 A 6 kilogram block is sliding down a 15 degree incline plane. It's lubricated by a 1 millimeter thick layer of SAE 30 weight oil at 20 degrees Celsius. The area of the bottom of the block, that is the side that is in contact with the oil, has a surface area of 35 square centimeters. I want to determine the terminal velocity of this block. So what we're going to be assuming here is that our velocity profile within the interface between the incline plane and the block looks the same as it did in the previous problem. On our stationary plate, we have no movement of the fluid and the block is moving down the plane and developing a linear profile within the oil. The difference between this analysis and the previous example is that the velocity of that moving block, which is the same as the moving plate earlier, is now unknown. We are going to be defining a y-axis normal to the incline plane for convenience and we are going to define an x-axis that is perpendicular to that and I'm going to call the positive direction of the x-axis down into the left just for fun. If the block is in equilibrium, which it would have to be for it to have reached terminal velocity, then the sum of forces in the x-direction are going to equal zero. Therefore, the weight of the block pulling it down the incline plane is going to equal the resistive force of the fluid. We are neglecting all air drag and any other drag forces that could be acting on the block. Our assumptions so far are that the fluid is Newtonian. We could also add the no slip assumption, but we will save that to talk about when it makes a little bit more sense. Like earlier, the shear stress is going to be the drag force divided by the area of effect here. Therefore, the force is going to be the shear stress multiplied by the area of effect. Then if we're assuming a Newtonian fluid, this is going to be dynamic viscosity multiplied by the derivative of u with respect to y times area and just like before, if we have a linear profile, we can describe that as the proportion of the way up u r between zero and one millimeter multiplied by the velocity of the block, which is the unknown. And if I rearrange this to write it as v over h times y, then when I derivate it, the y disappears and I have v over h. Therefore, this is dynamic viscosity multiplied by the velocity of the block over h times the area of effect. And that is the force going up and to the right. For the force pulling down into the left, let's look at a free body diagram. So theta in this circumstance is going to be this angle here as well as this angle over here. By the way, it's common for students to mix up this angle and this angle when they are assigning theta. An easy way to remember it, other than the trigonometric identities, is just to think about what would happen if the slope were very, very, very small. If theta were tiny like half a degree, then this line and this line are going to be basically on top of each other. But this angle is huge because they're so close to each other. So this one is not a tiny, tiny, tiny angle, but this one is. Anyway, then this x component leg of the weight force is going to be w times sine of theta. And that is going to be the force in the down into the left direction. I hope that this awkward way of drawing things is actually readable. Maybe if I draw it that way. Yeah, that's much more better. So since the force down into the left is going to equal the force up into the right, I can write these as being equal to one another. I will say the weight of the block times sine of theta is equal to mu multiplied by the velocity of the block divided by h multiplied by the area of effect. I'm solving for the velocity of the block. Therefore, I can write velocity of the block multiplied is equal to weight times sine of theta divided by mu times A and h goes up here. So if I start to plug in numbers here, the weight of the block is given as, I guess the mass is given as six kilograms. Therefore, the weight of the block is a mass times gravity. We are assuming standard gravity. I'll add that to my assumptions for good measure. I can say six kilograms multiplied by 9.81 meters per second squared multiplied by the sine of 15 degrees. I believe it was. Yeah, 15 degrees. The height of the fluid is one millimeter. And I'm going to change this to definitely a dot so it doesn't get misconstrued as a minus sign. And writing gets real poor when the iPad is real zoomed out. You know, as opposed to normal when it's perfect. The dynamic viscosity we had just looked up, but just for good measure, I will look it up again. In the appendix of our textbook, we want table A3. So if I pull up table A3, bring you guys over to it. I can see that the dynamic viscosity of SAE 30 weight oil at 20 degrees Celsius is 2.9 times 10 to the negative first kilograms per meter second. So that's 2.9 E negative one kilograms per meter second. Then the area is the only thing left and the area of effect is 35 square centimeters. Now, did it ask for a specific unit? It did not. So I will just calculate the answer in meters per second. I'm going to recognize that 100 centimeters are in one meter and I will square everything. And then there are 1000 millimeters in one meter. And now millimeters cancels millimeters kilograms cancels kilograms meters meters meters cancel meters squared. Excuse me. Two of the meters in the numerator are going to cancel the square meters in the denominator. And then the second in the numerator is going to cancel one of the seconds in the denominator. So I'm left with a meter per second calculator, if you would please. We have six times 9.81 times the sign of 15. And I have to add in the degree symbol for this particular calculator because it's in radians right now. And then I'm multiplying by 100 squared. And I'm dividing by 2.9 e to the negative one, multiplied by 35 multiplied by 1000. And I get a syntax error as per use. So six times 9.81 times sign of 15 degrees times 100 squared that matches what I'm expecting divided by 0.29. Yep, that's fine times 35 times 1000. So we feel pretty confident that the velocity that the block achieves is 15 meters per second. And we just kind of glossed over it. But I want to point out just in case it hasn't come up in any of your other courses, the terminal velocity is the maximum velocity achieved here. If you imagine the block just starting to move, it's not going to have much drag force because the drag force is a function of velocity right here. So there's not going to be much drag force, but the weight pulling it down is going to be the same as it always is. Therefore, it's going to have an unbalanced mechanical potential. It's going to be more force pulling it down into the left than up into the right. So it's going to accelerate the faster it moves the higher this velocity term, the higher the drag force, it will continue to accelerate until the force up into the right equals the force down into the left. That is the terminal velocity. It is the velocity at which acceleration stops.