4 In Fig. 6-39, a slab of mass m1 40 kg rests on a...

4 In Fig. 6-39, a slab of mass m1 40 kg rests on a frictionless floor, and a block of mass m2 12 kg rests on top of the slab. Between block and slab, the coefficient of static friction is 0.60, and the coefficient of kinetic friction is 0.40. A horizontal force of magnitude 120 N begins to pull directly on the block, as shown. In unit-vector notation, what are the resulting accelerations of (a) the block and (b) the slab?

4 In Fig. 6-39, a slab of mass
m1 40 kg rests on a frictionless
floor, and a block of mass m2 12 kg
rests on top of the slab. Between
block and slab, the coefficient of
static friction is 0.60, and the coefficient of kinetic friction is 0.40. A horizontal force of magnitude
120 N begins to pull directly on the block, as shown. In unit-vector
notation, what are the resulting accelerations of (a) the block and
(b) the slab?

Key Concepts

Newton's Second Law

This law states that the net force acting on an object is equal to the mass of that object multiplied by its acceleration. It is fundamental for determining the acceleration of each object when forces such as applied forces and frictional forces are involved.

Frictional Forces

Friction is the resistive force that occurs when two surfaces interact. It plays a critical role in problems involving motion, where it can either oppose the movement or prevent relative motion between objects. Understanding friction is crucial for setting up the equations that govern the motion of bodies in contact.

Static Versus Kinetic Friction

Static friction acts when there is no relative motion between surfaces and must be overcome to start sliding, while kinetic friction acts when there is relative motion. Differentiating between these two types of friction is essential to accurately predict when objects will begin to slide relative to each other.

Free-Body Diagrams

Drawing free-body diagrams helps in visualizing all the forces acting on a system. It is an indispensable tool in mechanics for breaking down complex problems into manageable parts, allowing one to apply Newton's laws to each individual object separately before considering their interaction.

Transcript

00:01 So here let's draw the free body diagram for the slab.

00:06 Here, going up would be the force normal of the slab.

00:11 Going down would be m sub sg, so that would be simply the weight of the slab.

00:21 On top of the slab would be the force normal for the block.

00:27 And going to the left would be the force of friction.

00:31 We can say that for the block, going up, force normal for the block, going down, of course, the weight of the block.

00:45 To the right would be the force of friction.

00:52 And to the left would be the applied force f.

00:56 And this is for the block.

01:00 So essentially, we can say that there is an applied force f of 100 newtons applied on the block.

01:08 We can say that here we're going to apply newton's second law for the s &y axis for the slab and the block.

01:18 So we're going to get four equations.

01:20 We can get the negative force of friction.

01:23 We'd be equal to the mass of the slab times the acceleration of the slab.

01:27 We can say the force normal of the slab minus the force normal of the block minus the mass of the slab times g would equal zero.

01:41 And then for the block, we have that the force of friction minus the applied force would be equal to the mass of the block times the acceleration of the block.

01:51 And then we can say that the force normal of the block minus the weight of the block would be equal to zero.

01:59 So we can then say that the maximum possible static friction magnitude would be the coefficient of static friction, multiplied by the force normal of the block.

02:12 This is going to give us the coefficient of static friction times the mass of the block times g.

02:18 And we can say that this is going to be equal to 0 .60 times 10 kilograms times 9 .8 meters per second squared.

02:30 And we find that this is equaling 59 neutens...

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