motion is fundamental to analyzing the dynamics of the blocks. It states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration (F = ma). This law is used to relate the applied force and frictional forces to the acceleration of each block in the problem.

Transcript

00:01 So here we have the free budget diagram for the two masses.

00:02 We're first going to apply newton's second law in the x direction for the first mass.

00:08 This would be m sub 1 times a sub x.

00:11 And here we have the force applied minus the frictional force equalling m sub 1 multiplied by a, the acceleration.

00:21 And we can say that then the force f would be equaling to the coefficient of static friction multiplied by n sub 1 plus m sub 1.

00:31 1 times the acceleration a.

00:33 We have then the sum of forces in the y direction for the first block.

00:41 This is going to be 0 in order to find say n sub 1 minus m sub 1 g, equaling 0, n sub 1 is then equal to m sub 1 g and we can plug this back into here.

01:02 And so f would then be equaling to the coefficient of static first.

01:08 Multiplied by m sub 1 multiplied by g plus m sub 1 multiplied by the acceleration a.

01:15 Of course then this is simply going to be equal to m sub 1 multiplied by the coefficient of static friction multiplied by g plus a.

01:26 So factoring out that first mass.

01:30 Now for block 2 we have the sum of forces in the x direction equaling m sub 2 a.

01:38 Now here, this would simply be equal to the frictional force.

01:43 So the coefficient of static friction, multiplied by n sub 1, would be equaling to m sub 2a.

01:52 This would, of course, be equal to mu sub s, m sub 1, g.

01:58 And so the acceleration a would be equalling to the coefficient of static friction times the mass of the first, multiplied by g.

02:08 This would be then divided by m sub 2 and we can then solve the acceleration would be equaling to 0 .456 multiplied by 2 .50 kilograms multiplied by 9 .81 meters per second squared and this would all be divided then by 3 .75 kilograms the acceleration is then equal to 2 .98 meters per second squared...

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