A uniform rod of weight 𝑊 is resting with one of its ends on a rough horizontal surface and the other on a rough inclined plane such that it lies in a plane perpendicular to the line of junction of the two planes. The tangent of the angle between the inclined plane and the horizontal is three-quarters. The coefficient of friction between the rod and the horizontal plane is one-third and between the rod and the inclined plane is one-fourth.
Okay, before we get to our question, let’s take stock of the information given so far. We have a uniform rod that’s in contact with a rough horizontal surface and a rough inclined plane. The rod is at rest, meaning it’s in equilibrium. And we’re told that the tangent of the angle between the inclined plane the rod rests on and the horizontal is three-quarters.
Before going on, let’s record some of this information. Let’s say that the point on the rod where it’s in contact with a rough horizontal surface is point one. And then we’ll say the other end of the rod, where it’s in contact with a rough inclined surface, is point two. Our problem statement tells us that the coefficient of friction at point one is equal to one-third and that at point two is one-quarter. And let’s also write down that the weight of our uniform rod is 𝑊. Okay, knowing all this, let’s continue on to our question.
Find the tangent of the angle that the rod makes with the horizontal in its limiting equilibrium.
We can draw that angle here, and we’ll call it 𝛼. Our goal then is to solve for the tan of 𝛼. Now, a very important point that our problem statement makes is that the rod is in limiting equilibrium. This means that it does satisfy the conditions of equilibrium, which we’ll name in a moment, but just barely. It’s just at the point of being in motion, sliding down and to the left. So that’s the condition of this uniform rod.
And to get started in solving for the tangent of this unknown angle 𝛼, let’s remind ourselves of the two conditions that equilibrium implies. When any object is in equilibrium, that means the net force acting on it is zero. And it also means that the net moment acting about any point along the object’s axis is zero. From these two conditions, we can generate a system of equations. And indeed, we can get three independent equations since forces in horizontal and vertical directions are independent of one another. It’s through solving this system of equations that we’ll generate that we hope to solve for the tan of 𝛼.
As a first step to writing these three equations out, let’s draw in all the forces that are acting on our uniform rod. Because this rod has mass, there’s a weight force that we can say acts directly downward from the midpoint of the rod. And there are also forces acting on the rod at points one and two. At point one, there’s a normal or reaction force acting vertically upward. We’ll call it 𝑅 one. And because we’re told that this horizontal surface is rough, there’s also a frictional force, we’ll call it 𝐹 one, acting to the right.
We know the force acts in this direction because the rod would tend to be moving to the left were it not for the resistive force of friction. Likewise, at location two, there’s a reaction force we’ll call 𝑅 two, which is perpendicular to the inclined plane. And there’s also a frictional force acting up the incline to resist the sliding down motion of the uniform rod.
With these five forces drawn in, we’ve completed the rod’s free body diagram. And we can remove the labels one and two for those points. The fact that our rod is in equilibrium means that if we add up all these forces in a horizontal direction, that sum will be zero, and likewise for all the forces in a vertical direction.
With forces going in so many different directions, it will be helpful to set up a sign convention for which directions are positive. Let’s call those directed vertically upward positive and those directed horizontally to the right also positive. This means a force to the left is negative or a force downward.
Knowing all this, let’s now apply our first condition of equilibrium to generate equations for forces in the horizontal and vertical directions. Considering horizontal forces first, we can see that the force 𝐹 one is directed entirely in what we’ve called the positive direction. Along with this, there’s a horizontal component of the force 𝐹 two, which is positive. For now, we’ll just call that 𝐹 two H for the horizontal component of the force 𝐹 two. And lastly, if we consider the reaction force 𝑅 two, we see that this has a horizontal component in the negative direction. For now, we’ll just call that 𝑅 two H for 𝑅 two in the horizontal direction. And because our rod is in equilibrium, all these forces add to zero.
Now let’s move on to consider the vertical forces. Here, we see the reaction force 𝑅 one acting purely in the positive direction. And then the weight force 𝑊 acts entirely in the negative. And then going back to the forces at point two, we see that 𝑅 two also has a vertical component. It will be positive. For now, we’ll just call that 𝑅 two V. And likewise, the frictional force 𝐹 two will also have a positive vertical component. We’ll label that 𝐹 two V. And these forces too all sum to zero.
We now have two of the three independent equations we can possibly generate from our two equilibrium conditions. To find our third equation, let’s apply this condition that the net moment about any point along the rod’s axis is zero. Because this condition applies for any point along the rod’s axis, we’re free to choose which point we want. Our answer will be the same no matter what. But let’s choose the point we’ve called point two, where the uniform rod meets the inclined plane. This means that when we write out our moment equation, we’ll be considering the tendencies of the other forces along the rod’s length — its weight force, 𝑅 one, and 𝐹 one — to create a rotation about the point we’ve picked.
And as we get to calculating moments, let’s also set up the sign convention that a moment creating a counterclockwise rotation about our point of interest is positive. This means that one leading to a rotation in the clockwise direction would be considered negative.
We’re going to calculate moments about our point of interest caused by these three forces: the weight force, the friction force 𝐹 one, and the reaction force 𝑅 one. In each case, we’ll be multiplying the component of that force that is perpendicular to the rod’s axis times the distance between where the force is applied and our point of rotation. So, for example, if we begin by calculating the moment caused by the weight force 𝑊, first, we recognize that there’s a component of this weight force that is perpendicular to the axis of our rod and that, because this angle here between the rod and the horizontal is 𝛼, so too is this angle right here, in this right triangle, given the components of the weight force.
All this means that the component of the weight force perpendicular to the rod, represented by this vector here, is equal to 𝑊 times the cos of 𝛼. But then we haven’t yet calculated the moment caused by 𝑊 because we need to multiply this value by the distance from where the weight force is applied to our rotation point. If we sketch that distance in on our diagram, we can see it’s equal to this length here, half the length of our rod.
This isn’t a value we’re given. But looking down at the lower left, let’s say that the overall length of our rod is 𝑙. Since the weight force 𝑊 acts at the midpoint along the rod’s length, that must mean it’s a distance of 𝑙 over two from our point of interest. And so this 𝑊 times the cos of 𝛼 times 𝑙 over two is the moment created by the weight force about the point we’ve picked. And note that because this force would tend to create a counterclockwise rotation about this point, it is indeed positive.
Okay, we’re now done with the moment caused by the weight force. And let’s move on to that caused by the friction force 𝐹 one. Just like for the weight force, this force too will break up into components, one of which, the pink component we’ve drawn in, is perpendicular to the rod’s axis. And because, as we’ve said, this angle is 𝛼, we see that perpendicular component is given by 𝐹 one times the sin of 𝛼. And then, in the case of this force, it’s not acting a distance of 𝑙 over two away from our point of interest, but rather a distance 𝑙, the whole length of the rod. The moment created by the force 𝐹 one then is 𝐹 one times the sin of 𝛼 times 𝑙. And this also is positive, tending to create a counterclockwise rotation.
Lastly, let’s look at the moment created by the force 𝑅 one. This force too can be broken up into components relative to the rod’s axis. We want the component of 𝑅 one drawn in pink here. And notice that because this angle is 90 degrees and this angle here is 𝛼, that tells us that this angle, in other words the interior of this right triangle here, must be equal to 90 degrees minus 𝛼. So we could start writing the moment created by 𝑅 one this way: 𝑅 one times the sin of 90 degrees minus 𝛼.
But we could then recall that the sine function is 90 degrees shifted from the cosine function. That means instead of writing the sin of 90 degrees minus 𝛼, we could instead write simply the cos of 𝛼. These two quantities are equal and so we’ll make that simplifying switch. Just like the friction force 𝐹 one, this acts at a distance of 𝑙 from our point of interest. But unlike the other two moments, this one is negative, tending to create a clockwise rotation about that point. Okay, so that’s it for our moments. And our equilibrium condition implies that all these add to zero. So we now have our system of three independent equations. And now we’ll start to work on these equations to eliminate unknowns.
Sliding these equations up, let’s first work on these horizontal and vertical components of the forces 𝑅 two and 𝐹 two. We’d like to replace these subscripts with information given to us in the problem statement. If we look at these two forces in our sketch, starting with the force 𝐹 two, we see that its horizontal component would look like this and its vertical component would look like this. And, moreover, the right triangle we’ve created by breaking this force up into its components is similar to this right triangle here. In other words, just as the tangent of this angle is three-quarters, so is the tangent of this angle.
And notice something else. This right triangle has sides of length three and four units. That tells us this is a specific kind of right triangle whose hypotenuse we then know has a length of five units. This is helpful because now, instead of just knowing the tangent of this angle, three divided by four, we can solve for the cosine or the sine. And indeed, the horizontal component of the force 𝐹 two is equal to 𝐹 two times the cosine of this angle, which according to this triangle we can see will be four divided by five.
That means we can replace every instance of 𝐹 sub two H with 𝐹 sub two times four-fifths. And likewise, if we consider the vertical component of 𝐹 two, given by this length right here, we can see that will be equal to 𝐹 two times the sine of this angle, whose tangent is three-quarters. And the sine of that angle then must be three divided by five. So wherever we see 𝐹 two V, we’ll replace that with 𝐹 two times three-fifths.
And now let’s work on the horizontal and vertical components of the reaction force 𝑅 two. Now, if we draw an expanded view of the place where our rod meets this inclined plane, we’ve said that the tangent of this angle here is equal to three-quarters. And we then know that the angle between 𝐹 two and 𝑅 two is 90 degrees. And we can also see that if we sketch in a dashed vertical line, the angle between this and the dashed horizontal line is 90 degrees, which means that this angle here, which we haven’t given a name to but it has a tangent of three-quarters, is identical to this angle here. And that’s good news because it means if we look at the vertical and horizontal components of the force 𝑅 two, where these components meet, making a right angle, this triangle overall is also a three-four-five triangle.
So the horizontal component of the force 𝑅 two, this component right here, is equal to 𝑅 two times the sine of this angle. That’s three divided by five. In other words, 𝑅 two H equals 𝑅 two times three-fifths. And in a similar way, the vertical component of 𝑅 two, here, would be equal to 𝑅 two times the cosine of this angle, that is, four divided by five. So 𝑅 two V is replaceable with 𝑅 two times four-fifths.
Alright, so we’re making good progress. But notice if we look at the variables in these three equations, we have more unknowns — 𝐹 one, 𝐹 two, 𝑅 two, 𝑅 one, and 𝑊 — than we’re able to solve for using three independent equations. But this is actually where our coefficients of friction, 𝜇 one and 𝜇 two, come into play.
Remember, we were told that this uniform rod is in limiting equilibrium. That means that the frictional forces 𝐹 one and 𝐹 two were at their maximum value to restrain the motion of the rod. And that tells us that, for example, 𝐹 one at its maximum value will equal 𝜇 one times the reaction force 𝑅 one. And likewise, the friction force 𝐹 two at its maximum value will be equal to 𝜇 two times 𝑅 two. Substituting in our given coefficient-of-friction values, we can say that 𝐹 one equals one-third 𝑅 one and 𝐹 two equals one-fourth 𝑅 two.
We can then substitute these expressions into our equations and replace every reference to 𝐹 one and 𝐹 two. So we do that, making those substitutions here, here, here, and here. And at this point, we can start working on these three equations separately. We’ll start with the horizontal force balance equation.
Copying that equation below first, notice that if we multiply these two fractions together, the four in numerator and denominator cancel. And then if we factor 𝑅 two out of these two terms, we get that 𝑅 one over three plus 𝑅 two times of the quantity one-fifth minus three-fifths equals zero. But then one-fifth minus three-fifths is negative two-fifths. And if we then add positive two-fifths 𝑅 two to both sides of this equation, we find that one-third 𝑅 one is equal to two-fifths 𝑅 two. Or, in other words, multiplying both sides by five divided by two, 𝑅 two is equal to five-sixths 𝑅 one.
We can now take this simplified result and replace our horizontal force balance equation with it. With that done, let’s now take a look at our vertical force balance equation. First, considering these two fractions, if we multiply them together, we get a result of three twentieths times 𝑅 two. Then, if we factor out 𝑅 two from each one of these two terms, we get that 𝑅 one minus 𝑊 plus 𝑅 two times the quantity four-fifths plus three twentieths equals zero. And adding four-fifths to three twentieths gives us a result of nineteen twentieths.
And at this point, let’s do this. Let’s replace this 𝑅 two in our expression with five-sixths 𝑅 one, which we know is equal to 𝑅 two. If we make that substitution and then multiply these two fractions together, we get a result of ninety-five one twentieths. And now we can factor 𝑅 one out of these two terms, leading us to this equation here. And if we add one to ninety-five one twentieths and if we also add the weight force 𝑊 to both sides of this equation, then we find that 215 divided by 120 times 𝑅 one is equal to 𝑊. And now let’s take this result and use it to replace our vertical force balance equation.
We’re now ready to apply this result to our moment’s balance equation and solve ultimately for the tan of 𝛼. Our first step will be to replace this instance of 𝑊 with 215 over 120 𝑅 one. Note that once we’ve done that, there will be a factor of 𝑅 one in all three terms on the left-hand side of the equation, as well as a factor of 𝑙. So we’ll factor out both 𝑙 and 𝑅 one. What we get is 𝑅 one times 𝑙 times the quantity 215 divided by 240, because of the factor of one-half here, multiplied by the cos of 𝛼 plus the sin of 𝛼 over three minus the cos of 𝛼 is all equal to zero.
And note that if we divide both sides of our equation by 𝑅 one times 𝑙, these factors cancel on the left, while dividing zero by them doesn’t affect that side. So we get 215 over 240 times the cos of 𝛼 plus the sin of 𝛼 over three minus the cos of 𝛼 equals zero.
And now let’s group together our two terms with cos 𝛼. That gives us this expression with cos 𝛼 multiplying 215 over 240 minus one. That comes out to negative 25 over 240. And we see now that we have a cos 𝛼 and sin 𝛼 term, but we want to solve for tan of 𝛼. We can recall though the trigonometric identity that the tangent of an angle is equal to the sine of that angle divided by the cosine of that same angle. And so if we divide both sides of our equation by the cos of 𝛼, what we’ll get is negative twenty-five two hundred and fortieths plus one-third times the tan of 𝛼 equals zero.
As a final step, we can add 25 over 240 to both sides of this equation and then multiply both sides by three. And when we do that, we find over here that the tan of 𝛼 equals 75 over 240, which reduces to five sixteenths. This is the tangent of the angle between our uniform rod and the rough horizontal plane.