A uniform cube of side ℓ rests on a rough floor. It is subject...

Question

A uniform cube of side ℓ rests on a rough floor. It is subjected to a steady horizontal pull F, exerted a distance h above the floor as shown in Fig. 12–85. As F is increased, the block will either begin to slide, or begin to tip over. Determine the coefficient of static friction μ_s so that: the block begins to slide rather than tip. [Hint: Where will the normal force on the block act if it tips?]

Accepted Answer

Hi everyone. Let's take a look at this practice problem dealing with forces and torques. This problem says in the illustrated scenario, a box sits atop a rough surface. A laborer applies a consistent horizontal force. F 1.3 m above its base to move the box as f steadily increases the box faces, the possibility of either sliding or tipping, calculate the coefficient of static friction needed for the box to slide rather than tip. Well, the question were given a diagram of what was described in the problem. We're also given four possible choices as our answers. For choice A we have 0.58. For choice B, we have 0.87. For choice C we have 1.2 and for choice D we have 1.7. Now, since we're dealing with forces, the first thing we want to do is draw a free body diagram. So we're gonna draw a square for our box and this box is going to have war forces acting on it. The first force is going to be the applied force which is acting near the top of the box just slightly below. But actually 1.3 m above the bottom of the box. And so we're going to have that force going to the left and we'll label that force as f the next force is going to be acting at the center of the box and that's gonna be its weight and that's going downwards. And that has a magnitude of MG. The next force is going to be the normal force which is acting on the bottom of the box. And that's going upwards, we'll label that force as FN and the final force is going to be the force of static friction that's also acting on the bottom of the box. And that's gonna be going to the right and we'll label that force as FFR. So we need to calculate the coefficient of static friction that is needed for the box to slide rather than tip. So let's look at what conditions we need in order to get the box to just be on the verge of tipping. And so that'll tell us our maximum applied force that we can actually have. So to do this, we're going to need torque. And so recall your definition of torque that is tau is equal to R cross F where Tau is the torque R is the distance from the axis of rotation to where the force is being applied and F is the force that's being applied. Now, we need to choose a point at which to calculate the torques. And we're actually going to choose the bottom left corner because if the box was getting right to the tip, that's gonna be the point where it actually rotates about. So we're going to do is add up all the torques about that point of rotation. And so to do that, we're gonna have the sum of our T torques and that's gonna be equal to. And we're actually gonna have two torques here. If we look at the normal force and the friction force, both of those are gonna be acting at that point of rotation. That's because that's gonna be the only point that's gonna be touching the ground. And therefore, the normal force is going to be acting at that point as well as the force of friction. But if I look at my definition of torque, the R value for both of those forces are going to be zero. So therefore, they apply no torque. So the only two forces that are gonna apply a torque are gonna be the applied force and the weight. So we're going to take the convention that if the torque is trying to rotate the object in a counterclockwise direction, it's gonna be a positive torque. And if it's trying to rotate the object in a clockwise direction, it's gonna be a negative torque. So let's look at our applied force. Um first. So for that, the force is acting a distance of 1.3 m from the point of rotation, that is 1.3 m and that's gonna be multiplied by our force f and those two vectors are actually perpendicular to each other. And that's what the cross product is actually doing is just picking up the perpendicular components between those two vectors. So I don't need to do anything else with this. Now, if I look at the weight, it's gonna be trying to rotate the box in a clockwise manner. So it's gonna be a negative torque. So I'll have minus. And here I actually want to use the perpendicular distance from the axis of rotation to where the force is acting. So that's going to be actually just half of the length of one of our sides of the boxes since the weight is acting at the center of the box. So the length of one side of the box were given in the diagram as 1.5 m. So half of that is going to be the 0.75 m and that's could be multiplied by the weight, which is MG. So here I've used the fact that the cross product give me just the perpendicular components acting on each other. So I've taken the perpendicular component of the distance that's going from the axis of rotation to the center of the box. Now I don't want the box to actually tip over. So I want this neck torque to actually equal zero. So this will now give me the maximum force that could be applied without tipping the box over. So I can solve this for F And so one way to do is add 0.75 m multiplied by MG to both sides and then divide both sides by 1.3 m. So then I'll get F equal to 0.75 m divided by 1.3 m. That fraction is multiplied by MG. So that's gonna be the maximum force that could be applied without the box tipping over. So now we can use that to calculate our coefficient of static friction to do this. We're going to have to look at Newton's second law. And so before I do that, if I come back up to my free body diagram, I'm gonna actually incorporate Accord system. So I'm gonna have the positive X going to the right and the positive Y going upwards. So one way to do is some of the forces in both directions using Newton's Second law, F is equal to ma. So let's start with the Y direction first. So we're gonna sum the forces in the Y direction and that's gonna be equal to in the positive direction. I'll have my normal force FN and in the downward direction, I'll have minus MG and I don't want any acceleration in the Y direction. So we're gonna set that sum of forces equal to zero. So that means that FN is equal to MG. I would do the same thing with the X direction. So when some of the forces in the X direction and that's gonna be equal to in deposit X direction, I'll have my force of fix. So I have FFR and going in the negative direction, I'll have my applied for. So I'll have minus F and again, I want this to be just right at the point where it begins to slip. So I want to set this force, this some of the force is equal to zero. So I still want no acceleration. And so that means that FFR is equal to F. However, I could write my force of friction in terms of the coefficient of static friction. So recall your formula or the static friction force that's gonna be muse of S multiplied by the normal force FN and that's gonna be equal to. And for F here, we'll plug in the expression that we found from our torque equation which is the 0.75 divided by 1.3. That quantity is multiplied by MG. And here have canceled out the units of meters in that coefficient. Now, from our Y um the sum of the forces in the Y direction, we see that the normal force is equal to MG. So I'll go ahead and plug that in. They'll have muse of S multiplied by MG is equal to the quantity of 0.75 divided by 1.3 in 20. That's multiplied by MG, the NGS will cancel. And I'm actually plug in that fraction into my calculator to get a value for M sub or MS we have M is equal to 0.58. So that's the answer that I'm looking for. And if I look at my answer choices, this corresponds to answer choice A. So just to recap of what we've done here, we start off by drawing a free by diagram. And then we looked at our net torque equation and set it equal to zero in order to solve for the maximum of five force that we can have without tipping the box. We then looked at Newton's second law in summing the forces in both the X and Y direction and studying those that sum of force is equal to zero since we want no acceleration. And so that's allowed us to write our normal force in terms of the weight as well as the friction force. In terms of the applied force, we could then use those expressions along with the expression that we found for the maximum applied force to have that we can have without tipping the box in order to solve warm use of which is the coefficient of static friction. So I hope that this has been useful. And I'll see you in the next video.

Key Concepts

Static Friction

Torque and Rotational Equilibrium

Normal Force

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