The 2.0 kg wood box in FIGURE P6.58 slides down a vertical wood wall while you push on it at a 45° angle. What magnitude of force should you apply to cause the box to slide down at a constant speed?

Question

Accepted Answer

Welcome back, everyone. We are making observations about the following system here. We have a steel box that is sliding down against a steel lubricated wall. Now, we also have a force applied by a boy who is pulling to the upper right direction from the corner of the box and we are tasked with finding what is the magnitude of that pole force being applied. Now, before we get started here, I do wish to acknowledge our multiple choice answers in the bottom left corner of our screen. Those are gonna be the values in which we strive for. So without further ado let us begin. I think the easiest way to approach this is to model our box as a dot And draw out a free body diagram. So what's at play here? Well, we have the whole force applied by the boy. Now the wall is going to push on the box and that is going to be a normal force. Now, what is acting downward? Well, we want to determine the force that will be applied by the boy so that the box just starts to slide upward, just starts to slide upward. So we are not only going to have a friction force between the steel and the lubricated steel, but we are also going to have the weight of our box or the force due to gravity here. Now, as you can see from our diagram, we have that this angle of the pulling force makes a 35 degree angle with the horizontal. Now, what we can do here is we can decompose our pulling force into FPX and FPY. It's X and Y components that is going to help us find our magnitude of our F pole. All right. Well, where do we begin from here? Let's do the sum of all forces use Newton's second law in the X direction. The box is not moving left or right. So we're gonna have zero acceleration, meaning the right hand side of our equation is zero. So what are the forces at play here? Well, we have a normal force in the positive X direction minus the X component of our pole force and this is equal to zero. What this means is that our normal force is equal to FP of X or that's just equal to the force of our pole times the cosine of our angle. Great. We can't really go anywhere from here. So now let's try the sum of all forces in the Y direction. Now we are looking just for enough force to be applied by the boy so that the box starts to slide up this can be a miniscule movement, it can move extremely slow. So what we can say is that acceleration is practically zero. So we can say the right hand side of our equation is zero. So what do we have at play here? Well, in the positive Y direction, we have the Y component of our pole force minus the kinetic friction force acting against motion minus the weight of our box. Now, this of course is gonna be equal to zero. Now, let's go ahead and plug in some values for FPY. We are just going to have the force of our pole times the sign of 35 minus our friction force. So we have a coefficient of friction times the normal force, but we solve for our normal force in the other dimension. So this is just going to be fe cosine of 35 minus the weight of our box, which is simply MG. Now, simplifying some things here, what we get is that FP is equal to or FP times sign of 35 minus UK cosine of 35 is equal to MG. Now, what I will do is I will divide both sides by our trigonometric arithmetic here. So sign 35 minus mute K cosine of 35 on the left hand side, let me go ahead and copy that over to the right hand side here. All right, you'll see on the left hand side, that large term cancels out and we get finally an equation for the magnitude of our pulling force. So let's go ahead and calculate that we have FP is equal to three, which is the mass, the box times acceleration due to gravity divided by the sign of our angle 35 minus mu K or coefficient of kinetic friction between steel and lubricated steel is going to be 0.05. And then multiply this by cosine of 35. When we plug all this into a calculator, what we get for a final answer is 55.2 newtons corresponding to our final answer. Choice of B Thank you all so much for watching. I hope this video helped. We will see you all in the next one.

The 2.0 kg wood box in FIGURE P6.58 slides down a vertical wood wall while you push on it at a 45° angle. What magnitude of force should you apply to cause the box to slide down at a constant speed?

Verified Solution

Key Concepts

Newton's First Law of Motion

Friction

Components of Forces