A cube of side ℓrests on a rough floor. It is subjected to a steady horizontal pull F, exerted a

step 1 Once again, welcome to a new problem. This time we have a cube that sits on a table. We have a c

A cube of side ℓrests on a rough floor. It is subjected to a...

Key Concepts

Static Friction

Static friction is the force that resists the initiation of sliding motion between two surfaces in contact. The coefficient of static friction (?_s) characterizes the ratio of this maximum resistive force to the normal force, and it determines the force required to overcome the grip between the surfaces before sliding occurs.

Equilibrium Analysis

Equilibrium analysis involves ensuring that the sum of all forces and torques (moments) acting on a body is zero. This approach is used to determine the conditions under which a body either remains at rest, begins to slide, or starts to tip, by accounting for both translational and rotational balances.

Rotational Equilibrium and Tipping

Rotational equilibrium refers to the state where the net moment about any pivot point is zero. In tipping problems, the applied forces create moments about the edge of the contact area, and tipping occurs when the moment due to the external force exceeds the restoring moment provided by the weight of the body acting through its center of mass.

Line of Action of the Normal Force

The line of action of the normal force is crucial in determining whether an object will slide or tip. Under increasing external forces, the distribution of the normal force shifts, and in the tipping scenario it becomes concentrated at a specific edge. Analyzing its location helps in setting up the moment balance equations to predict the onset of tipping.

Center of Mass

The center of mass is the point at which the mass of a body can be considered to be concentrated for the purpose of analysis of translational and rotational motion. Its position relative to the pivot point affects the moment due to gravity, playing a critical role in assessing the stability of the body against tipping.

Transcript

00:01 Once again, welcome to a new problem.

00:04 This time we have a cube that sits on a table.

00:11 We have a cube like that that sits on a flat table.

00:19 And it happens to have, you know, from a side view.

00:24 So assume you're looking at the cube like that.

00:29 You know, obviously it has a specific height.

00:35 L.

00:37 But, you know, just to see the side of the cube in two dimensions, we know it sits on a table and the cube itself happens to be pulled by a force that's a distance h from the ground level.

01:04 Okay, it's a distance h from the ground level.

01:13 L and this is l.

01:18 The block itself has a weight which we're going to call mg and it's facing two things.

01:30 The first one is the fact that it's being pulled by a force f and there is also a frictional force which we're calling fr, but remember friction is defined as mu s mg.

01:46 Or sometimes you could define friction as mu as fn excuse me where fn is the normal reaction fn is a normal reaction so there's a force pushing upwards right here which we're calling fn and since fn is the normal force we're the normal force we acts against the weight based on action and reaction forces, your frictional force will become mu s, m g.

02:37 Remember, mu s is the coefficient of friction, and that's one of the things we want to find.

02:46 In the first part of the problem, part a, we want to find the static friction.

02:56 We want to find the static friction or coefficient of static friction when the block begins to slide.

03:16 And then the second part of the problem, we're assuming that the block will begin to tip.

03:23 So meaning, you know, it's going to tip at this point.

03:27 Since the force is moving that way, it means if it tips, it's tipping clock.

03:33 In a clockwise direction.

03:36 So we want to compute the coefficient of static friction for these two cases.

03:42 For this block, we do have a weight and also we do have the force pulling the block itself.

03:54 The system is in equilibrium, so the sum of forces in the x is zero, because it's moving in the horizontal.

04:03 And we assume that in terms of the motion, positive x is in the right and then positive y is in the upwards.

04:18 So f minus the friction is going to be zero, frictional force because the f, f forces in the positive direction and the friction is in the negative direction.

04:36 So that means the force that's pulling is equivalent to the friction, or we can simplify it by saying it's new as mg.

04:46 And since the system is in equilibrium, if we take any torque around any specific point, so the sum of torque around a will be zero.

04:58 The force is moving towards the right, so the torque that it produces a is negative because that's clockwise, and then the weight is pulling downwards, so the talk it produces on a will be positive because it's counterclockwise...

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